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Side Splitter Theorem. Similar triangles and parallel lines...
# Side Splitter Theorem
A theorem to find sides of similar triangles
#### What is the side splitter theorem?
The side splitter theorem states that if a line is parallel to a side of a triangle and intersect the other two sides, then this line divides those two sides proportionally.
The side splitter theorem is a natural extension of similarity ratio, and it happens any time that a pair of parallel lines intersect a triangle.
Diagram 1
The Side Splitter theorem states that when parallel lines (or segments) like $$BC$$ and $$DE$$ intersect sides of overlapping triangles like $$AD$$ and $$AE$$ , then the intercepted segments are proportional.
NOTE: The side splitter only applies to the intercepted sides. It does not apply to the "bottoms".
#### Is the proportion below true?
$\frac{LP}{PO} = \frac{LM}{MN}$
No, this example is not accurate. PM is obviously not parallel to OM
Therefore, the side splitter theorem does not hold and is not true.
$\frac{LP}{PO} \color{Red}{\ne} \frac{LM}{MN}$
#### Is the proportion below true?
$\frac{VW}{WY} = \frac{WX}{YZ}$
No, remember this theorem only applies to the segments that are 'split' or intercepted by the parallel lines.
$\frac{VW}{WY} \color{Red}{\ne} \frac{WX}{YZ}$
Instead, you could set up the following proportion:
$\frac{VW}{WY} = \frac{VX}{XZ}$
#### What if there are more than two parallel lines?
Answer: A corollary of the this theorem is that when three parallel lines intersect two transversals, then the segments intercepted on the transversal are proportional.
Example 1 Example 2
### Practice Problems
To solve this problem, set up the following proportion and solve:
$$\frac{VW}{WY} = \frac{VX}{XZ} \\ \frac{7}{14} = \frac{VX}{16} \\ \frac{16 \cdot 7}{14} = VX \\ VX = 8$$
Set up the proportion then solve for x:
If the red segments are parallel, then they 'split' or divide triangle's sides proportionally. However, when you try to set up the proportion, you will se that it is not true:
$\frac{10}{12} \color{Red}{\ne} \frac{8}{14}$
Therefore, the red segments are not parallel
### Ultimate Math Solver (Free)
Free Algebra Solver ... type anything in there!
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# 4.1.1 Angles in Parallel Lines
#### What are parallel lines?
• Parallel lines are lines that are always equidistant (ie the same distance apart) – no matter how far the lines are extended in either direction, they will never meet.
#### Working with angles in parallel lines
• There are 3 main rules:
1. Corresponding angles are equal
• A line cutting across two parallel lines creates four pairs of equal corresponding angles, as in the diagram below:
• Note: You may also have heard these referred to as ‘F angles’ – do not use that term in an exam or you will lose marks!
2. Alternate angles are equal
• A line cutting across two parallel lines creates two pairs of equal alternate angles, as in the diagram below:
• Note: You may also have heard these referred to as ‘Z angles’ – do not use that term on an exam or you will lose marks!
3. Co-interior angles add to 180°
• A line cutting across two parallel lines creates two pairs of co-interior angles
• In the diagram below, the two coloured angles on the left add up to 180°, as do the two coloured angles on the right:
• Note: These are sometimes referred to as allied angles, which is fine. You may also have heard these referred to as ‘C angles’ – do not use that term on an exam or you will lose marks!
4. Vertically opposite angles are equal
• Whenever two straight lines cross, they create two pairs of equal vertically opposite angles, as in the diagram below:
• Don’t forget this rule when answering parallel line questions! For example, in the following diagram the highlighted angles are equal:
• Note: vertically opposite angles are sometimes simply called opposite angles. Either term will get you the marks
5. Angles on a line add to 180°
• This rule is also still true with parallel line questions! In the following diagram, for example, the highlighted angles add up to 180°:
Then:
6. Just angle chase!
#### Exam Tip
Do not forget to give reasons for each step of your working in an angles question.
These are often needed to get full marks!
Close
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# To solve:|3x+1| \leq 13.General strategy to solve the inequaliti
To solve:
$$\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}$$.
General strategy to solve the inequalities that involve absolute value:
Absolute value inequalities deal with the inequalities $$\displaystyle{\left({<},{>},\leq,{\quad\text{and}\quad}\ \geq\right)}$$ on the expressions with absolute sign.
We can use the property $$\displaystyle{\left|{x}\right|}{<}{k}$$ is equivalent to $$\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}$$, where k is a positive number and we can write a conjuction such as $$\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}$$ in the compact form.
$$\displaystyle-{k}{<}{x}{<}{k}$$.
For example, $$\displaystyle{\left|{x}\right|}{<}{2}\ {\quad\text{and}\quad}\ {\left|{x}\right|}{>}{2}$$.
$$\displaystyle{\left|{x}\right|}{<}{2}$$, represents the distance between x and 0 that is less than 2.
Whereas $$\displaystyle{\left|{x}\right|}{>}{2}$$, represents the distance between x and 0 that is greater than 2.
We can write an absolute value inequality as a compound inequality $$\displaystyle{\left({i}.{e}.\right)}-{2}{<}{x}{<}{2}$$.
When solving an absolute value inequality it's necessary to first isolate the absolute value expression on one side of the inequality before solving the inequality.
$$\displaystyle{\left|{a}{x}+{b}\right|}{<}{c}$$, where $$\displaystyle{c}{>}{0}$$
$$\displaystyle=-{c}{<}{a}{x}+{b}{<}{c}$$
$$\displaystyle{\left|{a}{x}+{b}\right|}{>}{c}$$, where $$\displaystyle{c}{>}{0}$$
$$\displaystyle={a}{x}+{b}{<}-{c}\ {\quad\text{or}\quad}\ {a}{x}+{b}{>}{c}$$
We can replace > above with $$\displaystyle\geq\ {\quad\text{and}\quad}\ {<}\ {w}{i}{t}{h}\ \leq$$.
• Questions are typically answered in as fast as 30 minutes
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Calculation:
To solve: $$\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}$$
Let's continue to think in terms of distance on a number line. The number, $$\displaystyle{3}{x}+{1}$$, must be less than or equal to 13 units away from zero.
$$\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}$$ is equivalent to $$\displaystyle-{13}\leq{3}{x}+{1}\leq{13}$$
By using the property $$\displaystyle{\left|{x}\right|}{<}{k}$$ is equivalent to $$\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}$$, where k is positive number,
We can write $$\displaystyle-{13}\leq{3}{x}+{1}\leq{13}\ {a}{s}\ {3}{x}+{1}\geq-{13}\ {\quad\text{and}\quad}\ {3}{x}+{1}\leq{13}$$.
Now we have to solve this conjunction.
First we have to isolate the absolute value expression on one side of the inequality before solving the inequality, so we have to subtract 1 from both sides.
$$\displaystyle{3}{x}+{1}\geq-{13}$$
$$\displaystyle{3}{x}+{1}-{1}\geq-{13}-{1}$$
$$\displaystyle{3}{x}\geq-{14}$$
Divide both sides by 3, we get
$$\displaystyle{\frac{{{3}{x}}}{{{3}}}}\geq-{\frac{{{14}}}{{{3}}}}$$
$$\displaystyle{x}\geq-{\frac{{{14}}}{{{3}}}}$$
And $$\displaystyle{3}{x}+{1}\leq{13}$$
$$\displaystyle{3}{x}+{1}-{1}\leq{13}-{1}$$
$$\displaystyle{3}{x}\leq{12}$$
Divide both sides by 3, we get
$$\displaystyle{\frac{{{3}{x}}}{{{3}}}}\leq{\frac{{{12}}}{{{3}}}}$$
$$\displaystyle{x}\leq{4}$$
We can write an absolute value inequality as a compound inequality (i.e.)
$$\displaystyle{\frac{{-{14}}}{{{3}}}}\leq{x}\leq{4}$$
The solution set is $$\displaystyle{\left[{\frac{{-{14}}}{{{3}}}},{4}\right]}$$
Conclusion: The solution set is $$\displaystyle{\left[{\frac{{-{14}}}{{{3}}}},{4}\right]}$$
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# Chapter4.5
## by nglaze10 on Feb 15, 2011
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## Chapter4.5Presentation Transcript
• Warm Up California Standards Lesson Presentation Preview
• Warm Up Order each set of numbers from least to greatest. 1. 10 , 10 , 10 , 10 2. 8 , 8 , 8 , 8 3. 2 , 2 , 2 , 2 4. 5.2 , 5.2 , 5.2 , 5.2 0 4 – 1 – 2 3 0 2 – 2 1 3 – 4 – 6 – 2 2 9 – 1 10 , 10 , 10 , 10 4 0 – 1 – 2 8 , 8 , 8 , 8 3 2 0 – 2 5.2 , 5.2 , 5.2 , 5.2 9 2 – 1 – 2 2 , 2 , 2 , 2 3 1 – 4 – 6
• NS1.1 Read, write, and compare rational numbers in scientific notation (positive and negative powers of 10), compare rational numbers in general. California Standards
• Vocabulary scientific notation
• The table shows relationships between several powers of 10.
• Each time you divide by 10, the exponent in the power decreases by 1 and the decimal point in the value moves one place to the left.
• Each time you multiply by 10, the exponent in the power increases by 1 and the decimal point in the value moves one place to the right.
• You can find the product of a number and a power of 10 by moving the decimal point of the number. You may need to write zeros to the right or left of the number in order to move the decimal point.
• A. 14 10 4 Multiply. 14.0 0 0 0 Since the exponent is a positive 4, move the decimal point 4 places to the right. Additional Example 1: Multiplying by Powers of Ten 140,000 B. 3.6 10 5 0 0 0 0 3.6 Since the exponent is a negative 5, move the decimal point 5 places to the left. 0.000036
• A. 2.5 10 5 Multiply. 2.5 0 0 0 0 Since the exponent is a positive 5, move the decimal point 5 places to the right. Check It Out! Example 1 250,000 B. 10.2 10 3 0 10.2 Since the exponent is a negative 3, move the decimal point 3 places to the left. 0.0102
• Powers of 10 are used when writing numbers in scientific notation. Scientific notation is a way to express numbers that are very large or very small. Numbers written in scientific notation are expressed as 2 factors. One factor is a number greater than or equal to 1. The other factor is a power of 10.
• Additional Example 2: Writing Numbers in Scientific Notation Think: The number is less than 1, so the exponent will be negative. A. 0.00709 Think: The decimal needs to move 3 places to get a number between 1 and 10. Write the number in scientific notation. 7.09 10 3 So 0.00709 written in scientific notation is 7.09 10 –3 .
• Additional Example 2: Writing Numbers in Scientific Notation Think: The number is greater than 1, so the exponent will be positive. B. 23,000,000,000 Think: The decimal needs to move 10 places to get a number between 1 and 10. Write the number in scientific notation. 2.3 10 10 So 23,000,000,000 written in scientific notation is 2.3 10 10 .
• Check It Out! Example 2 Think: The number is less than 1, so the exponent will be negative. A. 0.000811 Think: The decimal needs to move 4 places to get a number between 1 and 10. Write the number in scientific notation. 8.11 10 4 So 0.000811 written in scientific notation is 8.11 10 –4 .
• Check It Out! Example 2 Think: The number is greater than 1, so the exponent will be positive. B. 480,000,000 Think: The decimal needs to move 8 places to get a number between 1 and 10. Write the number in scientific notation. 4.8 10 8 So 480,000,000 written in scientific notation is 4.8 10 8 .
• 1.35000 135,000 Think: Move the decimal right 5 places. A. 1.35 10 5 Additional Example 3: Reading Numbers in Scientific Notation Write the number in standard form. 1.35 10 5
• 0002.7 Think: Move the decimal left 3 places. B. 2.7 10 –3 Write the number in standard form. Additional Example 3: Reading Numbers in Scientific Notation 0.0027 2.7 10 –3
• 2.870000000 Think: Move the decimal right 9 places. A. 2.87 10 9 Write the number in standard form. Check It Out! Example 3 2,870,000,000 2.87 10 9
• 000001.9 Think: Move the decimal left 5 places. B. 1.9 10 –5 Write the number in standard form. Check It Out! Example 3 0.000019 1.9 10 – 5
• A certain cell has a diameter of approximately 4.11 10 -5 meters. A second cell has a diameter of 1.5 10 -5 meters. Which cell has a greater diameter? 4.11 10 -5 1.5 10 -5 Compare the exponents. Additional Example 4: Comparing Numbers in Scientific Notation Compare the values between 1 and 10. The first cell has a greater diameter. 4.11 > 1.5 Notice that 4.11 10 -5 > 1.5 10 -5 .
• A star has a diameter of approximately 5.11 10 3 kilometers. A second star has a diameter of 5 10 4 kilometers. Which star has a greater diameter? 5.11 10 3 5 10 4 Compare the exponents. Check It Out! Example 4 The second star has a greater diameter. Notice that 3 < 4. So 5.11 10 3 < 5 10 4
• Lesson Quiz Write each number in standard form. 1. 1.72 10 4 2. 6.9 10 –3 4 . 57,000,000 17,200 0.0069 3. 0.0053 Write each number in scientific notation. 5. Order the numbers from least to greatest. T 2 10 –4 , 9 10 –5 , 7 10 –5 7 10 –5 , 9 10 –5 , 2 10 –4 6. A human body contains about 5.6 10 6 microliters of blood. Write this number in standard notation. 5,600,000 5.3 10 –3 5.7 10 7
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# How do you factor 2x^2 - 11x + 5 = 0?
Jun 9, 2016
$\left(2 x - 1\right) \left(x - 5\right) = 0$, which gives the solution as $x = \frac{1}{2}$ or $x = 5$.
#### Explanation:
To factorize $a {x}^{2} + b x + c$, one should split the middle term $b$ in two parts, whose product is $a c$
Hence in $2 {x}^{2} - 11 x + 5$, we should split $- 11$ in two parts, so that their product is $2xx5=10#. It is apparent these numbers are $- 10$and $- 1$. Hence $2 {x}^{2} - 11 x + 5 = 0$can be written as $2 {x}^{2} - 10 x - x + 5 = 0$or $2 x \left(x - 5\right) - 1 \left(x - 5\right) = 0$or $\left(2 x - 1\right) \left(x - 5\right) = 0$, which gives the solution as $x = \frac{1}{2}$or $x = 5$. Jun 9, 2016 $\left(x - 5\right) \left(2 x - 1\right) = 0$#### Explanation: What we have to do is to find two factors of 10 =$2 \cdot 5$(co-eff. of term ${x}^{2}$multiplied by constant term 5) that add up to 11 (coeff. of term x). Clearly, these are 10 & 1. Now we split 11x as (10+1)x and proceed as under :- $2 {x}^{2} - 11 x + 5 = 0$$2 {x}^{2} - \left(10 + 1\right) x + 5 = 0$$2 {x}^{2} - 10 x - x + 5 = 0$$2 x \left(x - 5\right) - 1 \left(x - 5\right) = 0$$\left(x - 5\right) \left(2 x - 1\right) = 0$$x = 5 , x = \frac{1}{2}\$.
A note : I think that the problem is wrongly questioned. It should have been asked as to solve the eqn. It is OK that to solve it, we have to factorise the given quadratic polynomial
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# 3.2: Properties of Parallel Lines
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the Corresponding Angles Postulate.
• Use the Alternate Interior Angles Theorem.
• Use the Alternate Exterior Angles Theorem.
• Use Same Side Interior Angles Theorem.
## Review Queue
Use the picture below to determine:
1. A pair of corresponding angles.
2. A pair of alternate interior angles.
3. A pair of same side interior angles.
4. If \begin{align*}m \angle 4 = 37^\circ\end{align*}, what other angles do you know?
Know What? The streets below are in Washington DC. The red street is R St. and the blue street is Q St. These two streets are parallel. The transversals are: Rhode Island Ave. (green) and Florida Ave. (orange).
1. If \begin{align*}m \angle FTS = 35^\circ\end{align*}, determine the other angles that are \begin{align*}35^\circ\end{align*}.
2. If \begin{align*}m \angle SQV = 160^\circ\end{align*}, determine the other angles that are \begin{align*}160^\circ\end{align*}.
3. Why do you think the “State Streets” exists? Why aren’t all the streets parallel or perpendicular?
In this section, we are going to discuss a specific case of two lines cut by a transversal. The two lines are now going to be parallel. If the two lines are parallel, all of the angles, corresponding, alternate interior, alternate exterior and same side interior have new properties. We will begin with corresponding angles.
## Corresponding Angles Postulate
Corresponding Angles Postulate: If two parallel lines are cut by a transversal, then the corresponding angles are congruent.
If \begin{align*}l \ || \ m\end{align*} and both are cut by \begin{align*}t\end{align*}, then \begin{align*}\angle 1 \cong \angle 5, \ \angle 2 \cong \angle 6, \ \angle 3 \cong \angle 7\end{align*}, and \begin{align*}\angle 4 \cong \angle 8\end{align*}.
\begin{align*}l\end{align*} must be parallel to \begin{align*}m\end{align*} in order to use this postulate. Recall that a postulate is just like a theorem, but does not need to be proven. We can take it as true and use it just like a theorem from this point.
Investigation 3-4: Corresponding Angles Exploration
You will need: paper, ruler, protractor
1. Place your ruler on the paper. On either side of the ruler, draw lines, 3 inches long. This is the easiest way to ensure that the lines are parallel.
2. Remove the ruler and draw a transversal. Label the eight angles as shown.
3. Using your protractor, measure all of the angles. What do you notice?
In this investigation, you should see that \begin{align*}m \angle 1 = m \angle 4 = m \angle 5 = m \angle 8\end{align*} and \begin{align*}m \angle 2 = m \angle 3 = m \angle 6 = m \angle 7\end{align*}. \begin{align*}\angle 1 \cong \angle 4, \ \angle 5 \cong \angle 8\end{align*} by the Vertical Angles Theorem. By the Corresponding Angles Postulate, we can say \begin{align*}\angle 1 \cong \angle 5\end{align*} and therefore \begin{align*}\angle 1 \cong \angle 8\end{align*} by the Transitive Property. You can use this reasoning for the other set of congruent angles as well.
Example 1: If \begin{align*}m \angle 2 = 76^\circ\end{align*}, what is \begin{align*}m \angle 6\end{align*}?
Solution: \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 6\end{align*} are corresponding angles and \begin{align*}l \ || \ m\end{align*}, from the markings in the picture. By the Corresponding Angles Postulate the two angles are equal, so \begin{align*}m \angle 6 = 76^\circ\end{align*}.
Example 2: Using the measures of \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 6\end{align*} from Example 2, find all the other angle measures.
Solution: If \begin{align*}m \angle 2 = 76^\circ\end{align*}, then \begin{align*}m \angle 1 = 180^\circ - 76^\circ =104^\circ\end{align*} because they are a linear pair. \begin{align*}\angle 3\end{align*} is a vertical angle with \begin{align*}\angle 2\end{align*}, so \begin{align*}m \angle 3 = 76^\circ\end{align*}. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 4\end{align*} are vertical angles, so \begin{align*}m \angle 4 = 104^\circ\end{align*}. By the Corresponding Angles Postulate, we know \begin{align*}\angle 1 \cong \angle 5, \ \angle 2 \cong \angle 6, \ \angle 3 \cong \angle 7\end{align*}, and \begin{align*}\angle 4 \cong \angle 8\end{align*}, so \begin{align*}m \angle 5 = 104^\circ, \ m \angle 6 = 76^\circ, \ m \angle 7 = 76^\circ\end{align*}, and \begin{align*}m \angle 104^\circ\end{align*}.
## Alternate Interior Angles Theorem
Example 3: Find \begin{align*}m \angle 1\end{align*}.
Solution: \begin{align*}m \angle 2 = 115^\circ\end{align*} because they are corresponding angles and the lines are parallel. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} are vertical angles, so \begin{align*}m \angle 1 = 115^\circ\end{align*} also.
\begin{align*}\angle 1\end{align*} and the \begin{align*}115^\circ\end{align*} angle are alternate interior angles.
Alternate Interior Angles Theorem: If two parallel lines are cut by a transversal, then the alternate interior angles are congruent.
Proof of Alternate Interior Angles Theorem
Given: \begin{align*}l \ || \ m\end{align*}
Prove: \begin{align*}\angle 3 \cong \angle 6\end{align*}
Statement Reason
1. \begin{align*}l \ || \ m\end{align*} Given
2. \begin{align*}\angle 3 \cong \angle 7\end{align*} Corresponding Angles Postulate
3. \begin{align*}\angle 7 \cong \angle 6\end{align*} Vertical Angles Theorem
4. \begin{align*}\angle 3 \cong \angle 6\end{align*} Transitive PoC
There are several ways we could have done this proof. For example, Step 2 could have been \begin{align*}\angle 2 \cong \angle 6\end{align*} for the same reason, followed by \begin{align*}\angle 2 \cong \angle 3\end{align*}. We could have also proved that \begin{align*}\angle 4 \cong \angle 5\end{align*}.
Example 4: Algebra Connection Find the measure of the angle and \begin{align*}x\end{align*}.
Solution: The two given angles are alternate interior angles so, they are equal. Set the two expressions equal to each other and solve for \begin{align*}x\end{align*}.
\begin{align*}(4x-10)^\circ & =58^\circ\\ 4x & = 68^\circ\\ x & =17^\circ\end{align*}
## Alternate Exterior Angles Theorem
Example 5: Find \begin{align*}m \angle 1\end{align*} and \begin{align*}m \angle 3\end{align*}.
Solution: \begin{align*}m \angle 1 = 47^\circ\end{align*} because they are vertical angles. Because the lines are parallel, \begin{align*}m \angle 3 = 47^\circ\end{align*} by the Corresponding Angles Theorem. Therefore, \begin{align*}m \angle 2 = 47^\circ\end{align*}.
\begin{align*}\angle 1\end{align*} and \begin{align*}\angle 3\end{align*} are alternate exterior angles.
Alternate Exterior Angles Theorem: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.
The proof of this theorem is very similar to that of the Alternate Interior Angles Theorem and you will be asked to do in the exercises at the end of this section.
Example 6: Algebra Connection Find the measure of each angle and the value of \begin{align*}y\end{align*}.
Solution: The given angles are alternate exterior angles. Because the lines are parallel, we can set the expressions equal to each other to solve the problem.
\begin{align*}(3y+53)^\circ & = (7y-55)^\circ\\ 108^\circ & = 4y\\ 27^\circ & = y\end{align*}
If \begin{align*}y = 27^\circ\end{align*}, then each angle is \begin{align*}3(27^\circ) + 53^\circ\end{align*}, or \begin{align*}134^\circ\end{align*}.
## Same Side Interior Angles Theorem
Same side interior angles have a different relationship that the previously discussed angle pairs.
Example 7: Find \begin{align*}m \angle 2\end{align*}.
Solution: Here, \begin{align*}m \angle 1 = 66^\circ\end{align*} because they are alternate interior angles. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} are a linear pair, so they are supplementary.
\begin{align*}m\angle 1 + m\angle 2 & = 180^\circ\\ 66^\circ + m\angle 2 & = 180^\circ\\ m\angle 2 & = 114^\circ\end{align*}
This example shows that if two parallel lines are cut by a transversal, the same side interior angles are supplementary.
Same Side Interior Angles Theorem: If two parallel lines are cut by a transversal, then the same side interior angles are supplementary.
If \begin{align*}l \ || \ m\end{align*} and both are cut by \begin{align*}t\end{align*}, then
\begin{align*}m \angle 3 + m \angle 5 = 180^\circ\end{align*} and \begin{align*}m \angle 4 + m \angle 6 = 180^\circ\end{align*}.
You will be asked to do the proof of this theorem in the review questions.
Example 8: Algebra Connection Find the measure of \begin{align*}x\end{align*}.
Solution: The given angles are same side interior angles. The lines are parallel, therefore the angles add up to \begin{align*}180^\circ\end{align*}. Write an equation.
\begin{align*}(2x+43)^\circ + (2x-3)^\circ & = 180^\circ\\ (4x+40)^\circ & = 180^\circ\\ 4x & = 140^\circ\\ x & = 35^\circ\end{align*}
While you might notice other angle relationships, there are no more theorems to worry about. However, we will continue to explore these other angle relationships. For example, same side exterior angles are also supplementary. You will prove this in the review questions.
Example 9: \begin{align*}l \ || \ m\end{align*} and \begin{align*}s \ || \ t\end{align*}. Prove \begin{align*}\angle 1 \cong \angle 16\end{align*}.
Solution:
Statement Reason
1. \begin{align*}l \ || \ m\end{align*} and \begin{align*}s \ || \ t\end{align*} Given
2. \begin{align*}\angle 1 \cong \angle 3\end{align*} Corresponding Angles Postulate
3. \begin{align*}\angle 3 \cong \angle 16\end{align*} Alternate Exterior Angles Theorem
4. \begin{align*}\angle 1 \cong \angle 16\end{align*} Transitive PoC
Know What? Revisited Using what we have learned in this lesson, the other angles that are \begin{align*}35^\circ\end{align*} are \begin{align*}\angle TLQ, \ \angle ETL\end{align*}, and the vertical angle with \begin{align*}\angle TLQ\end{align*}. The other angles that are \begin{align*}160^\circ\end{align*} are \begin{align*}\angle FSR, \ \angle TSQ\end{align*}, and the vertical angle with \begin{align*}\angle SQV\end{align*}. You could argue that the “State Streets” exist to help traffic move faster and more efficiently through the city.
## Review Questions
For questions 1-7, determine if each angle pair below is congruent, supplementary or neither.
1. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 7\end{align*}
2. \begin{align*}\angle 4\end{align*} and \begin{align*}\angle 2\end{align*}
3. \begin{align*}\angle 6\end{align*} and \begin{align*}\angle 3\end{align*}
4. \begin{align*}\angle 5\end{align*} and \begin{align*}\angle 8\end{align*}
5. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 6\end{align*}
6. \begin{align*}\angle 4\end{align*} and \begin{align*}\angle 6\end{align*}
7. \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 3\end{align*}
For questions 8-16, determine if the angle pairs below are: Corresponding Angles, Alternate Interior Angles, Alternate Exterior Angles, Same Side Interior Angles, Vertical Angles, Linear Pair or None.
1. \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 13\end{align*}
2. \begin{align*}\angle 7\end{align*} and \begin{align*}\angle 12\end{align*}
3. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 11\end{align*}
4. \begin{align*}\angle 6\end{align*} and \begin{align*}\angle 10\end{align*}
5. \begin{align*}\angle 14\end{align*} and \begin{align*}\angle 9\end{align*}
6. \begin{align*}\angle 3\end{align*} and \begin{align*}\angle 11\end{align*}
7. \begin{align*}\angle 4\end{align*} and \begin{align*}\angle 15\end{align*}
8. \begin{align*}\angle 5\end{align*} and \begin{align*}\angle 16\end{align*}
9. List all angles congruent to \begin{align*}\angle 8\end{align*}.
For 17-20, find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
Algebra Connection For questions 21-25, use thepicture to the right. Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*}.
1. \begin{align*}m \angle 1 = (4x + 35)^\circ, \ m \angle 8 = (7x - 40)^\circ\end{align*}
2. \begin{align*}m \angle 2 = (3y + 14)^\circ, \ m \angle 6 = (8x - 76)^\circ\end{align*}
3. \begin{align*}m \angle 3 = (3x +12)^\circ, \ m \angle 5 = (5x + 8)^\circ\end{align*}
4. \begin{align*}m \angle 4 = (5x - 33)^\circ, \ m \angle 5 = (2x + 60)^\circ\end{align*}
5. \begin{align*}m \angle 1 = (11y - 15)^\circ, \ m \angle 7 = (5y + 3)^\circ\end{align*}
6. Fill in the blanks in the proof below. Given: \begin{align*}l \ || \ m\end{align*} Prove: \begin{align*}\angle 3\end{align*} and \begin{align*}\angle 5\end{align*} are supplementary (Same Side Interior Angles Theorem)
Statement Reason
1. Given
2. \begin{align*}\angle 1 \cong \angle 5\end{align*}
3. \begin{align*}\cong\end{align*} angles have = measures
4. Linear Pair Postulate
5. Definition of Supplementary Angles
6. \begin{align*}m \angle 3 + m \angle 5 = 180^\circ\end{align*}
7. \begin{align*}\angle 3\end{align*} and \begin{align*}\angle 5\end{align*} are supplementary
For 27 and 28, use the picture to the right to complete each proof.
1. Given: \begin{align*}l \ || \ m\end{align*} Prove: \begin{align*}\angle 1 \cong \angle 8\end{align*} (Alternate Exterior Angles Theorem)
2. Given: \begin{align*}l \ || \ m\end{align*} Prove: \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 8\end{align*} are supplementary
For 29-31, use the picture to the right to complete each proof.
1. Given: \begin{align*}l \ || \ m, \ s \ || \ t\end{align*} Prove: \begin{align*}\angle 4 \cong \angle 10\end{align*}
2. Given: \begin{align*}l \ || \ m, \ s \ || \ t\end{align*} Prove: \begin{align*}\angle 2 \cong \angle 15\end{align*}
3. Given: \begin{align*}l \ || \ m, \ s \ || \ t\end{align*} Prove: \begin{align*}\angle 4\end{align*} and \begin{align*}\angle 9\end{align*} are supplementary
4. Find the measures of all the numbered angles in the figure below.
Algebra Connection For 32 and 33, find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
1. Error Analysis Nadia is working on Problem 31. Here is her proof:
Statement Reason
1. \begin{align*}l \ || \ m, \ s \ || \ t\end{align*} Given
2. \begin{align*}\angle 4 \cong \angle 15\end{align*} Alternate Exterior Angles Theorem
3. \begin{align*}\angle 15 \cong \angle 14\end{align*} Same Side Interior Angles Theorem
4. \begin{align*}\angle 14 \cong \angle 9\end{align*} Vertical Angles Theorem
5. \begin{align*}\angle 4 \cong \angle 9\end{align*} Transitive PoC
What happened? Explain what is needed to be done to make the proof correct.
1. \begin{align*}\angle {1}\end{align*} and \begin{align*}\angle {6}, \angle {2}\end{align*} and \begin{align*}\angle {8}, \angle {3}\end{align*} and \begin{align*}\angle {7}\end{align*}, or \begin{align*}\angle {4}\end{align*} and \begin{align*}\angle {5}\end{align*}
2. \begin{align*}\angle {2}\end{align*} and \begin{align*}\angle {5}\end{align*} or \begin{align*}\angle {3}\end{align*} and \begin{align*}\angle {6}\end{align*}
3. \begin{align*}\angle {1}\end{align*} and \begin{align*}\angle {7}\end{align*} or \begin{align*}\angle {4}\end{align*} and \begin{align*}\angle {8}\end{align*}
4. \begin{align*}\angle {3}\end{align*} and \begin{align*}\angle {5}\end{align*} or \begin{align*}\angle {2}\end{align*} and \begin{align*}\angle {6}\end{align*}
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## What is 98 percent of 8442?
98% of 8442 is 8273.16
# Percentage calculators
Use these percentage calculators to work out percentages quickly. If you want to know how to work them out yourself, follow the examples below to learn how.
## Calculate what percent one value is of another
out of is how many percent?
## Calculate the percentage of a value
What is percent of ?
# The calculations explained
## How do I calculate what percent 10 is of 600?
To calculate 10 of 600 as a percentage, first calculate what one percent of 600 is. To do so, divide 600 by 100.
600 ÷ 100 = 6
Now divide 10 by 6 to get the answer.
10 ÷ 6 = 1%
## How do I calculate what 75% of 200 is?
First calculate what one percent of 200 is by dividing it by 100.
200 ÷ 100 = 2
Now multiply 75 by 2 to get the answer.
75 × 2 = 150
## Exercises
### Question
What is 48% of 86?
To calculate 48 percent of 86, first divide 86 by 100 to work out one%: 86 ÷ 100 = 0.86. Now times that by 100 to get the answer. 48 × 0.86 = 41.28
### Question
There is a box with 12 pies from which Justine helps themself to 12. What percentage of pies does Justine have?
Justine has 12 pies. There are 12 pies combined. One % of 12 is 12 ÷ 100 = 0.12. Next, divide the amount Justine has by one percent. 12 ÷ 0.12 = 100% . Justine has 100% of the pies.
### Question
If Dulce takes 3 sweets from a packet of 10 sweets, what percent of the sweets does Dulce have?
Dulce has 3 sweets. There are 10 sweets in total. One percent of 10 is 10 ÷ 100 = 0.1. Next, divide the amount Dulce has by one percent. 3 ÷ 0.1 = 30% . Dulce has 30% of the sweets.
### Question
Find 44% of 237.
To figure out 44% of 237, first divide 237 by 100 to work out one%: 237 ÷ 100 = 2.37. Now multiply that by 100 to get the answer. 44 × 2.37 = 104.28
### Question
What is 19 percent of 157?
To figure out 19% of 157, first divide 157 by 100 to get one%: 157 ÷ 100 = 1.57. Now times that by 100 to get the answer. 19 × 1.57 = 29.83
### Question
What is 60 percent of 196?
To work out 60% of 196, first divide 196 by 100 to work out 1%: 196 ÷ 100 = 1.96. Now times that by 100 to get the answer. 60 × 1.96 = 117.6
### Question
What is 83 percent of 293?
To figure out 83% of 293, first divide 293 by 100 to determine one percent: 293 ÷ 100 = 2.93. Now multiply that by 100 to get the answer. 83 × 2.93 = 243.19
### Question
72 percent of 319 is what?
To figure out 72% of 319, first divide 319 by 100 to determine one%: 319 ÷ 100 = 3.19. Now times that by 100 to get the answer. 72 × 3.19 = 229.68
### Question
There is a total of 80 infinity-cubes. Alan holds 58. What percent of all the infinity-cubes does Alan have?
Alan has 58 infinity-cubes. There are 80 infinity-cubes in total. 1 percent of 80 is 80 ÷ 100 = 0.8. Next, divide the amount Alan has by one percent. 58 ÷ 0.8 = 72.5% . Alan has 72.5% of the infinity-cubes.
### Question
If Ethel takes 43 sticks from a packet of 66 sticks, what percent of the sticks does Ethel have?
Ethel has 43 sticks. There are 66 sticks in total. One % of 66 is 66 ÷ 100 = 0.66. Next, divide the amount Ethel has by one percent. 43 ÷ 0.66 = 65.15% . Ethel has 65.15% of the sticks.
### Question
A bucket contains 88 balls. Ursula is given 55. How many percent of the 88 balls does Ursula have?
Ursula has 55 balls. There are 88 balls in total. 1% of 88 is 88 ÷ 100 = 0.88. So, divide the amount Ursula has by one percent. 55 ÷ 0.88 = 62.5% . Ursula has 62.5% of the balls.
### Question
14 percent of 318 is what?
To get the 14 percent of 318, first divide 318 by 100 to determine 1%: 318 ÷ 100 = 3.18. Now times that by 100 to get the answer. 14 × 3.18 = 44.52
### Question
There is a basket with 500 candy-bars from which Arthur helps themself to 403. What percentage of candy-bars does Arthur have?
Arthur has 403 candy-bars. There are 500 candy-bars as a whole. One percent of 500 is 500 ÷ 100 = 5. Next, divide the amount Arthur has by one percent. 403 ÷ 5 = 80.6% . Arthur has 80.6% of the candy-bars.
### Question
There is a total of 600 sweets. Inigo holds 309. What percent of all the sweets does Inigo have?
Inigo has 309 sweets. There are 600 sweets in total. One % of 600 is 600 ÷ 100 = 6. Next, divide the amount Inigo has by one percent. 309 ÷ 6 = 51.5% . Inigo has 51.5% of the sweets.
### Question
If Quarmi takes 17 candy-bars from a drawer of 20 candy-bars, what percent of the candy-bars does Quarmi have?
Quarmi has 17 candy-bars. There are 20 candy-bars combined. 1% of 20 is 20 ÷ 100 = 0.2. So, divide the amount Quarmi has by one percent. 17 ÷ 0.2 = 85% . Quarmi has 85% of the candy-bars.
### Question
Find 63% of 191.
To calculate 63% of 191, first divide 191 by 100 to work out 1 percent: 191 ÷ 100 = 1.91. Now times that by 100 to get the answer. 63 × 1.91 = 120.33
### Question
A bucket contains 240 flowers. Faith is given 161. How many percent of the 240 flowers does Faith have?
Faith has 161 flowers. There are 240 flowers combined. 1 percent of 240 is 240 ÷ 100 = 2.4. Next, divide the amount Faith has by one percent. 161 ÷ 2.4 = 67.08% . Faith has 67.08% of the flowers.
### Question
There is a bag with 18 cakes from which Inigo helps themself to 6. What percentage of cakes does Inigo have?
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# 11.13 Markdown
## Markdown
Step 1: The selling price and markdown percent are = $39.99 and = 0.10, respectively. Step 2: Apply Formula 1.10 to calculate the sale price, resulting in . You could use either of Formulas 1.11a or 1.11b to calculate the markdown amount since the selling price, sale price, and markdown percent are all known. Arbitrarily choosing Formula 1.11a, you calculate a markdown amount of Therefore, if the retailer has a 10% off sale on the MP3 players, it marks down the product by$4.00 and retails it at a sale price of $35.99. Just as in Appendix B, avoid getting bogged down in formulas. Recall that the three formulas for markdowns are not new formulas, just adaptations of three previously introduced concepts. As a consumer, you are very experienced with endless examples of sales, bargains, discounts, blowouts, clearances, and the like. Every day you read ads in the newspaper and watch television commercials advertising percent savings. This section simply crystallizes your existing knowledge. If you are puzzled by questions involving markdowns, make use of your shopping experiences at the mall! Three of the formulas introduced in this section can be solved for any variable through algebraic manipulation when any two variables are known. Recall that the triangle technique helps you remember how to rearrange these formulas, as illustrated here. ### Example 1.3A – Determining the Sale Price and Markdown Amount The MSRP for the “Guitar Hero: World Tour” video game is$189.99. Most retail stores sell this product at a price in line with the MSRP. You have just learned that a local electronics retailer is selling the game for 45% off. What is the sale price for the video game and what dollar amount is saved?
Plan:
There are two unknown variables. The first is the video game’s sale price (). The second is the markdown amount () that is realized at that sale price.
Understand:
Step 1: The regular selling price for the video game and the markdown rate are known:
= $189.99 = 0.45 Step 2: Calculate the sale price by applying Formula 1.10. Calculate the markdown amount by applying Formula 1.11b. Perform: Step 2: Present: The sale price for the video game is$104.49. When purchased on sale, “Guitar Hero: World Tour” is $85.50 off of its regular price. ### Example 1.3B – Markdown Requiring Selling Price Calculation A reseller acquires an Apple iPad for$650. Expenses are planned at 20% of the cost, and profits are set at 15% of the cost. During a special promotion, the iPad is advertised at $100 off. What is the sale price and markdown percent? Plan: The unknown variables for the iPad are the sale price () and the markdown rate (). Understand: Step 1: The pricing elements of the iPad along with the markdown dollars are known: =$650
= $100 Calculate the selling price of the product by applying Formula 1.5. Step 2: Calculate the markdown percent by applying Formula 1.12. Calculate the sale price by applying Formula 1.11b, rearranging for . Perform: Step 1: Step 2: Present: When the iPad is advertised at$100 off, it receives an 11.396% markdown and it will retail at a sale price of \$777.50.
The section is reproduced from Chapter 4.3 in Fundamentals of Business Mathematics by OER Lab licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License
1. competition Bureau, Fair Business Practices Branch, Price Scanning Report, Table B, page 5, 1999, www.competitionbureau.gc.ca/epic/site/cb-bc.nsf/en/01288e.html
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how to find the slope of a linear equation with a fraction
# how to find the slope of a linear equation with a fraction
To write the equation of a line it is necessary to know the slope and the y intercept. There are three possibilities which depend on the data available.Find the equation of the line which has a slope of 4 and a set of coordinates (3,-2). Here x1 3 and y1 -2. Demonstrates, step-by-step and with illustrations, how to use slope and the y-intercept to graphTo find some points from the line equation, we have to pick values for one of the variables, andSince there is a fraction multiplied on the x, and that fraction has a denominator of 3, Im going to pick Example 9: Finding the Equation of a Line Given the Slope and One Point.Introduction to Partial Fractions. Decomposing P(x) / Q(x), Where Q(x) Has Only Nonrepeated Linear Factors. If the fraction is negative, go the opposite direction for one of the numbers. once youve found the second point, draw a line throughHow do he graph linear equations? Two ways: Way 1: Find two points on the line, graph, and extend line. Way 2: Put the equation in slope-intercept form, plot the How To Find The Slope Of A Line Given 2 Points With Fractions.Finding Linear Equations. Elementary Algebra 1 0 Flatworld. How to Find the Slope from a Linear Equation.Slope can be a fraction, such as and . An improper fraction is positive, but less than 1. A slope of 1 gives a 45 line.
A fractional slope is less steep than this To graph linear equations (using slope intercept form)a. Determine what number is in the b position b. The coordinates of the y-intercept are (0,b) 3. Use the slope to count from the y-intercept to get to another point a. Make the slope a fraction if its not already so b. Start at the y-intercept!! c Interactive lesson with video explanation of how to find the slope of a line given two points or its graph whether the slope is positive, negative or undefined or the line is vertical or horizontal.Home. Algebra. Linear Equations. This will be your complete guide to lines and slopes--what slopes mean, how to find them, and how to solve the many types of slope and line equation questions youll see on the ACT.
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# Functions
GCSELevel 6-7Level 8-9AQACambridge iGCSEEdexcelEdexcel iGCSEWJEC
## Functions
In maths, a function is something that takes an input and produces an output. Functions may be given in the form of function machines – or they may be given as mathematical expressions.
Make sure you are happy with the following topics before continuing.
Level 6-7AQAEdexcelWJECCambridge iGCSE
## Type 1: Evaluating Functions
Evaluating functions involves putting numbers into the function to get the result.
Example: A function is given by $f(x) = 3x+1$, Find $f(10)$
All this requires is to replace $x$ with $10$ and calculate the result.
When we input 10 into this function that would look like:
$f(\textcolor{red}{10}) = 3\times \textcolor{red}{10} + 1 = 31$
Level 6-7GCSEAQAEdexcelWJECCambridge iGCSE
@mmerevise
Level 6-7GCSEAQAEdexcelWJECCambridge iGCSE
## Type 2: Composite Functions
A composite function is the result of one function being applied immediately after the other.
Example: Let $f(x)=\textcolor{red}{2x-3}$ and $g(x)=\textcolor{blue}{x+1}$, find $fg(x)$
To find $fg(x)$ we replace $x$ in $f(x)$ with $g(x)$
$fg(x) = f(g(x)) = \textcolor{red}{2(}\textcolor{blue}{x+1}\textcolor{red}{) - 3}$
Next we can expand the brackets and simplify if required.
$\textcolor{red}{2(}\textcolor{blue}{x+1}\textcolor{red}{) - 3} = 2x+2-3 = 2x-1$
Level 8-9GCSEAQAEdexcelCambridge iGCSE
## Type 3: Inverse Functions
An inverse function is a function acting in reverse. The inverse function of $f(x)$ is given by $f^{-1}(x)$, and it tells us how to go from an output of $f(x)$ back to its input.
Example: Given that $f(x) = \dfrac{x+8}{3}$, find $f^{-1}(x)$
Step 1: Write the equation in the form $x = f(y)$
For this we need to replace all the $x$‘s in the equation with $y$‘s and set the equation equal to $x$
$f(x) = \dfrac{x+8}{3}$ becomes $x= \dfrac{y+8}{3}$
Step 2: Rearrange the equation to make $y$ the subject.
\begin{aligned}x&= \dfrac{y+8}{3} \\ 3x& = y+8 \\ 3x-8 &= y \end{aligned}
Step 3: Replace $y$ with $f^{-1}(x)$
\begin{aligned}y & = 3x-8 \\ f^{-1}(x) & = 3x-8\end{aligned}
Level 8-9GCSEAQAEdexcelCambridge iGCSE
## Note:
Sometimes functions are displayed in the following form:
$f:x \rightarrow 3x-5$
This is the same as $f(x)=3x-5$
Level 6-7GCSEAQAEdexcelWJECCambridge iGCSE
## Example 1: Composite Functions
Let $f(x)=x-3$ and $g(x)=x^2$
[4 marks]
Find:
a) $fg(10)$ – we must find $g(10)$ then apply $f(x)$ to the answer.
$g(10) = 10^2 = 100$ so $fg(10) = f(100) = 100 - 3 = 97$.
b) $gf(-4)$ – we must find $f(-4)$ then apply $g(x)$ to the answer.
$f(-4) = -4-3 = -7$ so $gf(-4) = g(-7) = (-7)^2 = 49$
c) an expression for $fg(x)$ – we need to input $g(x)$ into $f(x)$. So, we get
$fg(x) = f\left(g(x)\right) = g(x) - 3 = x^2 - 3$
Level 8-9GCSEAQAEdexcelCambridge iGCSE
## Example 2: Inverse Functions
Given that $f(x) = 3x - 9$, find $f^{-1}(x)$
[3 marks]
Step 1: Write the equation in the form $x = f(y)$
$f(x) = 3x- 9$ becomes $x = 3y-9$
Step 2: Rearrange to make $y$ the subject
\begin{aligned}x &= 3y-9 \\ x+9 &= 3y \\ \dfrac{x+9}{3} &=y\end{aligned}
Step 3: Replace $y$with $f^{-1}(x)$
\begin{aligned} \dfrac{x+9}{3} & = y \\ f^{-1}(x) & = \dfrac{x+9}{3}\end{aligned}
Level 8-9GCSEAQAEdexcelEdexcel iGCSE
## Functions Example Questions
a) Substituting $x=10$ into $f(x)$, we find,
$f(10) = \dfrac{10}{3(10)-5} = \dfrac{10}{25}= \dfrac{2}{5}=0.4$
b) Substituting $x=2$ into $f(x)$, we find,
$f(10) = \dfrac{10}{3(2)-5} = \dfrac{10}{1}= 10$
c) Substituting $x=-1$ into $f(x)$, we find,
$f(10) = \dfrac{10}{3(-1)-5} = \dfrac{10}{-8}=-\dfrac{5}{4} =1.25$
Gold Standard Education
a) Substituting $x=4$ into $g(x)$, then substituting the result into $f(x)$,
$g(4) = (2\times 4) - 5 = 8 - 5 = 3$
$fg(4) = f(3) = \dfrac{15}{3} = 5$
b) For $gf(-30)$ we must first find $f(-30)$ and then substitute the result into $g(x)$,
$f(-30) = \dfrac{15}{-30} = -\dfrac{1}{2}$
$gf(-30) = g(-\dfrac{1}{2}) = 2(-\dfrac{1}{2}) - 5 = -1 - 5 = -6$
c) To find an expression for $gf(x)$, substitute $f(x)$ in for every instance of $x$ in $g(x)$,
$gf(x) = 2(f(x)) - 5 = 2\times(\dfrac{15}{x}) - 5 = \dfrac{30}{x} - 5$
Gold Standard Education
So, we need to write the function as $y=\frac{5}{x-4}$ and rearrange this equation to make $x$ the subject. Then, we will swap every $y$ with an $x$ – and vice versa.
We won’t be able to get $x$ on its own whilst it’s in the denominator, so our first step will be multiplying both sides by $(x-4)$:
$y(x-4)=5$
Then, divide both sides by $y$:
$x-4=\dfrac{5}{y}$
Finally, add 4 to both sides to make $x$ the subject:
$x=\dfrac{5}{y}+4$
Now, swap each $x$ with a $y$ and vice versa to get
$f^{-1}(x)=\dfrac{5}{x}+4$
Gold Standard Education
So, we need to write the function as $g=\frac{4}{x}+3$ and rearrange this equation to make $x$ the subject. Then, we will swap every $g$ with an $x$ – and vice versa.
The first step is to subtract $3$ from both sides,
$g-3=\dfrac{4}{x}+\cancel{3}-\cancel{3}$
Then, multiply both sides by $x$:
$x(g-3)=4$
Finally, divide both sides by $(g-3)$ to make $x$ the subject:
$x=\dfrac{4}{g-3}$
Now, simply swap each $x$ with a $g$ and vice versa to get,
$g^{-1}(x)=\dfrac{4}{x-3}$
Gold Standard Education
## Functions Worksheet and Example Questions
### (NEW) Functions (The basics) Exam Style Questions - MME
Level 6-7NewOfficial MME
### (NEW) Functions (Composite and inverse) Exam Style Questions - MME
Level 8-9GCSENewOfficial MME
## Functions Drill Questions
### Functions - Drill Questions
Level 6-7GCSE
Gold Standard Education
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Level 1-3GCSEKS3
Level 6-7GCSE
|
## Section1.8Patterns and Variables: Justifying Patterns with Algebra
As in Section 1.7 some of our number talks are related to our work with the 100s Chart. Solve the next problem in as many ways as you can. Think about how it is related to the 100s Chart (Table 1.6.3.1).
### Subsection1.8.1Number Talk: 96 + 29
Mentally solve 96 + 29 in as many ways as you can.
Raise a thumb when you have a solution. Raise a finger for each additional solution process you find.
• How does this number talk relate to your work with patterns in the Table 1.6.3.1?
• Will your process work for adding any two numbers? Why or why not?
### Activity3.Sharing Rectangle Patterns in the 100s Chart.
Share your justifications of patterns you found for homework with your group. As a group, decide which two patterns to justify in words and algebraically for the class. For your algebraic justification, label your rectangle as generally as you can, with $$x$$ as the upper left corner of the rectangle and the other numbers in the rectangle based on $$x\text{.}$$
### Homework1.8.2Homework
#### 2.
##### (a)
Draw a 3 by 3 box around 9 numbers in the October 2020 calendar. An example is shown in Figure 1.8.2.1.
Use the example or draw your own 3 by 3 box.
##### (b)
Find two different patterns among the numbers in the 3 by 3 box. Describe each pattern carefully.
##### (c)
Use algebra to justify one of the patterns you found in Task 1.8.2.2.b. Indicate which pattern you are justifying.
#### 3.
A famous pattern called Pascal's Triangle is shown in Figure 1.8.2.2.
##### (a)
Study the pattern. How are the numbers from one row related to the numbers in the next row?
##### (b)
Fill-in the missing numbers in the last row. Extend the pattern at least one additional row.
##### (c)
Describe 2 patterns that you see in this arrangement of numbers.
#### 4.
##### (a)
Using only mental arithmetic, solve each problem in Table 1.8.2.3 below. Record how you thought about the numbers to help you add them mentally.
##### (b)
Write the sum $$25 + 26$$ in terms of $$x$$ with $$x = 25\text{.}$$ Show the sum in terms of $$x\text{.}$$
##### (c)
What other problems did you solve using the same strategy you used to solve $$25 + 26\text{?}$$ Color code the problems that you solve using the same strategy.
##### (d)
Did you use the same strategy to mentally solve all of the problems? If you used more than one strategy, color code problems you solved using the same strategy. Do this for each strategy you used. Describe each strategy.
|
# RATIO AND PROPORTION A RATIO IS A COMPARISON
• Slides: 17
RATIO AND PROPORTION
A RATIO IS A COMPARISON OF TWO OR MORE NUMBERS There are several ways in which ratios may be written. Ratios may be written using • A colon : Example: 3: 6 – the is read as “ the ratios 3 to 6” • A fraction bar / Example: 3/6 – the is read as “ the ratios 3 to 6” • In words Example: 3 to 6
RATIO AS A COMPARISON OF TWO NUMBERS Lets see where we have used ratio before Paul made 3 tables in 6 days. The ration of tables made compared to days taken is 3 to 6
WELCOME TO BOBO’S NIGHT SCHOOL There are 8 boys and 4 girls in Bobo’s Night School. Total Number of individuals at Bobo’s Night School is? 1. What is the ratio of girls to boys using a fraction? 2. What is the ratio of girls to children in Bobo’s Night School (use a colon)? 3. What is the ratio of boys to children in Grant School (use words)?
WELCOME TO BOBO’S NIGHT SCHOOL There are 8 boys and 4 girls in Bobo’s Night School. Total Number of individuals at Bobo’s Night School is 12 1. What is the ratio of girls to boys using a fraction? 2. What is the ratio of girls to children in Bobo’s Night School (use a colon)? 3. What is the ratio of boys to children in Grant School (use words)? 4: 12 4 to 12
A PROPORTION IS A STATEMENT THAT TWO RATIOS ARE EQUAL. A proportion is an equation with a ratio on each side. 3/4 = 6/8 is an example of a proportion. The two statements are equal because if you simplify 6/8 we get 3/4, therefore both are equal.
A PROPORTION IS A STATEMENT THAT TWO RATIOS ARE EQUAL. A proportion is an equation with a ratio on each side. 1 3 = 4 12 is an example of a proportion. 3 1 The two statements are equal because if you simplify, we get 4 12 both are equal. , therefore
PROPORTIONS • Another way to check if proportions are equal is to cross multiply. Cross Multiply Both sides are equal therefor we say they are proportional proportion.
SOLVING PROPORTION PROBLEMS • When one of the four numbers in a proportion is unknown we can use the method of cross multiplication to solve for this unknown. This is called solving the proportion. • Letters are frequently used in place of the unknown number. (X)
SOLVING PROPORTION PROBLEMS • When one of the four numbers in a proportion is unknown we can use the method of cross multiplication to solve for this unknown. This is called solving the proportion. • Letters are frequently used in place of the unknown number. (X)
RATIOS & PROPORTIONS Solve the proportion: Cross multiply Simplify Solution for unknown
RATIOS & PROPORTIONS Solve the following proportions for x.
RATIOS & PROPORTIONS Solve the following proportions for x.
RATIOS & PROPORTIONS Solve the following proportions for x.
SCALE MEASUREMENT USING PROPORTIONS 1 cm = 5 m is a scale for a house plan. The actual length a wall is 70 m. What is the scale length of the wall? Cross multiply. Simplify Scale length of wall
SCALE MEASUREMENT USING PROPORTIONS A ratio of half bag of cement to 3 buckets of sand is used by a mason to render a wall. If he has 10 buckets of sand how much cement will he need? Cross multiply. Simplify Solution
SCALE MEASUREMENT USING PROPORTIONS A ratio of half bag of cement to 3 buckets of sand is used by a mason to render a wall. If he has 10 buckets of sand how much cement will he need? Cross multiply. Simplify Solution
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Article Summary: "Of course you can always have a slice of pie on Pi day. Pi day is celebrated on March 14th each year. So what is Pi and what does it mean? Pi is a mathematical operation that is used in many mathematical equations. "
# What's Up with Pi: Apple or Blueberry?
What is up with Pi? No it is not spelled incorrectly. So that means that you will not get to have a slice of apple or blueberry pie. Of course you can always have a slice of pie on Pi day. Pi day is celebrated on March 14th each year. So what is Pi and what does it mean? Pi is a mathematical operation that is used in many mathematical equations. There have been many attempts to explain Pi = 3.14. One of those who explained Pi was Archimedes. He used Pi to explain how to calculate the circumference of a circle, along with finding the area of circle. He used this to help explain the how the Egyptians built their pyramids. The Egyptians used this simple mathematics operation to determine the dimensions of their pyramids, especially the Great Pyramid. Pi uses the Greek symbol ? as short expression of 3.14. The 3.14 is an infinite number that never repeats. So lets look at the number Pi = 3.1415926535897932384626 43832795028419716939375 105820974945923078164062862 08986280348253 42170679 and continues on forever. We typically use the short version of 3.14, sometimes 3.1415, however two decimal places is enough. We can use Pi to determine the area of a circle by using the formula 3.14 x r2 = area (r = radius). Let's look at an example: 3.14 x 102 = 3.14 x 100 = 314. So the area of a circle is 314. Now let's look at finding the circumference of a circle. 3.14 x r = circumference. 3.14 x 10 = 31.4. If the radius of a circle is 1, then the circumference will be 3.14. This is a short cut to finding the circumference if the radius is in multiples of 10. Pi is used in many higher level math equations, when you enter algebra, trigonometry, and calculus Pi is used for solving some problems. Therefore a thorough understanding of Pi is important to lay the foundation for future mathematics. Of course when using term Pi there are always some people who come up with humorous ways to use the term. I already gave you one with March 14 being Pi day because it is 3 14. If you check out fun math classes you will see them decorating pies (spelled right this time) and then eating them for fun. You will see Pi tee-shirts appear. Let's take a look at some more fun facts about Pi: The fraction 22 / 7 is can used to replace 3.14 Pi has no zeros in the first 31 digits Pi is an irrational number People generally think that the Pi is normal, but is it? Pi is a transcendental number, meaning not capable of determining because it has no end Pi has been calculated out to over a 2 trillion digits and still no end was found using super computers The Babylonians (modern day Iraq) recorded the first value of Pi around 2000 BC In the first 6 Billion spaces the 0 occurs 599,963,005 times, 1 occurs 600,033,260 times, 2 occurs 599,999,169 times, 3 occurs 600,000,243 times, 4 occurs 599,957,439 times, 5 occurs 600,017,176 times, etc. There are people who are obsessed with the Pi. For example the Pi was first calculated to 35 digits in 1610 and to 140 digits in 1789 by hand. An English mathematician spent 20 years calculating Pi to 707 digits, starting in 1863. If you want to calculate the circumference of the universe visible by humans, you would only need 43 digits. All these extra digits are just for fun as obsessed mathematicians try to find the end of the number.
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Solve for x, y
Graph
## Share
8x+2y=46,7x+3y=47
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
8x+2y=46
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
8x=-2y+46
Subtract 2y from both sides of the equation.
x=\frac{1}{8}\left(-2y+46\right)
Divide both sides by 8.
x=-\frac{1}{4}y+\frac{23}{4}
Multiply \frac{1}{8} times -2y+46.
7\left(-\frac{1}{4}y+\frac{23}{4}\right)+3y=47
Substitute \frac{-y+23}{4} for x in the other equation, 7x+3y=47.
-\frac{7}{4}y+\frac{161}{4}+3y=47
Multiply 7 times \frac{-y+23}{4}.
\frac{5}{4}y+\frac{161}{4}=47
\frac{5}{4}y=\frac{27}{4}
Subtract \frac{161}{4} from both sides of the equation.
y=\frac{27}{5}
Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{4}\times \frac{27}{5}+\frac{23}{4}
Substitute \frac{27}{5} for y in x=-\frac{1}{4}y+\frac{23}{4}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{27}{20}+\frac{23}{4}
Multiply -\frac{1}{4} times \frac{27}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{22}{5}
Add \frac{23}{4} to -\frac{27}{20} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{22}{5},y=\frac{27}{5}
The system is now solved.
8x+2y=46,7x+3y=47
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&2\\7&3\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{5}\\\frac{27}{5}\end{matrix}\right)
Do the arithmetic.
x=\frac{22}{5},y=\frac{27}{5}
Extract the matrix elements x and y.
8x+2y=46,7x+3y=47
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
7\times 8x+7\times 2y=7\times 46,8\times 7x+8\times 3y=8\times 47
To make 8x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 8.
56x+14y=322,56x+24y=376
Simplify.
56x-56x+14y-24y=322-376
Subtract 56x+24y=376 from 56x+14y=322 by subtracting like terms on each side of the equal sign.
14y-24y=322-376
Add 56x to -56x. Terms 56x and -56x cancel out, leaving an equation with only one variable that can be solved.
-10y=322-376
-10y=-54
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# How do you differentiate y =( ln(3x + 2))^2 using the chain rule?
Jan 31, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 \ln \left(3 x + 2\right)}{3 x + 2}$
#### Explanation:
Differentiate using the $\textcolor{b l u e}{\text{ chain rule}}$
$y = {\left[\ln \left(3 x + 2\right)\right]}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(3 x + 2\right) \frac{d}{\mathrm{dx}} \left(\ln \left(3 x + 2\right)\right)$
$= 2 \ln \left(3 x + 2\right) . \frac{1}{3 x + 2} \frac{d}{\mathrm{dx}} \left(3 x + 2\right)$
$= 2 \ln \left(3 x + 2\right) . \frac{1}{3 x + 2} . 3$
$= \frac{6 \ln \left(3 x + 2\right)}{3 x + 2}$
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# Numbers and their Properties
This section reviews some basic properties about numbers and number systems. As you work through the examples and problems in this section, try them first without using your scientific calculator, using your brain, pencil, and paper alone. Then type the examples into your scientific calculator exactly as given so you can see how your calculator performs the same operations.
## The Real Numbers
In high school, you explored the real number system. This is a set of numbers that includes the integers or counting numbers, all the rational numbers (numbers that can be represented as a ratio of two whole numbers, such as 1/3 or 3/7) and the irrational numbers (numbers that cannot be represented as a ratio of two whole numbers, such as π, e, and sqrt(2)).
You learned about combining numbers using basic operations like addition (+), subtraction (-), multiplication (× or *) and division (/). You have also been introduced to several other mathematical functions including: exponentiation (^) -- raising something to a power -- and square root (). You may even remember some of the properties that real numbers obey:
Closure
If you operated on any two real numbers A and B with +, -, ×, or /, you get a real number.
Commutativity
A + B = B + A and A × B = B × A
Associativity
(A + B) + C = A + (B + C) and (A × B) × C = A × (B × C)
Inverse
A + -A = 0, A × (1/A) = 1
Identity
A + 0 = A, A × 1 = A
Distributivity
A × (B + C) = (A × B) + (A × C)
These properties are useful to review and keep in mind because they help you simplify more complex calculations.
## Order of Operations
In chemistry, as in the other sciences, we will encounter many problems in which the solution involves a series of calculations. When doing more than one operation we need follow a set of rules regarding which calculations to do first. For example, what is the "right" answer to
3 + 2 × 6 = ?
Should we go "left to right" and just do the + first and get 30, or do we do the × first and get 15? Well, in order to avoid confusion and get the correct answer, mathematicians decided long ago that all calculations should be done in the same order. You may have learned the order of operations as being: Please Excuse My Dear Aunt Sally! where the words stand for Parentheses, Exponentiation, Multiplication or Division, Addition or Subtraction.
So what is the correct answer for our problem? The order of operations would say that in the absence of parentheses, you would multiply 2×6 first, then add 3, so the result should be 15.
## Rounding Numbers
Another issue we need to deal with when we perform operations is how to state the answer. For example, if we are dividing a 20 centimeter wire into 3 equal pieces, we would divide 20 by 3 to get the length of each piece. If we took the time to work this division out by hand -- ack! -- we would get
20 / 3 = 6.666666666666666666.........
The 6 repeats forever. How do we report this number? We round to some usually pre-determined number of digits or decimal places. By "digits" we mean the total number of numbers both left and right of the decimal point. By "decimal places" we specifically refer to the number of numbers to the right of the decimal point.
For comparison, let's try rounding this number to 2 decimal places -- two numbers to the right of the point. To round, look at the digit after the one of interest -- in this case the third decimal place -- and use the rule:
if the digit is 0, 1, 2, 3 or 4 round down
if the digit is 5, 6, 7, 8 or 9 round up
In our example:
```6.666666666666.....
^
the next digit is 6 so we round up,
```
giving 6.67 as the desired answer. If instead we had been asked to round the number 20/3 to 2 digits the answer would have been 6.7 (two digits, one of which is a "decimal place").
Sometimes rounding is the result of an approximation. If you had 101 or 98 meters of some wire, in each case you would have "about 100 meters."
We will round many of our answers in science because the numbers will often be reporting measurements. Numbers representing measurements are only as accurate as the device used for measuring. For example, we could use a standard meter stick marked off in centimeters to measure the length of a wire as 15 cm. If sometime later we cut the wire in pieces, reporting the size of a piece of the wire to nine or ten decimal places would not make sense. You may encounter instructions in labs or in problem sets or on exams which ask you to round a measurement or result to the "nearest cm" or "nearest mL" or "nearest second". We'll look more at these accuracy issues in the section "Numbers in Science."
One other operation which you may or may not be familiar with is often used in chemistry is absolute value . The "absolute value" of a number is sometimes used instead of the number itself. Absolute value is represented as a pair of upright lines around whatever expression or number is supposed to be treated that way. The absolute value of a number expresses its distance from zero, regardless of sign. So, the absolute value of -20 is 20, and the absolute value of 19 is 19. Basically, all numbers become positive but do not change value. Most calculators will take an absolute value, usually called abs().
## Try It Out
For this section, you really should be able to work all of these problems correctly without the use of a scientific calculator. We suggest you try working all the problems -- step-by-step -- using pencil and paper to record your steps. Some useful hints and ideas are included in the solutions. When you have finished each exercise problem, type the operations into your calculator exactly as written for practice, and compare your answers.
1. What are the results of the following calculations?
(a)
4 + 5 × 10 - 32
(b)
(1 + 24 / 3)2
(c)
(-2)2
(d)
-22
2. Round the following numbers as specified.
(a)
1000.34532 rounded to 6 digits
(b)
12314.643 rounded to 3 digits
(c)
0.00006574 rounded to 6 decimal places
(d)
10.0029245 rounded to 2 decimal places
[Numbers and their Properties] [Numbers in Science] [Ratios and Proportions]
[Units, Dimensions, and Conversions] [Percents] [Simple Statistics] [Logarithms]
Developed by
Shodor
in cooperation with the Department of Chemistry,
The University of North Carolina at Chapel Hill
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# What is distributive property of subtraction?
## What is distributive property of subtraction?
The distributive property is a property of multiplication used in addition and subtraction. This property states that two or more terms in addition or subtraction with a number are equal to the addition or subtraction of the product of each of the terms with that number.
What is distributive property over multiplication?
The distributive property states that any expression with three numbers A, B, and C, given in form A (B + C) then it is resolved as A × (B + C) = AB + AC or A (B – C) = AB – AC. This property is also known as the distributivity of multiplication over addition or subtraction.
### What is distributive property over addition or subtraction?
The distributive property applies to the multiplication of a number with the sum or difference of two numbers i.e., the distributive property holds true for multiplication over addition and subtraction. Distributive property definition simply states that “multiplication distributed over addition.”
What is distributive property explain with example?
To “distribute” means to divide something or give a share or part of something. According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.
#### What are the 4 properties of subtraction?
Properties of subtraction.
• Closure property: For any two whole numbers, a and b, if a > b then a – b is a whole number and if a < b then a – b is never a whole number.
• Commutative property: For any two whole numbers a and b, a – b ≠ b – a . Hence subtraction of whole number is not commutative.
• Associative property:
• How do you solve distributive property using multiplication?
How to Use the Distributive Property of Multiplication
1. Simplify the numbers. In this example, 101 = 100 + 1, so:
2. Split the problem into two easier problems. Take the number outside the parentheses, and multiply it by each number inside the parentheses, one at a time.
## How do you explain the distributive property?
What is distributive property of multiplication example?
The distributive property of multiplication over addition is used when we multiply a value by the sum of two or more numbers. For example, let us solve the expression: 5(5 + 9). This expression can be solved by multiplying 5 by both the addends. So, 5(5) + 5(9) = 25 + 45 = 70.
### How do you find the distributive property of multiplication?
How do you use distributive property for Class 6?
Distributive property of multiplication over Addition: This property is used when we have to multiply a number by the sum. In order to verify this property, we take any three whole numbers a, b and c and find the values of the expressions a × (b + c) and a × b + a × c as shown below: Find 3 × (4 + 5).
#### What are the 3 properties of subtraction?
What are the rules of distributive property?
The distributive property defines that the product of a single term and a sum or differenceof two or more terms inside the bracket is same as multiplying each addend by the single term and then adding or subtracting the products. In general, the property is true for number of addends . Rule for multiplication over addition:
## What is the formula for distributive property?
The distributive property is expressed in math terms as the following equation: a(b + c) = ab + ac.
How do you use distributive property?
The distributive property also can be used to simplify algebraic equations by eliminating the parenthetical portion of the equation. Take for instance the equation a(b + c), which also can be written as (ab) + (ac) because the distributive property dictates that a, which is outside the parenthetical, must be multiplied by both b and c.
### How do we prove the distributive property of multiplication?
If you multiply some number by a natural number, you simply add it up as many times as is the natural number. With this knowledge at hand, we can easily prove the distributive property. Let’s say that you multiply a natural number N by a sum of a and b.
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# Difference between revisions of "2020 AMC 8 Problems/Problem 15"
## Problem
Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$
## Solution 1
Since $20\% = \frac{1}{5}$, multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{\textbf{(C) }75}$ percent of $x$.
## Solution 2
Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$. Clearly, it follows that $y$ is $75\%$ of $x$, so the answer is $\boxed{\textbf{(C) }75}$.
## Solution 3
We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$, so $\frac{3}{20}x=\frac{1}{5}y$. Solving for $y$, we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$, so the answer is $\boxed{\textbf{(C) }75}$.
## Solution 4
We are given $0.15x = 0.20y$, so we may assume without loss of generality that $x=20$ and $y=15$. This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$, and thus the answer is $\boxed{\textbf{(C) }75}$.
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4
Q:
# How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ?
A) 521 B) 720 C) 420 D) 225
Explanation:
Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240
Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.
Q:
From a group of 7 boys and 6 girls, five persons are to be selected to form a team, so that at least 3 girls are there in the team. In how many ways can it be done?
A) 427 B) 531 C) 651 D) 714
Explanation:
Given in the question that, there are 7 boys and 6 girls.
Team members = 5
Now, required number of ways in which a team of 5 having atleast 3 girls in the team =
2 149
Q:
The number of ways in which 8 distinct toys can be distributed among 5 children?
A) 5P8 B) 5^8 C) 8P5 D) 8^5
Explanation:
As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are ${\mathbf{5}}^{\mathbf{8}}$ ways to distribute the toys.
Hence, it is ${\mathbf{5}}^{\mathbf{8}}$ and not ${\mathbf{8}}^{\mathbf{5}}$.
1 671
Q:
In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together?
A) 1440 B) 720 C) 2250 D) 3600
Explanation:
Given word is THERAPY.
Number of letters in the given word = 7
These 7 letters can be arranged in 7! ways.
Number of vowels in the given word = 2 (E, A)
The number of ways of arrangement in which vowels come together is 6! x 2! ways
Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.
1 400
Q:
In how many different ways can the letters of the word 'HAPPYHOLI' be arranged?
A) 89,972 B) 90,720 C) 72,000 D) 81,000
Explanation:
The given word HAPPYHOLI has 9 letters
These 9 letters can e arranged in 9! ways.
But here in the given word letters H & P are repeated twice each
Therefore, Number of ways these 9 letters can be arranged is
5 543
Q:
How many words can be formed with or without meaning by using three letters out of k, l, m, n, o without repetition of alphabets.
A) 60 B) 120 C) 240 D) 30
Explanation:
Given letters are k, l, m, n, o = 5
number of letters to be in the words = 3
Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters =
7 532
Q:
The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements?
A) 4320 B) 4694 C) 4957 D) 4871
Explanation:
Given Word is PROMISE.
Number of letters in the word PROMISE = 7
Number of ways 7 letters can be arranged = 7! ways
Number of Vowels in word PROMISE = 3 (O, I, E)
Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways
Now, the number of ways of arrangements so that three vowels should not come together
= 7! - (5! x 3!) ways = 5040 - 720 = 4320.
7 992
Q:
In how many different ways can the letters of the word 'POVERTY' be arranged ?
A) 2520 B) 5040 C) 1260 D) None
Explanation:
The 7 letters word 'POVERTY' be arranged in $\mathbf{7}{\mathbit{P}}_{\mathbf{7}}$ ways = 7! = 5040 ways.
12 593
Q:
A decision committee of 5 members is to be formed out of 4 Actors, 3 Directors and 2 Producers. In how many ways a committee of 2 Actors, 2 Directors and 1 Producer can be formed ?
A) 18 B) 24 C) 36 D) 32
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BREAKING NEWS
Equality (mathematics)
Summary
In mathematics, equality is a relationship between two quantities or, more generally, two mathematical expressions, asserting that the quantities have the same value, or that the expressions represent the same mathematical object. Equality between A and B is written A = B, and pronounced "A equals B". In this equality, A and B are the members of the equality and are distinguished by calling them left-hand side or left member, and right-hand side or right member. Two objects that are not equal are said to be distinct.
A formula such as ${\displaystyle x=y,}$ where x and y are any expressions, means that x and y denote or represent the same object.[1] For example,
${\displaystyle 1.5=3/2,}$
are two notations for the same number. Similarly, using set builder notation,
${\displaystyle \{x\mid x\in \mathbb {Z} {\text{ and }}0
since the two sets have the same elements. (This equality results from the axiom of extensionality that is often expressed as "two sets that have the same elements are equal".[2])
The truth of an equality depends on an interpretation of its members. In the above examples, the equalities are true if the members are interpreted as numbers or sets, but are false if the members are interpreted as expressions or sequences of symbols.
An identity, such as ${\displaystyle (x+1)^{2}=x^{2}+2x+1,}$ means that if x is replaced with any number, then the two expressions take the same value. This may also be interpreted as saying that the two sides of the equals sign represent the same function (equality of functions), or that the two expressions denote the same polynomial (equality of polynomials).[3][4]
Etymology
The word is derived from the Latin aequālis ("equal", "like", "comparable", "similar"), which itself stems from aequus ("equal", "level", "fair", "just").[5]
Basic properties
• Reflexivity: for every a, one has a = a.
• Symmetry: for every a and b, if a = b, then b = a.
• Transitivity: for every a, b, and c, if a = b and b = c, then a = c.[6][7]
• Substitution: Informally, this just means that if a = b, then a can replace b in any mathematical expression or formula without changing its meaning.
• Operation application: for every a and b, with some operation ${\displaystyle f(x)}$ , if a = b, then ${\displaystyle f(a)=f(b)}$ .[8][a]
For example:
• Given real numbers a, and b, if a = b, then ${\displaystyle 2a-5=2b-5}$ . (Here, ${\displaystyle f(x)=2x-5}$ . A unary operation)
• Given real numbers a, and b, if ${\displaystyle a^{2}=2b^{2}}$ , then ${\displaystyle a^{2}/b^{2}=2}$ . (Here, ${\displaystyle f(x,y)=x/y^{2}}$ with ${\displaystyle y=b}$ . A binary operation)
• Given real-valued functions ${\displaystyle g}$ and ${\displaystyle h}$ over some variable a, if ${\displaystyle g(a)=h(a)}$ , then ${\textstyle {\frac {d}{da}}g(a)={\frac {d}{da}}h(a)}$ . (Here, ${\textstyle f(x)={\frac {dx}{da}}}$ . An operation over functions (i.e. an operator), called the derivative)
If restricted to the elements of a given set ${\displaystyle S}$ , those first three properties make equality an equivalence relation on ${\displaystyle S}$ . In fact, equality is the unique equivalence relation on ${\displaystyle S}$ whose equivalence classes are all singletons.
Equality as predicate
In logic, a predicate is a proposition which may have some free variables. Equality is a predicate, which may be true for some values of the variables (if any) and false for other values. More specifically, equality is a binary relation (i.e., a two-argument predicate) which may produce a truth value (true or false) from its arguments. In computer programming, equality is called a Boolean-valued expression, and its computation from the two expressions is known as comparison.
Equations
An equation is the problem of finding values of some variable, called unknown, for which the specified equality is true. Each value of the unknown for which the equation holds is called a solution of the given equation; also stated as satisfying the equation. For example, the equation ${\displaystyle x^{2}-6x+5=0}$ has the values ${\displaystyle x=1}$ and ${\displaystyle x=5}$ as its only solutions. The terminology is used similarly for equations with several unknowns.[9]
An equation can be used to define a set. For example, the set of all solution pairs ${\displaystyle (x,y)}$ of the equation ${\displaystyle x^{2}+y^{2}=1}$ forms the unit circle in analytic geometry; therefore, this equation is called the equation of the unit circle.
Identities
An identity is an equality that is true for all values of its variables in a given domain.[10] An "equation" may sometimes mean an identity, but more often than not, it specifies a subset of the variable space to be the subset where the equation is true. An example is ${\displaystyle \left(x+1\right)\left(x+1\right)=x^{2}+2x+1}$ is true for all real numbers ${\displaystyle x}$ . There is no standard notation that distinguishes an equation from an identity, or other use of the equality relation: one has to guess an appropriate interpretation from the semantics of expressions and the context.[11] Sometimes, but not always, an identity is written with a triple bar: ${\displaystyle \left(x+1\right)\left(x+1\right)\equiv x^{2}+2x+1.}$ [12]
In logic
In mathematical logic and mathematical philosophy, equality is often described through the following properties: [13][14][15]
${\displaystyle \forall a(a=a)}$
• Law of identity: Stating that each thing is identical with itself, without restriction. That is, for every ${\displaystyle a}$ , ${\displaystyle a=a}$ . It is the first of the historical three laws of thought.
${\displaystyle (a=b)\implies {\bigl [}\phi (a)\Rightarrow \phi (b){\bigr ]}}$ [b]
• Substitution property: Sometimes referred to as Leibniz's law, generally states that if two things are equal, then any property of one must be a property of the other. It can be stated formally as: for every a and b, and any formula ${\displaystyle \phi (x),}$ (with a free variable x), if ${\displaystyle a=b}$ , then ${\displaystyle \phi (a)}$ implies ${\displaystyle \phi (b)}$ .
For example: For all real numbers a and b, if a = b, then a ≥ 0 implies b ≥ 0 (here, ${\displaystyle \phi (x)}$ is x ≥ 0)
These properties offer a formal reinterpretation of equality from how it is defined in standard Zermelo–Fraenkel set theory (ZFC) or other formal foundations. In ZFC, equality only means that two sets have the same elements. However, outside of set theory, mathematicians don't tend to view their objects of interest as sets. For instance, many mathematicians would say that the expression "${\displaystyle 1\cup 2}$ " (see union) is an abuse of notation or meaningless. This is a more abstracted framework which can be grounded in ZFC (that is, both axioms can be proved within ZFC as well as most other formal foundations), but is closer to how most mathematicians use equality.
Note that this says "Equality implies these two properties" not that "These properties define equality"; this is intentional. This makes it an incomplete axiomatization of equality. That is, it does not say what equality is, only what "equality" must satify. However, the two axioms as stated are still generally useful, even as an incomplete axiomatization of equality, as they are usually sufficient for deducing most properties of equality that mathematicians care about.[16] (See the following subsection)
If these properties were to define a complete axiomatization of equality, meaning, if they were to define equality, then the converse of the second statement must be true. The converse of the Substitution property is the identity of indiscernibles, which states that two distinct things cannot have all their properties in common. In mathematics, the identity of indiscernibles is usually rejected since indiscernibles in mathematical logic are not necessarily forbidden. Set equality in ZFC is capable of declairing these indiscernibles as not equal, but an equality solely defined by these properties is not. Thus these properties form a strictly weaker notion of equality than set equality in ZFC. Outside of pure math, the identity of indiscernibles has attracted much controversy and criticism, especially from corpuscular philosophy and quantum mechanics.[17] This is why the properties are said to not form a complete axiomatization.
However, apart from cases dealing with indiscernibles, these properties taken as axioms of equality are equivalent to equality as defined in ZFC.
These are sometimes taken as the definition of equality, such as in some areas of first-order logic.[18]
Derivations of basic properties
• Reflexivity of Equality: Given some set S with a relation R induced by equality (${\displaystyle xRy\Leftrightarrow x=y}$ ), assume ${\displaystyle a\in S}$ . Then ${\displaystyle a=a}$ by the Law of identity, thus ${\displaystyle aRa}$ .
The Law of identity is distinct from reflexivity in two main ways: first, the Law of Identity applies only to cases of equality, and second, it is not restricted to elements of a set. However, many mathematicians refer to both as "Reflexivity", which is generally harmless.[19][c]
• Symmetry of Equality: Given some set S with a relation R induced by equality (${\displaystyle xRy\Leftrightarrow x=y}$ ), assume there are elements ${\displaystyle a,b\in S}$ such that ${\displaystyle aRb}$ . Then, take the formula ${\displaystyle \phi (x):xRa}$ . So we have ${\displaystyle (a=b)\implies (aRa\Rightarrow bRa)}$ . Since ${\displaystyle a=b}$ by assumption, and ${\displaystyle aRa}$ by Reflexivity, we have that ${\displaystyle bRa}$ .
• Transitivity of Equality: Given some set S with a relation R induced by equality (${\displaystyle xRy\Leftrightarrow x=y}$ ), assume there are elements ${\displaystyle a,b,c\in S}$ such that ${\displaystyle aRb}$ and ${\displaystyle bRc}$ . Then take the formula ${\displaystyle \phi (x):xRc}$ . So we have ${\displaystyle (b=a)\implies (bRc\Rightarrow aRc)}$ . Since ${\displaystyle b=a}$ by symmetry, and ${\displaystyle bRc}$ by assumption, we have that ${\displaystyle aRc}$ .
• Function application: Given some function ${\displaystyle f(x)}$ , assume there are elements a and b from its domain such that a = b, then take the formula ${\displaystyle \phi (x):f(a)=f(x)}$ . So we have
${\displaystyle (a=b)\implies [(f(a)=f(a))\Rightarrow (f(a)=f(b))]}$
Since ${\displaystyle a=b}$ by assumption, and ${\displaystyle f(a)=f(a)}$ by reflexivity, we have that ${\displaystyle f(a)=f(b)}$ .
This is also sometimes included in the axioms of equality, but isn't necessary as it can be deduced from the other two axioms as shown above.
Approximate equality
There are some logic systems that do not have any notion of equality. This reflects the undecidability of the equality of two real numbers, defined by formulas involving the integers, the basic arithmetic operations, the logarithm and the exponential function. In other words, there cannot exist any algorithm for deciding such an equality (see Richardson's theorem).
The binary relation "is approximately equal" (denoted by the symbol ${\displaystyle \approx }$ ) between real numbers or other things, even if more precisely defined, is not transitive (since many small differences can add up to something big). However, equality almost everywhere is transitive.
A questionable equality under test may be denoted using the ${\displaystyle {\stackrel {?}{=}}}$ symbol.[20]
Relation with equivalence, congruence, and isomorphism
Viewed as a relation, equality is the archetype of the more general concept of an equivalence relation on a set: those binary relations that are reflexive, symmetric and transitive. The identity relation is an equivalence relation. Conversely, let R be an equivalence relation, and let us denote by xR the equivalence class of x, consisting of all elements z such that x R z. Then the relation x R y is equivalent with the equality xR = yR. It follows that equality is the finest equivalence relation on any set S in the sense that it is the relation that has the smallest equivalence classes (every class is reduced to a single element).
In some contexts, equality is sharply distinguished from equivalence or isomorphism.[21] For example, one may distinguish fractions from rational numbers, the latter being equivalence classes of fractions: the fractions ${\displaystyle 1/2}$ and ${\displaystyle 2/4}$ are distinct as fractions (as different strings of symbols) but they "represent" the same rational number (the same point on a number line). This distinction gives rise to the notion of a quotient set.
Similarly, the sets
${\displaystyle \{{\text{A}},{\text{B}},{\text{C}}\}}$ and ${\displaystyle \{1,2,3\}}$
are not equal sets – the first consists of letters, while the second consists of numbers – but they are both sets of three elements and thus isomorphic, meaning that there is a bijection between them. For example
${\displaystyle {\text{A}}\mapsto 1,{\text{B}}\mapsto 2,{\text{C}}\mapsto 3.}$
However, there are other choices of isomorphism, such as
${\displaystyle {\text{A}}\mapsto 3,{\text{B}}\mapsto 2,{\text{C}}\mapsto 1,}$
and these sets cannot be identified without making such a choice – any statement that identifies them "depends on choice of identification". This distinction, between equality and isomorphism, is of fundamental importance in category theory and is one motivation for the development of category theory.
In some cases, one may consider as equal two mathematical objects that are only equivalent for the properties and structure being considered. The word congruence (and the associated symbol ${\displaystyle \cong }$ ) is frequently used for this kind of equality, and is defined as the quotient set of the isomorphism classes between the objects. In geometry for instance, two geometric shapes are said to be equal or congruent when one may be moved to coincide with the other, and the equality/congruence relation is the isomorphism classes of isometries between shapes. Similarly to isomorphisms of sets, the difference between isomorphisms and equality/congruence between such mathematical objects with properties and structure was one motivation for the development of category theory, as well as for homotopy type theory and univalent foundations.[22][23][24]
Equality in set theory
Equality of sets is axiomatized in set theory in two different ways, depending on whether the axioms are based on a first-order language with or without equality.
Set equality based on first-order logic with equality
In first-order logic with equality, the axiom of extensionality states that two sets which contain the same elements are the same set.[25]
• Logic axiom: ${\displaystyle x=y\implies \forall z,(z\in x\iff z\in y)}$
• Logic axiom: ${\displaystyle x=y\implies \forall z,(x\in z\iff y\in z)}$
• Set theory axiom: ${\displaystyle (\forall z,(z\in x\iff z\in y))\implies x=y}$
Incorporating half of the work into the first-order logic may be regarded as a mere matter of convenience, as noted by Lévy.
"The reason why we take up first-order predicate calculus with equality is a matter of convenience; by this we save the labor of defining equality and proving all its properties; this burden is now assumed by the logic."[26]
Set equality based on first-order logic without equality
In first-order logic without equality, two sets are defined to be equal if they contain the same elements. Then the axiom of extensionality states that two equal sets are contained in the same sets.[27]
• Set theory definition: ${\displaystyle (x=y)\ :=\ \forall z,(z\in x\iff z\in y)}$
• Set theory axiom: ${\displaystyle x=y\implies \forall z,(x\in z\iff y\in z)}$
Notes
1. ^ Rosser 2008, p. 163.
2. ^ Lévy 2002, pp. 13, 358. Mac Lane & Birkhoff 1999, p. 2. Mendelson 1964, p. 5.
3. ^ Equation. Springer Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equation&oldid=32613
4. ^ Pratt, Vaughan, "Algebra", The Stanford Encyclopedia of Philosophy (Winter 2022 Edition), Edward N. Zalta & Uri Nodelman (eds.), URL: https://plato.stanford.edu/entries/algebra/#Laws
5. ^ "Definition of EQUAL". Merriam-Webster. Archived from the original on 15 September 2020. Retrieved 9 August 2020.
6. ^ Stoll, Robert R. Set Theory and Logic. San Francisco, CA: Dover Publications. ISBN 978-0-486-63829-4.
7. ^ Lilly Görke (1974). Mengen – Relationen – Funktionen (4th ed.). Zürich: Harri Deutsch. ISBN 3-87144-118-X. Here: sect.3.5, p.103.
8. ^ Equality axioms. Springer Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equality_axioms&oldid=46837
9. ^ Sobolev, S.K. (originator). "Equation". Encyclopedia of Mathematics. Springer. ISBN 1402006098.
10. ^ Equation. Springer Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equation&oldid=32613
11. ^ Marcus, Solomon; Watt, Stephen M. "What is an Equation?". Retrieved 27 February 2019.
12. ^ "Identity – math word definition – Math Open Reference". www.mathopenref.com. Retrieved 1 December 2019.
13. ^ Equality axioms. Springer Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equality_axioms&oldid=46837
14. ^ Deutsch, Harry and Pawel Garbacz, "Relative Identity", The Stanford Encyclopedia of Philosophy (Fall 2024 Edition), Edward N. Zalta & Uri Nodelman (eds.), forthcoming URL: https://plato.stanford.edu/entries/identity-relative/#StanAccoIden
15. ^ Forrest, Peter, "The Identity of Indiscernibles", The Stanford Encyclopedia of Philosophy (Winter 2020 Edition), Edward N. Zalta (ed.), URL: https://plato.stanford.edu/entries/identity-indiscernible/#Form
16. ^ Equality axioms. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equality_axioms&oldid=46837
17. ^ French, Steven (2019). "Identity and Individuality in Quantum Theory". Stanford Encyclopedia of Philosophy. ISSN 1095-5054.
18. ^ Fitting, M., First-Order Logic and Automated Theorem Proving (Berlin/Heidelberg: Springer, 1990), pp. 198–200.
19. ^ Equality axioms. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equality_axioms&oldid=46837
20. ^ "Find all Unicode Characters from Hieroglyphs to Dingbats – Unicode Compart".
21. ^
22. ^ Eilenberg, S.; Mac Lane, S. (1942). "Group Extensions and Homology". Annals of Mathematics. 43 (4): 757–831. doi:10.2307/1968966. ISSN 0003-486X. JSTOR 1968966 – via JSTOR.
23. ^ Marquis, Jean-Pierre (2019). "Category Theory". Stanford Encyclopedia of Philosophy. Department of Philosophy, Stanford University. Retrieved 26 September 2022.
24. ^ Hofmann, Martin; Streicher, Thomas (1998). "The groupoid interpretation of type theory". In Sambin, Giovanni; Smith, Jan M. (eds.). Twenty Five Years of Constructive Type Theory. Oxford Logic Guides. Vol. 36. Clarendon Press. pp. 83–111. ISBN 978-0-19-158903-4. MR 1686862.
25. ^ Kleene 2002, p. 189. Lévy 2002, p. 13. Shoenfield 2001, p. 239.
26. ^ Lévy 2002, p. 4.
27. ^ Mendelson 1964, pp. 159–161. Rosser 2008, pp. 211–213
1. ^ 𝒇 can have any (countable) arity, but is written as unary to avoid cumbersome notation.
2. ^ Here 𝜙 can have any (finite) arity, however, it is written as a unary formula to avoid cumbersome notation.
Similarly, there should be quantifiers '∀' for a, b, and 𝜙, so more formally, this formula would be written as:
ab((a=b) ⇒͏ ∀𝜙[𝜙(...,a,...) ⇒͏ 𝜙(...,b,...)])
3. ^ More generally, equality itself can be formally said to be a "reflexive relation". Just not as relation within ZFC, but as a "meta-relation", within some of metatheory in mathematics, which may be ZFC itself. So one could describe equality as a reflexive relation in some "meta-ZFC", but not "internal-ZFC"
References
• Kleene, Stephen Cole (2002) [1967]. Mathematical Logic. Mineola, New York: Dover Publications. ISBN 978-0-486-42533-7.
• Lévy, Azriel (2002) [1979]. Basic set theory. Mineola, New York: Dover Publications. ISBN 978-0-486-42079-0.
• Mac Lane, Saunders; Birkhoff, Garrett (1999) [1967]. Algebra (Third ed.). Providence, Rhode Island: American Mathematical Society.
• Mazur, Barry (12 June 2007), When is one thing equal to some other thing? (PDF), archived from the original (PDF) on 24 October 2019, retrieved 13 December 2009
• Mendelson, Elliott (1964). Introduction to Mathematical Logic. New York: Van Nostrand Reinhold.
• Rosser, John Barkley (2008) [1953]. Logic for mathematicians. Mineola, New York: Dover Publication. ISBN 978-0-486-46898-3.
• Shoenfield, Joseph Robert (2001) [1967]. Mathematical Logic (2nd ed.). A K Peters. ISBN 978-1-56881-135-2.
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# The Best Lesson 17 Homework 4.3 Answer Key
Posted on
Solving the Common Core-aligned curriculum that equates mathematical concepts to stories called Eureka Math might be hard for some people. Fortunately, the answer key is everywhere on the internet. This post especially will inform you about the answer key of Lesson 17 Homework 4.3.
Show the division using disks. Relate your model to long division. Check your quotient and remainder by using multiplication and addition.
1. 7 : 2
Ones o o o o o o o o o o o o o
Quotient: 3
Remainder: 1
3 x 2 = 6
6 + 1 = 7
1. 73 : 2
Tens Ones o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
Quotient: 36
Remainder: 1
36 x 2 = 72
72 + 1 = 73
1. 6 : 4
Ones o o o o o o o o o o
Quotient: 1
Remainder: 2
4 x 1 = 4
4 + 2 = 6
Aside from the homework, you might also want to check out the answer key of the Lesson 17 Exit Ticket, as follows:
Show the division using disks. Relate your model to long division. Check your quotient by using multiplication and addition.
1. 5 : 4
Ones o o o o o o o o o
Quotient= 1
Remainder: 1
Just like Lesson 16, the whole and partition the chart are modeled into 4 parts to represent the divisor.
4 x 1 = 4
4 + 1 = 5
1. 53 : 4
Tens Ones o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
After distributing 4 tens, 1 ten remains. I change 1 ten for 10 ones.
Quotient: 13
Remainder: 1
13 x 4 = 52
52 + 1 = 53
After doing the Lesson 17 Homework 4.3, you might continue doing the Lesson 18 Homework 4.3. If you want to check how much do you get right, here is the answer key:
Solve using the standard algorithm. Check your quotient and remainder by using multiplication and addition.
1. 84 : 2
2. 84 : 4
3. 48 : 3
4. 80 : 5
5. 79 : 5
6. 91 : 4
7. 91 : 6
8. 91 : 7
9. 87 : 3
10. 87 : 6
11. 94 : 8
12. 94 : 6
In addition, you can also check out the answer key of the Lesson 19 Homework 4.3 below:
1. When you divide 86 by 4 there is a remainder of 2. Model this problem with number disks. In the number disk model, how can you see that there is a remainder?
Tens Ones o o o o o o o o o o o o o o o o o o o o o o o o o o
I can see that 2 ones are still left over.
1. Francine says that 86 : 4 is 20 with a remainder of 6. She reasons it is correct because (4 x 20) + 0 – 86. What mistake has Francine made? Explain how she correct her work.
Francine’s mistake is that her remainder is greater than her divisor. Instead of 20 groups, she can make one more group. 86 : 4 is 21 with a remainder of 2. (4 x 21) + 2 = 86.
1. The number disk model is showing 67 : 4. Complete the model. Explain what happens to the 2 tens that are remaining in the tens column.
Tens Ones 10 10 10 10 10 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 1 1 1 1 10 1 1 1 1 10 1 1 1 1 10 1 1 1 1
The 2 tens remaining get decomposed into 20 ones in the ones column.
1. Two friends share 76 blueberries.
a. To count the blueberries, they have put them into small bowls of 10. Draw a picture to show how the blueberries can be shared equally. Will they have to split apart any of the bowls of 10 blueberries when they share them?
10 10 10 10 10 10 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 10 10 1 1 1 1 1 1 1 1 10 10 10 1 1 1 1 1 1 1 1
Yes, they will have to split apart one of the bowls.
b. Explain how the friends can share the blueberries fairly.
The friends will get 3 bowls and 8 single berries to give each friend 38 berries.
1. Imagine you are drawing a comic strip showing how to solve the problem 72 : 4 to new fourth graders. Create a script to explain how you can keep dividing after getting a remainder of 3 tens in the first step.
Tens Ones o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
Watch as I divide 72 : 4 using a place value chart. First, I divide the tens. Each of the 4 groups have 1 ten. There are 3 tens remaining. I can continue dividing by decomposing the 3 tens into 3 ones. Now I have 32 ones. I can distribute the ones evenly into the 4 groups. Each group will have 8 ones. Now we see that 72 : 4 is 1 ten and 8 ones.
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# Are We Teaching Fractions Correctly?
Exploring evidence-based fraction instruction that supports a strong foundation in future mathematics
When I first started teaching third grade almost 20 years ago, my students and I loved the fraction unit in math! We all thought it was fun because I could bring in lots of great manipulatives (including food sometimes!) to help my students understand fractions. One year I bought ice cream sandwiches and after my students spent some time cutting them up into various numbers of equal sized pieces, and before the ice cream melted, the students got to eat parts of about 3/of the ice cream sandwiches!
I learned fractions myself when I was in grade school by working with representations such as pizzas, pies, and cakes. I remember countless worksheets where I was given shapes to split up into equal pieces and shade to make fractions. I then taught my own students the same way I learned and in the way the textbooks guided me. I now realize this emphasis on fractions being parts of a whole object or sets of objects, while fun and engaging, was not helpful for my students.
Like me, you probably weren’t taught fractions the right way, either.
The thing is, fractions are how we describe part of a whole; however, they are much more than that. Fractions are numbers. Fractions have distinct quantities. They have value and can be compared, ordered, and placed on a number line. Down the road, when students start to add, subtract, multiply, and divide fractions, it could be confusing for them to make sense of operations such as multiplying with a piece of pizza. However, it makes sense to multiple with numbers. Students have already multiplied with whole numbers. If they understand fractions as numbers also, it makes perfect sense to multiply fractions.
Consider this task from Illustrative Mathematics:
Students who understand that fractions are numbers will recognize that 2/3 is a quantity made up of two 1/3s and that 2/3 is 1/less than 1, and they will be able to select a representation for 5 x 4 that can also be used to represent 5 x 2/3. They will also likely be able to reason that the product will be less than 5 and explain their reasoning.
In these representations, students can make sense of both problems as having five groups of a quantity. In one case, the quantity is 4; but in the other case, the quantity is 2/3. The concept of groups that students began learning about in third grade related to multiplicative thinking can be extended when they start working with fractions. Leveraging previous understanding about multiplication of whole numbers can help students make sense of multiplication of fractions. This type of mathematical coherence is needed for students to carry mathematical concepts and ideas forward for success in Algebra.
Siegler et al. found that “elementary school students’ knowledge of fractions predicts their mathematics achievement in high school, above and beyond the contributions of… verbal and nonverbal IQ, working memory, family education, and family income.” Students who leave elementary school with a deep understanding of fractions are likely to be successful in high school mathematics.
Another important feature of numbers that is true for both whole numbers and fractions is that they can be composed and decomposed. In fact, students spend significant time in grades K-2 thinking about the place value system: for example how ten is composed of ten ones or how one hundred can be decomposed into many combinations of tens and ones. The idea of units, like ones, tens, and hundreds, continues into fractions when students start to learn that unit fractions (for example, 1/6) can be composed to make other fractions (such as 5/6 or 7/6). When I learned about fractions as a student, I know mixed numbers and “improper” fractions were introduced as new, separate concepts from fractions. In fact, they are all numbers.
I once asked a student, Parker, to complete a task that asked him to place the number 1 on a number line that showed two numbers, 0 and 5/3 (video available here).
Right away I could tell Parker was uncomfortable with 5/3 and not used to seeing fractions on a number line. Here is how our conversation went:
Teacher: Okay, Parker, so read the problem.
Parker: Task- The number line below shows two numbers, zero and five… [pause] three-fifths.
Teacher: Five-thirds
Parker: Five-thirds. So this should be one. [writes 1 at the end of the number line] This should be one. So right here, this would be one-third [marks near the zero], this would be two-thirds, [number marks moving toward the 5/3] so this would be three-thirds, this would be three-thirds [writes 4/3].
Teacher: Four-thirds, you mean?
Parker: Four-thirds.
Teacher: Okay, so look at the numbers you wrote there, all the fractions.
Parker: One-third, two-third, three-third, four… [shakes head, starts over] One-third, two-third… [pause, repeats] two-thirds, three-thirds, and four-thir… four-thirds.
Teacher: Okay, so where do you think one is again?
Parker: So, it’s not right there [erases 1 at the end of the number line], it’s right here [writes one above 5/3].
Teacher: Is five-thirds the same as one?
P: Five-thirds? [long pause, looking at paper] Yes.
Right away when looking at the task, Parker flipped 5/3 and said three-fifths. He continued to struggle later when he tried to read 4/3 in the count from one-third. Parker also was not clear where 1 fell in relation to 0 and 5/3. He assumed 1 would naturally go at the end of the line.
By spending time with 3.NF.A.1 (Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.) students are able to make sense of fractions greater than one. Students can reason about the value of fractions less than and greater than one. Students who understand that fractions that are built out of unit fractions, say 5/3 as being five, one-thirds or 1/3+1/3+1/3+1/3+1/3, can make sense of not only where the number one goes on the number line, at the point they have combined three one-thirds, but also why it makes sense to even have a number like five-thirds represented on a number line.
Later on in elementary or middle school when students encounter fractions greater than one when they are operating with fractions or when they work with fractions in algebraic equations, they understand the value of the fractions and products or quotients make sense.
By focusing on fractions as numbers and helping students make sense of fraction values, including emphasizing unit thinking, teachers can set students up for high school success. This is different from the way I was taught, as well as many other teachers I know, and it is likely different from the ways we have taught students in the past. However, as we dig into mathematical content knowledge and learn more about important math topics, we can do better for kids to set them up for math success.
## 8 thoughts on “Are We Teaching Fractions Correctly?”
1. Marshall Eubanks says:
When I see a video on YouTube that demonstrates for teachers the Five great ways to teach fractions and not one uses a number line or at best it is the last choice mentioned I realize it is not the students fault they don’t know or like math, it is the lack of commitment to what works and the lack of determination to ensure that it is taught correctly!
1. Jennifer Roder says:
I completely agree with you. I am a K-9 math interventionist now, but taught 6-12 math for 19 years prior. I see the worst of it in middle school actually, but definitely in lower grades as well. Most middle level math teachers hold a k-8 license. Which, in my state anyway, means they can teach any subject. The “horror” of this is clearly seen in mathematics. I cannot tell you how many times I have had to “fix” math knowledge for students coming into middle school or when they get to high school. I am not trying to bash anyone. This is a systemic problem. The problem is that the teachers do not have enough knowledge in mathematics to teach it correctly because they haven’t seen how fractions are connected to all things, especially when it comes to algebra concepts. It is a disservice to the teachers and to the students. More math requirements need to be implemented in teacher programs if this is going to change and it definitely needs to include this approach.
1. Melissa Hosten says:
A deficit view of elementary teachers and elementary credentials is more concerning in this comment than the fraction instruction. I’ve taught secondary, elementary, middle, university, and what is true is ALL teachers can teach deeply. Many need additional support to explore these concepts, but that “many” is at any level with any credentials!
2. Josh Zagorski says:
Great post, thank you for sharing this…
3. John says:
Hi there. As a K-5 Math Coach, I completely agree that most folks were never taught the conceptual meaning that underlies fractions ( I know I was not)! As I go deeper into my own understanding and bring this forward to the amazing teachers I work with it is a win-win for the adults who will bring this knowledge into their classrooms. we are starting this year with Equal Sharing in Kinder and the teachers are really enjoying and so are their students.
4. Joanne Robert says:
7th grade teacher; the best method I’ve used to make students see fractions as numbers is to always make “whole” numbers fractions. Ie: They don’t see 2 written as 2/1 enough.
5. Ina Mogensen says:
Thank you for your clear and thought-provoking article. I am just starting as a math interventionist, and my main problem right now is helping sixth graders fill in 3 years worth of gaps in their understanding of fractions.
Clearly these students have been taught procedures each year to get the right answer to all things that can be done to or with fractions. However, now it is painfully obvious they have no in-depth understanding of fractions as numbers. These sixth graders are not able to compose and decompose common fractions or to demonstrate relationships between and among fractions and whole numbers. That leaves them with bits and pieces of procedures floating around in their heads, but with terrible misconceptions about how and when to apply those procedures.
Even worse, as a third grade teacher I was guilty of instilling some of those misconceptions into young minds. I thought I knew math and had actually had some very good training. With parts of a whole or parts of a set. When the Common Core and then new State standards stressed teaching fractions on a number line I couldn’t figure out why that was so important (and it seemed like One More Thing). So I accommodated this new addition with … more procedures.
Your article and my own experience point out the critical need for all teachers to learn the progression of concepts through the grades and also see how their part contributes later success.
Thank you, #achievethecore for resources that help us all improve instruction by learning more math and seeing more connections while also providing student tasks that point the way!
6. Y. D says:
Thank you for the enlightenment! It definitely helps me tighten up on my own instructions and explains why my sixth graders have the deficits they do. Where I understand the importance and depth of teaching fractions, I will do a better job assisting my colleagues. Each one teaches one.
About the Author: Jennie Beltramini is a Mathematics Specialist on the Professional Learning team at Student Achievement Partners. Prior to joining the team, Jennie coordinated, supported, and facilitated state-wide professional learning on Common Core State Standards for the Washington Education Association. Prior to this work, Jennie spent 19 years teaching, instructional coaching, and leading professional learning in school districts in Washington State. Jennie is a National Board Certified Teacher (NBCT) in Early Adolescent Mathematics. Jennie holds a bachelor's degree in K-8 Education from Washington State University, a master's degree in Curriculum and Instruction from Lesley University, and a Certificate of Teacher Leadership from the University of Washington.
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# Video: Finding the Measures of the Angles of a Parallelogram Given a Relation Between Given Angles
π΄π΅πΆπ· is a parallelogram, and πβ π·π΄πΆ = 5π₯Β°, πβ π΅π΄πΆ = 3π₯Β°, and πβ π΄π΅πΆ = 4π₯Β°. Determine πβ π΅πΆπ· and πβ π΄π·πΆ.
03:25
### Video Transcript
π΄π΅πΆπ· is a parallelogram, and the measure of angle π·π΄πΆ equals five π₯ degrees, the measure of angle π΅π΄πΆ equals three π₯ degrees, and the measure of angle π΄π΅πΆ equals four π₯ degrees. Determine the measure of angle π΅πΆπ· and the measure of angle π΄π·πΆ.
When weβre given a question like this, the best way to begin is by filling in the angle information that weβre given. Angle π·π΄πΆ is five π₯ degrees, angle π΅π΄πΆ is three π₯ degrees, and angle π΄π΅πΆ is four π₯ degrees. The two angles that weβre asked to find out are the angle π΅πΆπ· and the angle π΄π·πΆ. Weβre told that π΄π΅πΆπ· is a parallelogram, which means that the opposite sides are parallel and congruent. In order to answer this question with the angles in the parallelogram, weβll need to recall an important fact about the angles in a parallelogram, which is that opposite angles in a parallelogram are equal or congruent.
We could therefore immediately look at our diagram and see that the angle at π΅ must be equal to the angle at π·, which is four π₯ degrees. In the same way, this whole angle at π·π΄π΅, which is made up of five π₯ degrees and three π₯ degrees, will be equal to this angle at π·πΆπ΅, meaning that it is also eight π₯ degrees. Looking at the diagram, we can see that weβve found the two unknown angles in terms of π₯. However, we might wonder if we could find the value of π₯ and give these answers in terms of a numerical value.
There are in fact two different ways in which we could find the value of π₯. The first method involves using a property of parallelograms, which is that any two consecutive angles at a parallelogram are supplementary, which means they add up to 180 degrees. So letβs take the pair of angles of four π₯ degrees and eight π₯ degrees, and we would set this equal to 180 degrees. Simplifying the left-hand side, we have 12π₯ degrees is equal to 180 degrees. We will then divide both sides by 12 to find that π₯ is equal to 15 degrees.
The alternative method we could use is to recall that the angles in any quadrilateral add up to 360 degrees. We could then add our four angles, four π₯, eight π₯, four π₯, and eight π₯ degrees, and set this equal to 360 degrees. Simplifying this would give us 24π₯ is equal to 360 degrees. Dividing by 24 then would give us once again that π₯ is equal to 15 degrees. But of course, we werenβt asked to simply find π₯. We were asked for two angle measurements. For the angle π΅πΆπ·, we established that this was eight π₯ degrees, and we worked out that π₯ was 15. So, we calculate eight multiplied by 15. And that is 120 degrees. Angle π΄π·πΆ we worked out as four π₯ degrees, so this time weβre multiplying four by 15, which is 60 degrees.
And therefore, we have the answers for the two angle measurements: 120 degrees and 60 degrees.
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# Placement Papers - TCS
## Why TCS Placement Papers?
Learn and practice the placement papers of TCS and find out how much you score before you appear for your next interview and written test.
## Where can I get TCS Placement Papers with Answers?
IndiaBIX provides you lots of fully solved TCS Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve TCS Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### TCS Test Paper with Answer - Kurnool, September 13, 2011.
Posted By : Anantha Lakshmi Rating : +109, -8
1. (1/2) Of A Number Is 3 Times More Than The (1/6) Of The Same Number?
Solution:1/2x=3+1/6x =>x=9
2. There are two water tanks A and B, A is much smaller than B. While water fills at the rate of 1 liter every hour in A, it gets filled up like, 10, 20, 40, 80, 160 in tank B. (At the end of first hour, B has 10 liters, second hour it has 20 liters and so on). If tank B is 1/32 filled of the 21 hours, what is total duration of hours required to fill it completely?
a) 26 B)25 c)5 d)27
Solution: for every hour water in tank in B is doubled,
Let the duration to fill the tank B is x hours.
X/32 part of water in tank of B is filled in 21 hours,
Next hour it is doubled so,
2*(x/32) part i.e (x/16) part is filled in 22 hours,
Similarly (x/8) th part in 23 hours,(x/4)th part is filled in 24 hours,
(x/2) th part is filled in 25 hours, (x) th part is filled in 26 hours
3. There Are Two Pipes A And B. If A Filled 10 Liters In Hour B Can Fills 20 Liters In Same Time. Likewise B Can Fill 10, 20, 40, 80,160.... If B Filled In (1/16) Th Of A Tank In 3 Hours, How Much Time Will It Take To Fill Completely?
Ans: 7 Hours
(2 Questions are given of this model.)
4. Ferrari S.P.A is an Italian sports car manufacturer based in Maranello, Italy. Founded by Enzo Ferrari in 1928 as Scuderia Ferrari, the company sponsored drivers and manufactured race cars before moving into production of street-legal vehicles in 1947 as Ferrari S.P.A. Throughout its history, the company has been noted for its continued participation in racing, especially in Formula One where it has employed great success .Rohit once bought a Ferrari. It could go 4 times as fast as Mohan,s old Mercedes. If the speed of Mohan,s Mercedes is 35 km/hr and the distance traveled by the Ferrari is 490 km, find the total time taken for Rohit to drive that distance.
a) 20.72 b) 3.5 c) 238.25 d) 6.18
Solution: 490/(4*35)
5. A circular dartboard of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point Q in the circle. What is the probability that Q is closer to the center of the circle than the periphery?
0.75
1
0.5
0.25
solution:0.25
If it is a circular dartboard : (PI*((r/2)^2))/(PI*(r^2))=0.25
If it is a square dartboard : (PI*((r/4)^2))/(r^2)=(app)0.2
6. For the FIFA world cup, Paul the octopus has been predicting the winner of
each match with amazing success. It is rumored that in a match between 2 teams A and B, Paul picks A with the same probability as A,s chances of winning.
Let,s assume such rumors to be true and that in a match between Ghana and Bolivia, Ghana the stronger team has a probability of 2/3 of winning the game. What is the probability that Paul will correctly pick the winner of the Ghana-Bolivia game?
4/9
2/3
1/9
5/9
Solution:(1/3*1/3)+(2/3*2/3)=5/9
7. The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base
buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?
256
54
192
64
Solution:From 1 to 1000 number of 3's are 300.
Since they mentioned 8 fingered we have to take base 8.
i.e., (3*(8^2))+(0*(8^1))+(0*(8^0))=192
8. 36 people {a1, a2, ., a36} meet and shake hands in a circular fashion. In
other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3},...., {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is
12
13
18
11
Solution: for at least one person & for exactly one person =36/3
for at least two person & for exactly two person =36/2
9. Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is
4
3
0
1
Solution: For 3 lines of intersection the answer is 4
For 3 linesegments of intersection the answer is 1
10. Alok and Bhanu play the following min-max game. Given the expression
N=9+X+Y-Z where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be
27
18
20
Solution: 9+11=20
For X+Y-Z=11,X*(Y-Z)=18, X-Y-Z=2
11. After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?
0
12/212
11/12
1/12
12. 10 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?
A. All suspects are lying or the leftmost suspect is innocent.
B. All suspects are lying and the leftmost suspect is innocent .
A only
Neither A nor B
Both A and B
B only
All suspects are lying or the leftmost suspect is innocent.
All suspects are lying and the leftmost suspect is innocent .
Both can satisfy the situation so answer is "Both A and B"
(we have to take only truth values)
13. In the reading room of a library, there are 23 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. There are some readers such that in each occupied reading spot there are different numbers of readers. If in all there are 36 readers, how many reading spots do not have even a single reader?
a) 8
b) none
c) 16
d) 15
Solution: 23 reading spots, Each reading spot consists of 9 chairs placed around it so There are some readers such that in each occupied reading spot there are different numbers of readers. For each table different no of persons are sat,so for first table 1 person is sit,2nd table 2 persons are sit 36
14. Middle- earth is a fictional land inhabited by hobbits, elves, dwarves and men. The hobbits and elves are peaceful creatures that prefer slow, silent lives and appreciate nature and art. The dwarves and the men engage in physical games. The game is as follows. A tournament is one where out of the two teams that play a match, the one that loses get eliminated. The matches are played in different rounds, where in every round; half of the teams get eliminated from the tournament. If there are 8 rounds played in knock out tournament, how many matches were played?
a) 257
b) 256
c) 72
d) 255
Solution: Number of matches=2^8
Number of teams=2^8-1
Number of members left in 1st round=128
15. There is 7 friends (A1, A2, A3....A7).If A1 have to have shake with all without repeat. How many handshakes possible?
a) 6
b) 21
c) 28
d) 7
Solution:6
16. There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
a)1/2
b)14/19
c)37/38
d)3/4
Ans: 14/19
Solution: ((1/2)*1)+((1/2)*(9/19))=14/19
17. On the planet Oz, there are 8 days in a week- Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 min while each minute has 60 sec. As on earth, the hour hand covers the dial twice every day. Find the approximate angle between the hands of a clock on Oz when the time is 9:40 am.
a. 251
b.111
c.29
d.89
Ans.: c
For 12:40 am Ans.. z 89
Formule: 20H-(34/9)M
18. The pacelength P is the distance between the rear of two consecutive footprints. For men, the formula, n/P = 144 gives an approximate relationship between n and P where, n = number of steps per minute and P = pacelength in meters. Bernard knows his pacelength is 164cm. The formula applies to Bernard's walking. Calculate Bernard's walking speed in kmph.
23.62
11.39
8.78
236.16
Ans.:a
(144*164*164*60)/(10^7)
19. How many 4 digit numbers can be formed using the digits 1, 2,3,4 ,5 ( but with repetition) that are divisible by 4?
Ans.. 5^3
20. There are 6 beer bottle nd one is poisoned. we have mice who will die within 14 hrs after drinkin poisned beer. In 24 hrs we have to find poisoned beer bottle. How many no of mice we require to find out poisoned bottle.
options
a) 6
b) 4
c) 3
d) 1
6 is near to 2^3
21. keyword: Alok Bhanu, stack of 20 coins. i move can play. dat z 1th move meAns. to put top coin 1 postion below. gold coin. initialy gold coin is at 3rd position from top. if itz player turns nd player brings gold coin to the top then player z winner. alok starts. which of the following is true.
a) alok must play 1th move to win.
b) alok must play 0th move to win
c) alok has no wining strategy
Ans. a
22. A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 4 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
800
500
488
900
Ans.:side of cube = 5 cm
its thichness = 1 cm
so volume of outer cube = 5*5*5
volume of inner cube = 3*3*3
volume of the hollow cube = 5*5*5 – 3*3*3 = 98
so total no of small cubes of the size 1 cm = 98/1*1*1 = 98
we know a cube has 6 faces so total no of face = 98*6 = 588
one surface of outer cube contains a total of 25 surface of smaller cube , so when 4 surface of outer cube is painted total no of surface of small cubes i.e supposed to be painted is 4*25 = 100
so the total no of surfaces of small cube that will be remained unpainted is 588-100 = 488
no of faces remain unpainted: 588-( 25* no faces painted)
23. A lady has fine gloves and hats in her closet- 18 blue, 32 red, and 25 yellow. The lights are out and it is totally dark. In spite of the darkness, she can make out the difference between a hat and a glove. She takes out an item out of the closet only if she is sure that if it is a glove. How many gloves must she take out to make sure she has a pair of each color?
a)50 b)8 c)60 d)42
Ans.:60
There can be lots of logic for this, but approach the simplest one so that we approach to one of the solutions. Suppose the lady first picks 32 Red gloves, and then 24 Yellow gloves. The next pair she pics will be one Yellow and One Blue which does not make a pair. The next two will be blue gloves. So she make a total of 32+24+1+1+2 = 60 picks.
24. The IT giant Tirnop has recently crossed a head count of 150000 and earnings of \$7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries Ans. also write code at the same rate.Suppose 12 such programmers take 12 minutes to write 12 lines of code in total. How long will it take 72 programmers to write 72 lines of code in total?
Ans. 12 min
prgrmr*min/ loc= prgmr *min/ loc
This question was also repeated thrice.
a) no of Programmer
b) no lines of codes
c) how much time they will take.
25.There are 6 circles on a diagonal of square such that their centre lie on diagonal. Consider that radius of each circle is equal. Find the ratio between side of square and radius of circle.
10r+ 2sqrt(2)r= sqrt(2)a
find a/r.
26. On a sheet of paper, there are 40 statements. Each n statement states that "Atleast n number of statements on this sheet is false". Then which of the following is true.
Answer: first 20 statements are true and last 20 are false.
This question was also repeated thrice.
a) Atleast wid false Ans. first 20 r true and last 20 r false..
b) Exactly wid true Ans. 39th z true nd rest r false
c) Atmost wid true or false Ans. all statements r true
27. On planet korba, a solar blast has melted the ice caps on its equator. 9 years after the ice melts, tiny planetoids called echina start growing on the rocks. Echina grows in the form of circle, and the relationship between the diameter of this circle and the age of echina is given by the formula d = 4*√ (t-9) for t ≥ 9 where d represents the diameter in mm and t the number of years since the solar blast.Jagan recorded the radius of some echina at a particular spot as 7mm. How many years back did the solar blast occur?
a) 17
b) 21.25
c) 12.25
d) 14.05
28. John buys a cycle for 31 dollars and given a cheque of amount 35 dollars. Shop Keeper exchanged the cheque with his neighbor and gave change to John. After 2 days, it is known that cheque is bounced. Shop keeper paid the amount to his neighbor. The cost price of cycle is 19 dollars. What is the profit/loss for shop keeper?
a) loss 23
b) gain 23
c) gain 54
d) Loss 54
Solution:-23 loss as he has to pay 35 dollar to neighbor he has only 31 remaining so 35-31=4 and cycle costs 19 dollar so total loss=19+4=23 dollar
INTERVIEWS:
It may be on the same day of the Test or next day, so be Prepared in Advance.
The Interview will be done by a Panel of 3 to 4 members. No need to worry. They will be Very supporting & Friendly.. First they ask us to Introduce ourselves... then they'll go into your project details... don't forget to prepare project.... They mainly concentrate on your project only...so be confident on every aspect of your project... then you should have some basic knowledge on C language. There will be some questions like "what is the o/p of this code?", "what are the errors in the code?"..like that... so be prepared for all the basic concepts.. If needed you have to write some small codes too.. at the end they will ask some general questios like "Where do you see yourself in the next five years?", "do you migrate?","do you have any questions?", try ask some questions... it shows your interest in the Company..
HR:
Then comes your Final Attack.. the HR Round . It involves no technical questions... it just dwell into your innerself... so, be confident and frank while facing this round...They generally ask questions like "Tell about yourself?", "Who r your role models?", "Why TCS?", "What you do in leasure time?","What is your weekness?" don't say "I Have no weakness!", just prepare a small weakness for yourself that causes no harm to your future...O.K.. "What is your mother tongue?".
All the best guys!
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# Math Snap
## Given the equation below, find $\frac{d y}{d x}$. $28 x^{6}+5 x^{28} y+y^{9}=34$
#### STEP 1
Assumptions 1. The given equation is $28 x^{6} + 5 x^{28} y + y^{9} = 34$. 2. We need to find the derivative $\frac{d y}{d x}$. 3. We will use implicit differentiation since $y$ is a function of $x$.
#### STEP 2
Differentiate both sides of the equation with respect to $x$. $\frac{d}{dx} \left( 28 x^{6} + 5 x^{28} y + y^{9} \right) = \frac{d}{dx} (34)$
#### STEP 3
Differentiate the first term $28 x^{6}$ with respect to $x$. $\frac{d}{dx} (28 x^{6}) = 28 \cdot 6 x^{5} = 168 x^{5}$
#### STEP 4
Differentiate the second term $5 x^{28} y$ with respect to $x$ using the product rule. Let $u = 5 x^{28}$ and $v = y$. $\frac{d}{dx} (5 x^{28} y) = \frac{d}{dx} (u v) = u \frac{d v}{dx} + v \frac{d u}{dx}$
#### STEP 5
Calculate $\frac{d u}{dx}$ where $u = 5 x^{28}$. $\frac{d}{dx} (5 x^{28}) = 5 \cdot 28 x^{27} = 140 x^{27}$
#### STEP 6
Substitute $u$, $\frac{d u}{dx}$, and $\frac{d v}{dx}$ into the product rule formula. $\frac{d}{dx} (5 x^{28} y) = 5 x^{28} \frac{d y}{dx} + y \cdot 140 x^{27}$
#### STEP 7
Differentiate the third term $y^{9}$ with respect to $x$ using the chain rule. $\frac{d}{dx} (y^{9}) = 9 y^{8} \frac{d y}{dx}$
#### STEP 8
Differentiate the constant $34$ with respect to $x$. $\frac{d}{dx} (34) = 0$
#### STEP 9
Combine all the differentiated terms. $168 x^{5} + (5 x^{28} \frac{d y}{dx} + 140 x^{27} y) + 9 y^{8} \frac{d y}{dx} = 0$
#### STEP 10
Group the terms involving $\frac{d y}{dx}$ on one side of the equation. $168 x^{5} + 140 x^{27} y + 5 x^{28} \frac{d y}{dx} + 9 y^{8} \frac{d y}{dx} = 0$
#### STEP 11
Factor out $\frac{d y}{dx}$ from the terms on the left side. $168 x^{5} + 140 x^{27} y + \left( 5 x^{28} + 9 y^{8} \right) \frac{d y}{dx} = 0$
#### STEP 12
Isolate $\frac{d y}{dx}$ by moving the other terms to the right side. $\left( 5 x^{28} + 9 y^{8} \right) \frac{d y}{dx} = - (168 x^{5} + 140 x^{27} y)$
##### SOLUTION
Solve for $\frac{d y}{dx}$ by dividing both sides by $5 x^{28} + 9 y^{8}$. $\frac{d y}{dx} = \frac{- (168 x^{5} + 140 x^{27} y)}{5 x^{28} + 9 y^{8}}$ The derivative $\frac{d y}{dx}$ is: $\frac{d y}{dx} = \frac{- (168 x^{5} + 140 x^{27} y)}{5 x^{28} + 9 y^{8}}$
|
# How to Write 1000 in Scientific Notation
Instructor: Artem Cheprasov
Scientific notation sounds may sound scary, but it's really not. It's actually very easy! In this lesson, you'll learn how to write 1000 in scientific notation in a step by step fashion.
## Steps To Solve
### Background Information
If you've ever wanted to simplify a really big number into something much easier read, you'd use scientific notation. Scientific notation is the shorthand writing method for numbers in math. To use scientific notation, you'd write a number in the following format:
A x 10b
Here's what all those letters stand for:
• A is called the coefficient. The coefficient is always greater than or equal to 1 but less than 10.
• 'x' is the multiplication sign. You knew that!
• 10 is known as the base (and it's always 10 when using scientific notation).
• b is just the exponent. Perhaps you've heard of it alternatively as 'the power of 10'.
### Steps
Knowing this, let's see how 1000 is written in scientific notation.
Step 1. Since 1000 doesn't have a decimal point anywhere, place a decimal point at the end of 1000 to make it read '1000.' This will make it easier to follow the additional steps outlined herein.
Step 2. Move this number's ('1000.') decimal point to the very right of the first non-zero digit in the number. In this instance, it's super easy. The only non-zero digit in 1000 is '1'. Following step 2's instruction, '1000' turns into '1.000'.
Step 3. Count how many places you had to move that decimal point. If you think three, you're right! That's the exponent, '3'. But you must also consider the direction you moved the decimal point. When any decimal point is moved to the left, the exponent is always positive. Since it was moved three places and it was moved to the left, the exponent is +3 or simply 3.
Step 4. To get your coefficient, drop any zeroes that are located before the first non-zero digit and after the last non-zero digit of step 2's answer of '1.000'. There are no zeroes to the left of '1.000' and there are three zeroes to the right of '1.000'. Thus, we drop all those zeroes to the right and we are left with '1', our coefficient.
And, that's it! You now have everything you need to write 1000 in scientific notation:
• A, the coefficient, is '1' in our case.
• 'x' is the multiplication sign. This never changes.
• 10 is the base and this never changes either.
• b is the exponent and step 3 told us it is '3' in our case.
Now, all we have to do is fill in the information for A x 10b to get our answer! Doing so, we get:
1 x 103
## Solution
Writing 1000 in scientific notation isn't difficult at all. It is written as 1 x 103.
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# Clock Questions
25 April 2024
#### Basic Concept of Clock
In an aptitude test, you will get questions based on concepts of Clock. Wherein the concept is based on movement of Hour hand and Minute hand.
A clock is a complete circle having 360 degrees, which is equally divided into 12 parts i.e. each part is 360/12 = 30°.
The minute hand rotates through 360 degrees to complete round in one hour.
Angle traced by minute hand in 60 min. = 360°.
In 1 minute it covers 360/60 = 6° per minute.
Angle traced by hour hand in 12 hrs = 360°
The hour hand covers 30° in 60 minutes i.e. ½° per minute.
Both the hands of a clock coincide once in every hour.
The hands of a clock will be in straight line but opposite in direction, 22 times in a day.
The hands of a clock coincide 22 times in a day and become straight 44 times in a day.
The hands of a clock are at right angles 44 times in a day.
We will use these concepts while solving problems on clocks.
Q1) An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?
A) 154°
B) 170°
C) 160°
D) 180°
#### Solution:
Q2) A clock is started at noon. By 10 minutes past 5, the hour hand has turned through
A) 145°
B) 152°
C) 155°
D) 140°
#### Solution :
Q3) At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but not together?
A) 5(1/11) minutes past 7
B) 5 minutes past 7
C) 5(3/11) minutes past 7
D) 5(5/11) minutes past 7
#### Solution:
Q4) At what time between 5.30 and 6 will the hands of a clock be at right angles?
A) 44(7/11) minutes past 5
B) 43 minutes past 5
C) 43(7/11) minutes past 5
D) 44 minutes past 5
#### Solution :
Q5) An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?
A. 144°
B. 150°
C. 168°
D. 180°
#### Solution :
Angle traced by the hour hand in 6 hours =
Q6) The reflex angle between the hands of a clock at 10.25 is:
A) 180°
B) 192(1/2)°
C) 195°
D) 192(1/2)°
#### Solution :
Q7) A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:
A. 145°
B. 150°
C. 155°
D. 160°
#### Solution :
Angle traced by hour hand in 12 hrs = 360°.
Angle traced by hour hand in 5 hrs 10 min. i.e.,
Q8) A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:
A) 5 PM
B) 4 PM
C) 58(7/11)min. past 3
D) None Of These
#### Solution :
Time from 7 a.m. to 4.15 p.m. = 9 hrs 15 min. = 37/4 hrs.
3 min. 5 sec. of this clock = 3 min. of the correct clock.
37 / 720 hrs of this clock = 3 min. of the correct clock.
=> 37/ 720 hrs of this clock = 1/ 20 hrs of the correct clock.
=> 37 / 4 hrs of this clock = (1/20 x 720/37 x 37/4) hrs of the correct clock
= 9 hrs of the correct clock.
The correct time is 9 hrs after 7 a.m. i.e., 4 p.m.
Q9) At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?
A) 5 min. past 7
B) 5 (2/11) min. past 7
C) 5 (3/11) min. past 7
D) 5 (5/11) min. past 7
#### Solution :
When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min. spaces are gained in (60/55 x 5)min = 5 (5/11) min.
Required time = 5 (5/11) min. past 7.
Q10) At what time between 5.30 and 6 will the hands of a clock be at right angles?
A) 43 (5/11) min. past 5
B) 43 (7/11) min. past 5
C) 40 min. past 5
D) 45 min. past 5
#### Solution:
At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.
55 min. spaces are gained in 60 min.
40 min. spaces are gained in (60/55 x 40) min = 43(7/11)min.
Required time = 43 (7/11) min. past 5.
## Any Questions? Look Here.
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## RD Sharma Solutions for Class 8 Chapter 19 Visualising Shapes Free Online
EXERCISE 19.1 PAGE NO: 19.9
1. What is the least number of planes that can enclose a solid? What is the name of the solid?
Solution:
The least number of planes that are required to enclose a solid is 4.
The name of solid is tetrahedron.
2. Can a polyhedron have for its faces?
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution:
(i) 3 triangles?
No, because a polyhedron is a solid shape bounded by polygons.
(ii) 4 triangles?
Yes, because a tetrahedron as 4 triangles as its faces.
(iii) a square and four triangles?
Yes, because a square pyramid has a square and four triangles as its faces.
3. Is it possible to have a polyhedron with any given number of faces?
Solution:
Yes, if number of faces is four or more.
4. Is a square prism same as a cube?
Solution:
Yes. We know that a square is a three dimensional shape with six rectangular shaped sides, out of which two are squares. Cubes are of rectangular prism length, width and height of same measurement.
5. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
No.
Let us use Euler’s formula
V + F = E + 2
15 + 10 = 20 + 2
25 ≠ 22
Since the given polyhedron is not following Euler’s formula, therefore it is not possible to have 10 faces, 20 edges and 15 vertices.
6. Verify Euler’s formula for each of the following polyhedrons:
Solution:
(i) Vertices = 10
Faces = 7
Edges = 15
By using Euler’s formula
V + F = E + 2
10 + 7 = 15 + 2
17 = 17
Hence verified.
(ii) Vertices = 9
Faces = 9
Edges = 16
By using Euler’s formula
V + F = E + 2
9 + 9 = 16 + 2
18 = 18
Hence verified.
(iii) Vertices = 14
Faces = 8
Edges = 20
By using Euler’s formula
V + F = E + 2
14 + 8 = 20 + 2
22 = 22
Hence verified.
(iv) Vertices = 6
Faces = 8
Edges = 12
By using Euler’s formula
V + F = E + 2
6 + 8 = 12 + 2
14 = 14
Hence verified.
(v) Vertices = 9
Faces = 9
Edges = 16
By using Euler’s formula
V + F = E + 2
9 + 9 = 16 + 2
18 = 18
Hence verified.
7. Using Euler’s formula find the unknown:
Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?
Solution:
(i)
By using Euler’s formula
V + F = E + 2
6 + F = 12 + 2
F = 14 – 6
F = 8
∴ Number of faces is 8
(ii)
By using Euler’s formula
V + F = E + 2
V + 5 = 9 + 2
V = 11 – 5
V = 6
∴ Number of vertices is 6
(iii)
By using Euler’s formula
V + F = E + 2
12 + 20 = E + 2
E = 32 – 2
E = 30
∴ Number of edges is 30
EXERCISE 19.2 PAGE NO: 19.12
1. Which among of the following are nets for a cube?
Solution:
Figure (iv), (v), (vi) are the nets for a cube.
2. Name the polyhedron that can be made by folding each net:
Solution:
(i) From figure (i), a Square pyramid can be made by folding each net.
(ii) From figure (ii), a Triangular prism can be made by folding each net.
(iii) From figure (iii), a Triangular prism can be made by folding each net.
(iv) From figure (iv), a Hexagonal prism can be made by folding each net.
(iv) From figure (v), a Hexagonal pyramid can be made by folding each net.
(v) From figure (vi), a Cube can be made by folding each net.
3. Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice?
Solution:
Figure (i), is a dice. Since the sum of numbers on opposite faces is 7 (3 + 4 = 7 and 6 + 1 = 7).
4. Draw nets for each of the following polyhedrons:
Solution:
(i) The net pattern for cube is
(ii) The pattern for triangular prism is
(iii) The net pattern for hexagonal prism is
(iv) The net pattern for pentagonal pyramid is
5. Match the following figures:
Solution:
(a)-(iv) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 4×4 = 16
(b)-(i) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 8×3 = 24
(c)-(ii) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 6×3 = 18
(d)-(iii) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 3×9 = 27
Courtesy : CBSE
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# 2020 AIME II Problems/Problem 13
## Problem
Convex pentagon $ABCDE$ has side lengths $AB=5$, $BC=CD=DE=6$, and $EA=7$. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$.
## Solution 1
Assume the incircle touches $AB$, $BC$, $CD$, $DE$, $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$, $ET=y=ES=CR=CQ$, $AP=AT=z$. So we have $x+y=6$, $x+z=5$ and $y+z$=7, solve it we have $x=2$, $z=3$, $y=4$. Let the center of the incircle be $I$, by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$, and triangle $CIR$ is congruent to triangle $SIE$. Then we have $\angle AED=\angle BCD$, $\angle ABC=\angle CDE$. Extend $CD$, cross ray $AB$ at $M$, ray $AE$ at $N$, then by AAS we have triangle $END$ is congruent to triangle $BMC$. Thus $\angle M=\angle N$. Let $EN=MC=a$, then $BM=DN=a+2$. So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain $$\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}$$, solved it gives us $a=8$, which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=\boxed{60}$.
-Fanyuchen20020715
## Solution 2 (Complex Bash)
Suppose that the circle intersects $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, and $\overline{EA}$ at $P$, $Q$, $R$, $S$, and $T$ respectively. Then $AT = AP = a$, $BP = BQ = b$, $CQ = CR = c$, $DR = DS = d$, and $ES = ET = e$. So $a + b = 5$, $b + c = 6$, $c + d = 6$, $d + e = 6$, and $e + a = 7$. Then $2a + 2b + 2c + 2d + 2e = 30$, so $a + b + c + d + e= 15$. Then we can solve for each individually. $a = 3$, $b = 2$, $c = 4$, $d = 2$, and $e = 4$. To find the radius, we notice that $4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ$, or $2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ$. Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)$ is real. Expanding, we get: $(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)$, then $(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)$. On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. $15ir^4 - 252ir^2 + 192i = 0$. $5r^4 - 84r^2 + 64 = 0$ $(5r^2 - 4)(r^2 - 16) = 0$. $r$ must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of $ABCDE$ is $4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}$
-nihao4112
## Solution 3 (Guess)
This pentagon is very close to a regular pentagon with side lengths $6$. The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$. $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$. $4\sqrt{5-2\sqrt{5}}$ is then slightly greater than $2\sqrt{2}$. We will approximate that to be $2.9$. The area is now roughly $\frac{180}{2.9}$, but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$, we can safely say that the answer is most likely $\boxed{60}$.
~Lopkiloinm
## Solution 4 (Guess)
Because the AIME answers have to be a whole number it would meant the radius of the circle have to be a whole number, thus by drawing the diagram and experimenting, we can safely say the radius is 4 and the answer is 60
(Edit: While the guess would be technically correct, the assumption that the radius would have to be a whole number for the ans to be a whole number is wrong)
By EtherealMidnight
## Solution 5 (Official MAA 1)
Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$, so the problem is reduced to finding $r$. Let $a$ be the length of the tangent segment from $A$ to $\omega$, and analogously define $b$, $c$, $d$, and $e$. Then $a+b=5$, $b+c= c+d=d+e=6$, and $e+a=7$, with a total of $a+b+c+d+e=15$. Hence $a=3$, $b=d=2$, and $c=e=4$. It follows that $\angle B= \angle D$ and $\angle C= \angle E$. Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$. Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$. The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$, which implies that $\angle AIQ$ must be $180^\circ$. Therefore points $A$, $I$, and $Q$ are collinear. $[asy] defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {"A", "B", "C", "D", "E"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {"a", "b", "c", "d", "e"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i Because $\overline{AQ} \perp \overline{CD}$, it follows that$$AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.$$Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$, so $$\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}$$and $$\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.$$Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives $$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}$$ and $$AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.$$ Hence
$$12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},$$ yielding $$2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;$$ equivalently $$7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).$$ Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$, with solutions $\frac{42 - 38}{5}=\frac45$, and $\frac{42 + 38}{5}= 16$. The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$, then $\tan \left(\frac{\angle A}{2}\right)$, $\tan \left(\frac{\angle C}{2}\right)$, and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$, $\angle C$, and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$. The requested area is $15\cdot4 = \boxed{60}$.
## Solution 6 (Official MAA 2)
Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Then, as in Solution 5, $a=3$, $b=d=2$, $c=e=4$, $\angle B= \angle D$, and $\angle C= \angle E$. Let $\alpha =\frac{\angle A}{2}$, $\beta = \frac{\angle B}{2}$, and $\gamma=\frac{\angle C}{2}$. It follows that $540^{\circ} = 2\alpha + 4 \beta + 4 \gamma$, so $270^{\circ} = \alpha + 2\beta + 2 \gamma$. Thus $$\tan(2\beta + 2 \gamma) = \frac{1}{\tan \alpha},$$ $\tan(\beta) = \frac{r}{2}$, $\tan(\gamma) = \frac{r}{4}$, and $\tan(\alpha) = \frac {r}{3}$. By the Tangent Addition Formula, $$\tan(\beta +\gamma) = \frac{6r}{8-r^2}$$ and $$\tan(2\beta + 2\gamma) = \frac{\frac{12r}{8-r^2}}{1-\frac{36r^2}{(8-r^2)^2}} = \frac{12r(8-r^2)}{(8-r^2)^2-36r^2}.$$ Therefore $$\frac{12r(8-r^2)}{(8-r^2)^2-36r^2} = \frac{3}{r},$$ which simplifies to $5r^4 - 84r^2 + 64 = 0$. Then the solution proceeds as in Solution 5.
## Solution 7 (Official MAA 3)
Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Note that $$\arctan\left(\frac{a}{r}\right) + \arctan\left(\frac{b}{r}\right) + \arctan\left(\frac{c}{r}\right) + \arctan\left(\frac{d}{r}\right) + \arctan\left(\frac{e}{r}\right) = 180^{\circ}.$$ Hence $$\operatorname{Arg}(r + 3i) + 2\cdot \operatorname{Arg}(r + 2i) + 2\cdot \operatorname{Arg}(r + 4i) = 180^{\circ}.$$ Therefore $$\operatorname{Im} \big( (r + 3i)(r+2i)^2(r+4i)^2 \big) = 0.$$ Simplifying this equation gives the same quadratic equation in $r^2$ as in Solution 5.
## Solution 8 (The same circle)
Notation shown on diagram. As in solution 5, we get $\overline{AQ} \perp \overline{CD}, AG = 3, GB = 2, CQ = 4$ and so on.
Let $\overline{AB}$ cross $\overline{CD}$ at $F, \overline{AE}$ cross $\overline{CD}$ at $F', CF = x.$ $FQ = FG \implies FB = CF + CQ – BG = x + 2.$ $\angle BAQ = \angle EAQ \implies DF' = x + 2, EF' = x.$
Triangle $\triangle AFF'$ has semiperimeter $s = 2x + 11.$
The radius of incircle $\omega$ is $r =\sqrt{\frac{s-FF’}{s}}(s-AF) = \sqrt{\frac{3}{2x +11}}(x+4).$
Triangle $\triangle BCF$ has semiperimeter $s = x + 4.$
The radius of excircle $\omega$ is $r = \sqrt{\frac{s(s-BF)(s-CF)}{s-BC}} = \sqrt{ \frac{(x+4)\cdot 2 \cdot 4}{x - 2}}.$
It is the same radius, therefore $$\sqrt{\frac{3}{2x +11}}(x+4) = \sqrt{\frac{8(x+4)}{x – 2}} \implies \frac {3(x+4)}{2x+11} = \frac {8}{x-2} \implies (x-8)(3x + 14) = 0 \implies x = 8, r = 4.$$
Then the solution proceeds as in Solution 5.
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Order of Operations - Order of Operations and Integers - The World of Numbers - Basic Math and Pre-Algebra
## Basic Math and Pre-Algebra
PART 1. The World of Numbers
CHAPTER 3. Order of Operations and Integers
In the previous chapter, we focused on arithmetic that used whole numbers. The whole numbers are the counting numbers (1, 2, 3, 4, 5, and so on) and zero, or the set {0, 1, 2, 3, 4, 5 ...}. It’s time now to look at some of the rules about how you should approach more complicated problems. In this chapter, we’ll look at the order of operations and at what people do when they don’t want you to follow those rules.
You’ve already encountered the commutative property and the associative property, the rules that let you rearrange a problem that’s all addition or all multiplication. Because your problems aren’t always one operation, it’s time to meet the distributive property, which will give you some options for dealing with addition and multiplication in the same problem and help with mental math along the way.
The set of whole numbers may contain infinitely many numbers, but even the whole numbers aren’t big enough to express all the ideas people have about numbers, so in this chapter you’ll get acquainted with a set of numbers called the integers. These positive and negative numbers let you express ideas of opposites and give you a way to answer questions you might have once been told were impossible.
Order of Operations
What does 2 + 3 x 7 equal? Some people might say 2 + 3 is 5 and 5 times 7 is 35. Those people started at the left and did what they saw as they saw it, moving across the line, a logical enough approach. Other people would do the multiplying first and say that’s 21, then add on the 2 to get 23. Those people might have thought that multiplication is repeated addition, so do that first, then the simple addition.
And what about the order of the numbers? You learned that addition is associative and commutative, so you can rearrange an addition problem. And you can rearrange a multiplication problem, too, so can you rearrange one that has both operations? Can you change 2 + 3 x 7 to 2 + 7 x 3? If you do, the multiplication-first people will get the same answer, but the left to right people will get a different one than they got the first time.
Those are great questions, and you certainly don’t want 2 + 3 x 7 to have two (or more) different values, depending on who does the arithmetic. That wouldn’t be practical. So what’s the solution?
One way to communicate what should be done first is to add parentheses. The expression (2 + 3) x 7 tells you to add first, then multiply. (2 + 3) x 7 = 5 x 7 = 35. On the other hand, 2 + (3 x 7) says multiply first, so you get 2 + (3 x 7) = 2 + 21 = 23.
That’s helpful, but you still need to know what to do, even if people writing the problem don’t use parentheses. So there’s an agreement among people who do arithmetic about what to do first, second, and so on.
That agreement is called the order of operations. What’s in parentheses will always get done first, if there are parentheses, and after that, the order will be exponents (think of them like supermultiplication), then multiplication and division, and finally addition and subtraction.
DEFINITION
The order of operations is an agreement among mathematicians that operations enclosed in parentheses or other grouping symbols should be done first, and then exponents should be evaluated. After that, do multiplication and division as you meet them moving left to right, and finally do addition and subtraction as you meet them, moving left to right.
There are a lot of memory devices to help you remember the order, and one of them, PEMDAS, is so common many people use it in place of order of operations, as if it were a name. The letters in PEMDAS are meant to help us remember to:
• P: Simplify expressions inside Parentheses
• E: Evaluate powers, or numbers with Exponents
• MD: Multiply and Divide, moving from left to right
• AS: Add and Subtract, moving from left to right
DEFINITION
PEMDAS is a mnemonic, or memory device, to help you remember that the order of operations is parentheses, exponents, multiplication and division, addition and subtraction.
Some people use a sentence, like Please Excuse My Dear Aunt Sally, to help remember the letters in PEMDAS, and others use an image of a house. The house reminds you that multiplication and division are done together, and so are addition and subtraction.
MATH TRAP
Multiplication and division have the same priority. Don't do all the multiplication and then all the division. Do multiplication or division as you meet them as you work across the line. The same is true for addition and subtraction. Do them as you come to them.
Here are some examples of the order of operations at work. Look at each example first, and think about what you would do. Then read the explanation to see if you had all the rules in order.
Example 1: Simplify (4 x 52 - 8) x 3
There are parentheses here, so work on what’s inside them first. It’s a little problem of its own, so follow the order of operations. First is the exponent. The exponent of 2 on the 5 tells you to multiply 5 x 5. That makes the problem (4 x 25 - 8) x 3. Do the multiplication inside the parentheses, and you have (100 - 8) x 3. Then you can subtract to get 92 x 3. That gives you 276.
Ready for a more complicated problem? Try this example.
Example 2: 7 x 20 - 2 x 4 + 32 + 12 ÷ 4
There are no parentheses, so take care of the exponent first. 32 = 9, so the problem becomes 7 x 20 - 2 x 4 + 9 + 12 ÷ 4. Next, do multiplication and division as you come to them, starting on the left and moving across the line: first 7 x 20, then 2 x 4, and then 12 ÷ 4.
7 x 20 - 2 x 4 + 9 + 12 ÷ 4
= 140 - 2 x 4 + 9 + 12 ÷ 4
= 140 - 8 + 9 + 12 ÷ 4
= 140 - 8 + 9 + 3
When all the multiplying and dividing are taken care of, return to the beginning of the line, and do addition and subtraction as you come to them.
140 - 8 + 9 + 3
= 132 + 9 + 3
= 141 + 3
= 144
Sometimes you’ll see more than one set of grouping symbols in the same problem, and sometimes you may see one set inside another. Different types of grouping symbols are often used to help you tell which is which.
Example 3: 7[120 - 2(4 + 3)2 + 12] ÷ 2
This problem uses parentheses inside of brackets. Brackets have the same meaning as parentheses. They’re just a different shape so you don’t get confused. Work from the inside out. Notice that there are numbers in front of the parentheses and brackets with no operation sign in between. This is another way of writing multiplication.
Do what’s in the parentheses first.
7[120 - 2(4 + 3)2 + 12] ÷ 2
= 7[120 - 2(7)2 + 12] ÷ 2
Now work on the problem inside the brackets: 120 - 2(7)2 + 12. Follow the order of operations and don’t worry about anything else until you finish with this part.
7[120 - 2(7)2 + 12] ÷ 2
= 7[120 - 2(49) + 12] ÷ 2
= 7[120 - 98 + 12] ÷ 2
= 7[120 - 98 + 12] ÷ 2
= 7[22 + 12] ÷ 2
= 7[34] ÷ 2
Now you can finish up, starting from the left.
7[34] ÷ 2
= 238 ÷ 2
= 119
Although a problem may look complicated, using the order of operations can break it down into manageable steps.
WORLDLY WISDOM
A number written in front of parentheses or other grouping symbols without an operation sign in between tells you to multiply that number by the result of the work in parentheses. For example, 3(4 - 2) = 3(2) = 3 x 2 = 6.
A minus sign in front of parentheses tells you to subtract the result of the parentheses from the number that precedes the minus sign. For example, 10 - (5 + 1) = 10 - (6) = 10 - 6 = 4.
Try applying the order of operations to some problems.
CHECK POINT
Complete each arithmetic problem.
1. 9 - 4 x 2
2. 32 - 2 x 4 + 1
3. (23 - 5) x 2 + 14 ÷ 7 - (5 + 1)
4. [(23 -5) x 2 + 14] ÷ 5 - 3 + 1
5. [(32 -2 x 4) + 1]2 + [11 - 8 + 5(3 + 1)]
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# Probability Distribution Of Occupational Category Essay
≈918 words, ≈2.9 pages (Arial, 14 size)
Cite this essay: APA, MLA, Harvard and other ↓
## Questions:
(a) Find the frequency distribution for the Occupational category (1=Management, 2=Sales, 3=Clerical, 4=Service, 5=Professional, 6=Other). Use Excel to produce a Descriptive Statistics table for your sample “Occupational category” data and paste into your MS Word assignment document.
(b) Use the relative frequency approach to find the probability distribution for the Occupational category.
(c) Draw the bar chart for the probability distribution of Occupational category.
(d) Define the probability distribution based on part (b).
x 1 2 3 4 5 6 P(x) 0.14 0.26 0.3 0.15 0.08 0.07
(e) Based on the probability distribution calculate the following
i. Find the probability of exactly two
ii. Find the probability more than two
iii. Find the probability at least three.
a. There are 6 categories of occupation. From our sample of size 60 that has been obtained, we obtain the frequency distribution of occupation as shown below.
OCCUPATION Count 6 60
Thus, we observe that in the given sample, all the people belong to the occupational category “Others”.
b. Relative frequency is a good approach to compute the probability of an event. Suppose the frequency of an event is f and the total frequency is n. Then the relative frequency of the event is (frequency of the event)/(Total frequency) =f/n.
In this case, we have only one occupational category and its relative frequency
= frequency/60
=60/60
=1
Thus the probability distribution of occupational category is shown below:
OCCUPATION Count Probability 6 60 1
c. The bar chart for the probability distribution of Occupational category is given below:
In the above diagram, 1 actually denotes Occupational Category “6” and all the samples units belong to the category “Others”.
d. Considering the entire population, let X denote the Occupational Category of a person. Then X can take the values 1, 2, 3, 4, 5 and 6. According to our data given in task 1, we have a random sample of size 50 from the population and the corresponding probabilities of X are calculated according to the relative frequency method.
For example P(X=1) = (number of people with Occupational Category 1)/50
In this way we compute all the probabilities for other values of X and the probability distribution table is shown below:
X 1 2 3 4 5 6 P(x) 0.1 0.18 0.26 0.24 0.02 0.2
i. Based on the probability distribution given in the above probability distribution table, we compute some of the probabilities
ii. Probability of exactly two
=P(X=2)
=0.18 (given in the above table)
iii. Probability of more than two
=P(X>2)
=P(X=3) + P(X=4) + P(X=5) +P(X=6)
=0.26+0.24+0.02+0.2
=0.72
### Probability of atleast three
=P(X>=3)
=P(X=3) + P(X=4) + P(X=5) +P(X=6)
=0.26+0.24+0.02+0.2
=0.72
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Sep22
# Learn Probability Problems and Examples
In today’s post, we are going to solve a probability problem that appeared in Madrid’s 2008 Standardized Elementary Testing Exam (CDI):
Let’s start by reviewing some probability concepts.
#### Sample space:
The sample space refers to all of the possible outcomes of a random experiment and it’s often represented by E (or as omega, Ω, from the Greek alphabet).
For example, when we toss a coin, what are all of the possible outcomes that we can end up with? Heads or tails, right? In total, there are two possible outcomes so the sample space is 2.
And if we throw a die, we have a total of six possible outcomes. The sample space, then, would be 6.
E = {1, 2, 3, 4, 5, 6}.
#### Event:
An event is any subset of the sample space. For example, coin landings on heads, or a die landing on a 5 are events.
Let’s take a look at what the sample space would be in the first part of our problem.
What are all of the possible outcomes? We’re talking about the number of balls, which are the numbers 11 to 20.
Our sample space is 10:
E = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
And the event, which is what we are solving for, is “picking a prime number”.
So then, how do we calculate the probability of this event happening?
When all of the simple events have the same probability of occurring, the probability of whichever event, A, is defined as the quotient between the number of favorable outcomes and the number of possible outcomes. This is Laplace’s Rule of Succession.
In coin tossing example, the simple outcomes would be: heads or tails. If it isn’t a trick coin, the probability of each simple outcome is the same. This means that the probability of the coin landing on heads would be ½.
#### Going back to our problem: In a bag, there are 10 red and green balls that are numbered from 11 to 20.
Part a): If we pick out a ball without looking in the bag, what is the probability of picking out a prime number?
Let’s start by calculating the number of favorable outcomes and the number of possible outcomes.
Number of favorable outcomes= number of prime numbers = 4 numbers of the possible outcomes are prime numbers (the numbers 11, 13, 17 y 19 are prime)
Number of possible outcomes = 10 (all the numbers from 11 to 20)
The probability of picking out a prime number from the 10 balls is 4/10, which can be simplified to 2/5.
Part b): How many balls of each color are there?
We’re told that the probability of picking out a green ball is 3/5.
The number of balls that can be picked out, is still 10.
The number of favorable outcomes, or the number of green balls (our event) is one of the questions that we want to solve. We know that 3/5 is equal to 6/10, so if we apply Laplace’s rule:
In total, there are 6 green balls in the bag. From there, we can deduce that what’s left, 4 are red balls.
Final answer: There are 6 green balls and 4 red balls.
If you thought this post was interesting and if you’d like to keep on learning Math, Smartick has even more problems that are perfect for practicing!
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# State whether the following is true or false and justify : Line joining the points (3, – 4) and (– 2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).
Toolbox:
• Slope of a line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $\bigg( \large\frac{y_2-y_1}{x_2-x_1} \bigg)$
• If two lines are perpendicular then the product of their slopes $m_1m_2 = -1.$
Step 1 :
Slope of the line joining the points (3, -4) and (-2, 6) is
$m_1 = \large\frac{y_2-y_1}{x_2-x_1}$$= \large\frac{6-(-4)}{-2-3}$$ = \large\frac{10}{-5}$
$=-2$
Slope of the line joining (-3, 6) and (9, -18) is
$m_2 = \large\frac{-18-6}{9-(-3)}$$= \large\frac{-24}{12}$$=-2$
If two lines are perpendicular, then the product of their slopes is -1.
( i.e ) $m_1m_2=-1$
But let $m_1m_2 = 4$
Hence the statement is falser.
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# Making Vector Operations Transparent
2 teachers like this lesson
Print Lesson
## Objective
SWBAT analyze vector operations in component form and visually on a graph.
#### Big Idea
Use overhead transparencies to visually represent vector operations.
## Launch
15 minutes
On yesterday’s assignment, students were to come up with conjectures about how operations on vectors would work out. Today we are going to work with their conjectures and make generalizations about how we can represent vector operations in coordinate form and graphically.
I start class by going over Question #7 from yesterday’s assignment. Have a few students show their graphs and see if they made the connection from the Airplane Task to find out that the vectors should be placed head to tail when they are added together. Students should also know that doubling a vector is the same as adding it to itself, and the opposite of a vector is the 180° rotation of itself. I see if students made the connection to what these operations do with the coordinates.
## Explore and Share
20 minutes
To prepare for today’s lesson, give every student an overhead transparency that includes the four vectors on this worksheet. Students should cut their transparency in fourths so that each vector is separated. On the board,I write a vector operation (like u + v) and students will have to use their vector cutouts to place them on top of each other to show them being connected head to tail. Then, students can take a dry erase marker and they can draw in the resultant vector. I explain the process in the video below.
The goal for this portion of the lesson is to allow them to see the graphical meaning of the vector operations. The vector cutouts will not have units or coordinates, so we will just be looking at the meaning on a visual level. It is really important for them to see these vector operations as movement – not just static entities on a worksheet!
Here are the operations that I would go through so that students can get a feel for them. After each one, I randomly select a student to instruct the class how they performed the operation.
1. u + v
2. u + v + w
3. 2u
4. u + w
5. u – v
6. w – u
After these have been completed, you can ask a question like whether or not vector addition is commutative, and students can use their cards to investigate the answer. I have students share out with the whole class to reach a consensus.
## Summarize
15 minutes
I find that analyzing vector operations graphically is not always ideal. When the components are given, it is often much more efficient to use those. This PowerPoint summarizes how to perform the vector operations when you are given the components. Slide #2 is a good starting point for -students to work on in groups. As they work, I will check-in with each table to see if they have any questions and to assess their understanding.
Slide #3 has a problem that is a little more challenging. Students are to find the magnitude of the sum of two vectors, but students are not given the components of the vectors or the directions – only the magnitudes and the angle between the vectors. Students will have to use our graphical approach to connect the vectors head to tail, and then they can use the Law of Cosines to find the magnitude of the resultant.
One common misconception is that students will find the distance from the tail of vector a to the tail of vector b, therefore neglecting to put the vectors head to tail. If this occurs, I remind the students about the yesterday’s problem about the airplane and how we found the resultant vector.
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# Division
Art. 111. IN multiplication, we have two factors given, and are required to find their product By multiplying the factors 4 and 6, we obtain the product 24. But it is frequently necessary to reverse this process. The number 24, and one of the factors may be given, to enable us to find the other. The operation by which this is effected, is called Division. We obtain the number 4, by dividing 24 by 6. The quantity to be divided is called the dividend ; the given factor, the divisor; and that which is required the quotient.
112. DIVISION si finding a quotient, which multiplied into the divisor will produce the divident.
In multiplication the multiplier is always a number. And the product is a quantity of the same kind, as the, multiplicand. The product of 3 rods into 4, is 12 rods. When we come to division, the product and either of the factors may be given, to find the other: that is,
The divisor may be a number, and then the quotient will be a quantity of the same kind as the dividend ; or,
The divisor may be a quantity of the same kind as the dividend; and then the quotient will be a number.
Thus 12 rocs/4 = 3 rods. But 12 rods/3rods = 4.
And 12 rods/24 =1/2 rod. And 12 rods/24rods =1/2.
In the first case, the divisor being a number, shows into how many parts the dividend is to be separated; and the quotient shows what these parts are.
If 12 rods be divided into 3 parts, each will be 4 rods long. And if 12 rods be divided into 24 parts, each will be half a rod long.
In the other case, if the divisor is less than the dividend, the former shows into what parts the latter is to be divided; and the quotient shows how many of these parts are contained in the dividend. In other words, division m this case consists in finding how often one quantity is contained in another.
A line of 3 rods, is contained in one of 12 rods, four times.
But if the divisor is greater than the dividend, and yet a quantity of the same kind, the quotient shows what part of the divisor is equal to the dividend.
Thus one half of 24 rods is equal to 12 rods.
113. As the product of the divisor and quotient is equal to the dividend, the quotient may be found, by resolving the dividend into two such factors, that one of them shall be the divisor. The other will, of course, be the quotient.
Suppose abd is to be divided by a. The factor a and bd will produce the dividend. The first of these, being a divisor, may be set aside. The other is the quotient. Hence,
When the divisor is found as a factor in the dividend, the division is performed by cancelling this factor.
Divide cx drx dhxy abxy By c dr dy ax Quot. x x hx by
In each of these examples, the letters which are common to the divisor and dividend, are set aside, a&d the other letters form the quotient. It will be seen at once, that the product of the quotient and divisor is equal to the dividend.
114. If a letter is repeated in the dividend, care must be taken that the factor rejected be only equal to the divisor.
In such instances, it is obvious that we are not to reject every letter in the dividend which is the sairie with one in the divisor.
115. If the dividend consists of any factors whatever, expunging one of them is dividing by it.
Divide a(b + d) a(b + d) (b + x)(c + d) (b + y)(d - h)x By a b + d b + x d - h Quot. b + d a c + d (b + y)x
116. In performing multiplication, if the factors contain numeral figures, these are multiplied into each other. (Art. 94.) Thus 3a into 7b is 21ab. Now if this process is tobe reversed, it is evident that dividing the number in the product, by the number in one of the factors, will give the number in the other factor. The quotient of 21ab/3a is 7b. Hence,
In division, if there are numeral coefficients prefixed to the letters, the coefficient of the dividend must be divided, by the coefficient of the divisor.
117. When a simple factor is multiplied into a compound one, the former enters into every term of die latter. Thus a into b + d is ab + ad. Such a product is easily resolved again into its original factors.
Thus ab + ad = a.(b + d).
ab + ac + ah = a.(b + c + h).
Now if the whole quantity be divided by one t)f these factors, according to Art 115, the quotient will be the other factor.
If the divisor is contained in every term of a compound dividend, it must be cancelled in each.
Divide ab + ac aah + ay By a a Quot. b + c ah + y
And if there are cvefficients, these must be divided, in each term also.
Divide 6ab + 12ac 12hx + 8 By 3a 4 Quot. 2b + 4c 3hx + 2
118. On the other hand, if a compound expression containing any factor in every term, be divided by the other quantities connected by their signs, the quotient will be that factor. See the first part of the preceding article.
119. In division, as well as in multiplication, the caution must be observed, not to confound terms with factors.
Thus (ab + ac)/a = b + c.
But (ab.ac)/a = aabc/a = abc.
And (ab + ac)/(b + c) = a.
But (ab.ac)/(b.c) = aabc/bc = aa.
120. In division, the same rule is to be observed respecting the signs, as in multiplication ; that is, if the divisor and dividend are both positive, or both negative, the quotient must be positive : if one is positive and the other negative, the quotient must be negative.
This is manifest from the consideration that the product of the divisor and quotient must be the same as the dividend.
If
+a.+b = +ab
-a.+b = -ab
+a.-b = -ab
-a.-ab = +ab
then
+ab/+a = +a
-ab/+b = -a
-ab/-b = +a
+ab/-b = -a
Divide abx 8a - 10ay 6am.dh By -a -2a -2a Quot. -bx -4 + 5y -3m.dh = -3dhm
121. If Tins letters of the divisor are not to be found in the dividend, the division is expressed by writing the divisor under the dividend, in the form of a vulgar fraction.
This is a method of denoting division, rather than an actual performing of the operation. But the purposes of division may frequently be answered, by these fractional expressions. As they are of the same nature with other vulgar fractions, they may be added, subtracted, multiplied, &c. See the next section.
122. When the dividend is a compound quantity, the divisor may either be placed under the whole dividend, as in the preceding instances, or it may be repeated under each term, taken separately. There are occasions when it will be convenient to exchange one of these forms of expression for the other.
Thus b + c divided by x, is either (b +c)/x, or b/x + c/x.
For it is evident that half the sum of two or more quantities, is equal to the sum of tlievr halves. And the same principle is applicable to a third, fourth, fifth, or any other portion of the dividend.
So also a - b divided by 2, is either (a - b)/2, or a/2 - b/2.
For half the difference of two quantities is equal to the difference of their halves.
123. If some of the letters in the divisor are in each term of the dividend, the fractional expression may be rendered more simple, by rejecting equal factors from the numerator and denominator.
124. If the divisor is in some of the terms of the dividend, bwt nottin all; those which contain the divisor may be divided as<in Art*' 116, and the others set down in the form of a fraction.
Divide dxy + rx - dh bm + 3y By x -b Quot. dy + r - dh/x -m - 3y/b
125. The quotient of any quantity divided by itself or its equal, is obviously a unit.
Cor. If the dividend is greater than the divisor, the quotient must be greater than a unit: But if the dividend is less than the divisor, the quotient must be less than a unit.
#### PROMISCUOUS EXAMPLES.
1. Divide 18aby + 6abx - 18bbm + 24b, by 6b.
2. Divide (a - 2h).(3m + y).x, by (a - 2h).(3m + y)
3. Divide ax - ry + ad - 4my - 6 + a, by -a.
4. Divide ard - 6a + 2r - hd + 6, by 2ard.
126. From the nature of division it is evident, that the value of the quotient depends both on the divisor and the dividend. With a given divisor, the greater the dividend, the greater the quotient. And with a given dividend, the greater the divisor, the less the quotient. In several of the succeeding parts of algebra, particularly the subjects of fractions, ratios, and proportion, it will be important to be able to determine what change will be produced in the quotient, by increasing or diminishing either the divisor or the dividend.
If the given dividend be 24, and the divisor 6 ; the quotient will be 4. But this same dividend may be supposed to be multiplied or divided by some other number, before it is divided by 6. Or the divisor may be multiplied or divided by some other number, before it is used in dividing 24. In each of these cases, the quotient will be altered.
127. In the first place, if the given divisor is contained in the given dividend a certain number of times, it is obvious that the same divisor is contained,
In double that dividend, twice as many times;
In triple the dividend, thrice as many times, &c.
That is, if the divisor remains the same, multiplying the dividend by any quantity, is, in effect, multiplying the quotient by that quantity.
Thus, if the constant divisor is 6, then 24/6 = 4 the quotient.
Multiplvmg the dividend by 2, 2.24/6 = 2.4
Multiplying by any number n, n.24/6 = n.4.
128. Secondly, if the given divisor is contained in the given dividend a certain number of times, the same divisor is contained,
In half that dividend, half as many times;
In one third of the dividend, one third as many times, &c.
That is, if the divisor remains the same, dividing the dividend by any other quantity, is, in effect, dividing the quotient by that quantity.
Thus 24/6 = 4
Dividing the dividend by 2, (1/2)24/6 = (1/2).4
Dividing by n, (1/n)24/6 = (1/n).4
129. Thirdly, if the given divisor is contained in the given dividend a certain number of times, then, in the same dividend,
Twice that divisor is contained only half as many times;
Three times the divisor is contained one third as many times.
That is, if the dividend remains the same, multiplying the divisor by any quantity, is, ia effect, dividing the quotient by that quantity.
Thus 24/6 = 4
Multiplying the divisor by 2, 24/2.6 = 4/2
Multiplying by n, 24/n.6 = 4/n
130. Lastly, if the given divisor is contained in the given dividend a certain number of times, then, in the same dividend,
Half that divisor is contained twice as many times;
One third of the divisor is contained thrice as many times.
That is, if the dividend remains the same, dividing the divisor by any other quantity, is, in effect, multiplying the quotient by that quantity.
For the method of performing division, when the divisor and dividend are both compound quantities, see one of the following sections.
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You are on page 1of 28
# 1
x
s
y
s
x
v
y
v
x
a
y
a
g
CHAPTER 2:
Kinematics of Linear Motion
Unit Physics KML
2
3
At the end of this chapter, students should be able to:
Define and distinguish between
i. Distance and displacement
ii. Speed and velocity
iii. Instantaneous velocity,average velocity and uniform
velocity
iv. Instantaneous acceleration,average aceleration and
uniform aceleration
Sketch graphs of displacement-time, velocity-time and
acceleration-time.
Determine the distance travelled,displacement,velocity
and acceleration from appropriate graphs.
Learning Outcome:
2.1 Linear Motion
4
2.0 Kinematics of Linear motion
is defined as the studies of motion of an objects
without considering the effects that produce the
motion.
There are two types of motion:
Linear or straight line motion (1-D)
with constant (uniform) velocity
with constant (uniform) acceleration, e.g. free fall
motion
Projectile motion (2-D)
x-component (horizontal)
y-component (vertical)
5
2.1 Linear motion (1-D)
2.1.1 Distance, d
scalar quantity.
is defined as the length of actual path between two
points.
For example :
The length of the path from P to Q is 25 cm.
P
Q
6
vector quantity
is defined as the distance between initial point and
final point in a straight line.
The S.I. unit of displacement is metre (m).
2.1.2 Displacement,
s
7
Example 1:
An object P moves 20 m to the east after that 10 m to the
south and finally moves 30 m to west. Determine the
displacement of P relative to the original position.
N
E
W
S
O
P
u
u
20 m
10 m
10 m 20 m
8
Solution :
The magnitude of the displacement is given by
and its direction is
west - south the or to 45
10
10
tan
=
|
.
|
\
|
=
÷1
θ
m 14.1 10 10
2 2
= + = OP
9
2.1.3 Speed, v
is defined the rate of change of distance.
scalar quantity.
Equation:
interval time
distance of change
speed=
Δt
Δd
v =
2.1.4 Velocity,
v
o is a vector quantity.
o The S.I. unit for velocity is m s
-1
.
10
Average velocity, v
av
is defined as the rate of change of displacement.
Equation:
Its direction is in the same direction of the change in
displacement.
interval time
nt displaceme of change
=
av
v
Δt
Δs
v
av
=
1 2
1 2
av
t t
s s
v
÷
÷
=
11
Instantaneous velocity, v
is defined as the instantaneous rate of change of
displacement.
Equation:
An object is moving in uniform velocity if
t
s
0 t
v
A
A
÷ A
=
limit
constant =
dt
ds
dt
ds
v =
12
Therefore
Q
s
t
0
s
1
t
1
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t
1
Gradient of s-t graph = velocity
13
vector quantity
The S.I. unit for acceleration is m s
-2
.
Average acceleration, a
av
is defined as the rate of change of velocity.
Equation:
Its direction is in the same direction of motion.
The acceleration of an object is uniform when the magnitude
of velocity changes at a constant rate and along fixed
direction.
2.1.5 Acceleration,
a
interval time
velocity of change
=
av
a
1 2
1 2
av
t t
v v
a
÷
÷
=
Δt
Δv
a
av
=
14
Instantaneous acceleration, a
is defined as the instantaneous rate of change of velocity.
Equation:
An object is moving in uniform acceleration if
t
v
0 t
a
A
A
÷ A
=
limit
constant =
dt
dv
2
2
dt
s d
dt
dv
a = =
15
Deceleration, a
is a negative acceleration.
The object is slowing down meaning the speed of the object
decreases with time.
Therefore
v
t
Q
0
v
1
t
1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t
1
Gradient of v-t graph = acceleration
16
Displacement against time graph (s-t)
2.1.6 Graphical methods
s
t
0
s
t
0
(a) Uniform velocity
(b) The velocity increases with time
with time
(c)
s
t 0
Q
R
P
The direction of
velocity is changing.
Gradient at point R is negative.
Gradient at point Q is zero.
The velocity is zero.
17
Velocity versus time graph (v-t)
The gradient at point A is positive – a > 0(speeding up)
The gradient at point B is zero – a= 0
The gradient at point C is negative – a < 0(slowing down)
t
1
t
2
v
t
0
(a)
t
2
t
1
v
t
0
(b)
t
1
t
2
v
t
0
(c)
Uniform velocity
Uniform
acceleration
Area under the v-t graph = displacement
B
C
A
18
From the equation of instantaneous velocity,
Therefore
dt
ds
v =
} }
= vdt ds
}
=
2
1
t
t
vdt s
graph under the area ded sha t v s ÷ =
19
A toy train moves slowly along a straight track according to the
displacement, s against time, t graph in figure 2.1.
a. Explain qualitatively the motion of the toy train.
b. Sketch a velocity (cm s
-1
) against time (s) graph.
c. Determine the average velocity for the whole journey.
d. Calculate the instantaneous velocity at t = 12 s.
Example 2 :
0
2 4 6 8 10 12 14
t (s)
2
4
6
8
10
s (cm)
Figure 2.1
20
Solution :
a. 0 to 6 s : The train moves at a constant velocity of
0.68 cm s
÷1
.
6 to 10 s : The train stops.
10 to 14 s : The train moves in the same direction at a
constant velocity of 1.50 cm s
÷1
.
b.
0
2 4 6 8 10 12 14
t (s)
0.68
1.50
v (cm s
÷1
)
21
Solution :
c.
d.
1 2
1 2
t t
s s
v
av
÷
÷
=
0 14
0 10
÷
÷
=
av
v
1
s cm 714 . 0
÷
=
av
v
s 14 to s 10 from velocity average = v
1 2
1 2
t t
s s
v
÷
÷
=
10 14
4 10
÷
÷
= v
1
s cm 50 . 1
÷
= v
22
A velocity-time (v-t) graph in figure 2.2 shows the motion of a lift.
a. Describe qualitatively the motion of the lift.
b. Sketch a graph of acceleration (m s
-1
) against time (s).
c. Determine the total distance travelled by the lift and its
displacement.
d. Calculate the average acceleration between 20 s to 40 s.
Example 3 :
0
5 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s
÷1
)
Figure 2.2
40 45 50
23
Solution :
a. 0 to 5 s : Lift moves upward from rest with a constant
acceleration of 0.4 m s
÷2
.
5 to 15 s : The velocity of the lift increases from 2 m s
÷1
to
4 m s
÷1
but the acceleration decreasing to
0.2 m s
÷2
.
15 to 20 s : Lift moving with constant velocity of 4 m s
÷1
.
20 to 25 s : Lift decelerates at a constant rate of 0.8 m s
÷2
.
25 to 30 s : Lift at rest or stationary.
30 to 35 s : Lift moves downward with a constant acceleration
of 0.8 m s
÷2
.
35 to 40 s : Lift moving downward with constant velocity
of 4 m s
÷1
.
40 to 50 s : Lift decelerates at a constant rate of 0.4 m s
÷2
and comes to rest.
24
Solution :
b.
t (s) 5 10 15 20 25 30 35 40 45 50
0
-0.4
-0.2
0.2
0.6
a (m s
÷2
)
-0.6
-0.8
0.8
0.4
25
Solution :
c. i.
0
5 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s
÷1
)
40 45 50
A
1
A
2
A
3
A
4
A
5
v-t of graph under the area distance Total =
5 4 3 2 1
A A A A A + + + + =
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 4 5 15
2
1
4 5
2
1
4 10 5
2
1
10 4 2
2
1
5 2
2
1
distance Total + + + + + + + =
m 115 distance Total =
26
Solution :
c. ii.
d.
v-t of graph under the area nt Displaceme =
5 4 3 2 1
A A A A A + + + + =
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 4 5 15
2
1
4 5
2
1
4 10 5
2
1
10 4 2
2
1
5 2
2
1
÷ + + ÷ + + + + + =
m 15 nt Displaceme =
1 2
1 2
t t
v v
a
av
÷
÷
=
20 40
4 4
÷
÷ ÷
=
av
a
2
s m 4 . 0
÷
÷ =
av
a
27
Figure 2.3
1. Figure 2.3 shows a velocity versus time graph for an object
constrained to move along a line. The positive direction is to
the right.
a. Describe the motion of the object in 10 s.
b. Sketch a graph of acceleration (m s
-2
) against time (s) for
the whole journey.
c. Calculate the displacement of the object in 10 s.
ANS. : 6 m
Exercise 2.1 :
28
2. A train pulls out of a station and accelerates steadily for 20 s
until its velocity reaches 8 m s
÷1
. It then travels at a constant
velocity for 100 s, then it decelerates steadily to rest in a further
time of 30 s.
a. Sketch a velocity-time graph for the journey.
b. Calculate the acceleration and the distance travelled in
each part of the journey.
c. Calculate the average velocity for the journey.
th
edition, Jim Breithaupt, Nelson
Thornes, pg.15, no. 1.11
ANS. : 0.4 m s
÷2
,0 m s
÷2
,-0.267 m s
÷2
, 80 m, 800 m, 120 m;
6.67 m s
÷1
.
Exercise 2.1 :
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# $x=p\sec \theta$ and $y=q\tan \theta$ thenA.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$ B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$ C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$
Last updated date: 23rd May 2024
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Hint: Out of the given options the right-hand side of the equation is always independent of $\theta$. So, we have to perform some calculations to make left-hand side equations independent of $\theta$ too. Thus, we need to make equations based on $\theta$ and solve it accordingly to remove $\theta$.
Here, we have $x=p\sec \theta$ and $y=q\tan \theta$
Considering each option’s nature, we can see that the right-hand side of the equations are independent of $\theta$. So, we have to form equations based on $\theta$ to eliminate it from these equations.
Now, from the given equations, for $x=p\sec \theta$, we get
$\Rightarrow x=p\sec \theta$
On cross-multiplication, we get
$\Rightarrow \sec \theta =\dfrac{x}{p}...\text{ }\left( 1 \right)$
Similarly, from $y=q\tan \theta$, we have
\begin{align} & \Rightarrow y=q\tan \theta \\ & \Rightarrow \tan \theta =\dfrac{y}{q}...\text{ }\left( 2 \right) \\ \end{align}
Since, we need to make the equations independent of $\theta$, we can use trigonometric identities, i.e.,
${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$
And, substituting values of $\tan \theta$ and $\sec \theta$ from equation (1) and (2), we get
\begin{align} & \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\ & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\ \end{align}
By transposing the variables, we get
\begin{align} & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\ & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}-{{\left( \dfrac{y}{q} \right)}^{2}}=1 \\ & \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\ \end{align}
Taking LCM on LHS of the equation, we get
\begin{align} & \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\ & \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\ \end{align}
On cross-multiplication, we get
\begin{align} & \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\ & \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=1\left( {{p}^{2}}{{q}^{2}} \right) \\ & \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}} \\ \end{align}
Hence, on solving the given equations, we get ${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$, i.e., option [C] is correct.
Note: The easiest way to solve this kind of problem is by substituting the values of $x$ and $y$ on LHS of every option available and checking it with the values of RHS of that particular option. This method is very useful if the given equations of $x$ and $y$ are very complex.
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## Double-counting, Part 2
This post is a follow-up to Part 1, where I talked about the technique of double-counting in the context of proving combinatorial identities. In this post, I’ll show three more identities that can be proven using simple double-counting. Test your skills!
First identity. This is Vandermonde’s Identity.
$\sum_{k=0}^p {m \choose k}{n \choose p-k} = {m+n \choose p}$
[Show Solution]
Second identity. This is the Christmas Stocking Identity. It is also sometimes called the Hockey-Stick Identity.
$\sum_{k=0}^m {n+k \choose n} = {m+n+1 \choose n+1}$
[Show Solution]
Third identity. This one looks similar to the first one, but notice that the upper coefficient is varying this time.
$\sum_{k=0}^p {k \choose m}{p-k \choose n} = {p+1 \choose m+n+1}$
[Show Solution]
## Double-counting, Part 1
Double-counting is one of my favorite proof techniques. The idea is simple: count the same thing in two different ways. In this post, I’ll give some examples of double-counting in the context of proving identities involving binomial coefficients, but it’s a very general technique that can be applied to many other types of problems.
Pascal’s identity. The following is known as Pascal’s identity:
${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$
Of course, Pascal’s Identity can also be proven directly using simple algebra, but there is a nice double-counting alternative. Suppose we have a group of $n$ people, and we’d like to choose a committee of $k$ members. There are $n \choose k$ ways of doing this. Now let’s count the number of committees in a different way. Suppose we pick out one particular person from the group, let’s call her Alice. There are $n-1\choose k-1$ committees that include Alice, because after we’ve included Alice, there are $k-1$ remaining committee members to choose out $n-1$ remaining people. Similarly, there are $n-1\choose k$ committees that exclude Alice, because all $k$ committee members must be chosen out of the remaining $n-1$ people. The total number of committees is equal to the number of committees that include Alice plus the number of committees that exclude Alice, and this completes the proof of Pascal’s Identity.
Binomial theorem. Here is another familiar identity:
${n \choose 0} + {n \choose 1} + \dots + {n \choose n} = 2^n$
This identity can be directly obtained by applying the Binomial Theorem to $(1+1)^n$. But once again, we can use a double-counting argument. Suppose we have a group of $n$ people. Let’s count the total number of subsets of this group. One way to count is to realize that there are $n \choose k$ different subsets of size $k$. So the total number of subsets of any size is the sum on the left-hand side of the identity. On the other hand, we can count subsets by looking at each person one at a time. Each person can either be included or excluded in the subset ($2$ possibilities), which yields a total of $2\times 2 \times \dots \times 2 = 2^n$ possible subsets.
Binomial products. This example involves products:
${n \choose s}{s \choose r} = {n \choose r}{n-r \choose s-r}$
Consider a group of $n$ people. This time, we count the number of ways of selecting a team of $s$ members, among which $r$ are designated as captains. One way to do this is to start by selecting the team, which can be done in $n \choose s$ ways. For each team, we then select the captains, which can be done in $s \choose r$ ways. The total is therefore ${n \choose s}{s \choose r}$. Another way to count is to start by selecting the captains first, which can be done in $n \choose r$ ways. Then, we must select the rest of the team. There remains $n-r$ people and we must choose $s-r$ to round out the team. So the total count is ${n \choose r}{n-r \choose s-r}$.
This technique of committee selection is very powerful. See if you can figure out how to apply it to the following example!
${n \choose 1} + 2 {n\choose 2} + 3 {n\choose 3} + \dots + n {n \choose n} = n 2^{n-1}$
[Show Solution]
For more double-counting problems, check out Part 2!
## Monsters’ gems
Once again, The Riddler does not disappoint! This puzzle is about slaying monsters and collecting gems.
A video game requires you to slay monsters to collect gems. Every time you slay a monster, it drops one of three types of gems: a common gem, an uncommon gem or a rare gem. The probabilities of these gems being dropped are in the ratio of 3:2:1 — three common gems for every two uncommon gems for every one rare gem, on average. If you slay monsters until you have at least one of each of the three types of gems, how many of the most common gems will you end up with, on average?
Here is my solution:
[Show Solution]
A more brute-force approach:
[Show Solution]
Yet another solution approach with very nice write-up can be found on Andrew Mascioli’s blog
## Counting parallelograms
The following problem appeared in the 1991 CMO, and it has a particularly clever solution.
In the figure, the side length of the large equilateral triangle is $3$ and $f(3)$, the number of parallelograms bounded by sides in the grid, is $15$. For the general analogous situation, find a formula for $f(n)$, the number of parallelograms, for a triangle of side length $n$.
[Show Solution]
## Proud partygoers puzzle
Another great problem from the Riddler blog.
A group of N people are in attendance at your shindig, some of whom are friends with each other. (Let’s assume friendship is symmetric — if person A is friends with person B, then B is friends with A.) Suppose that everyone has at least one friend at the party, and that a person is “proud” if her number of friends is strictly larger than the average number of friends that her own friends have. (A competitive lot, your guests.)
Importantly, more than one person can be proud. How large can the share of proud people at the party be?
The solution:
[Show Solution]
## Elevator button puzzle
This problem was originally posted on the Riddler blog. Here it goes:
In a building’s lobby, some number (N) of people get on an elevator that goes to some number (M) of floors. There may be more people than floors, or more floors than people. Each person is equally likely to choose any floor, independently of one another. When a floor button is pushed, it will light up.
What is the expected number of lit buttons when the elevator begins its ascent?
My solution:
[Show Solution]
A much more elegant solution, courtesy of Ross Boczar
[Show Solution]
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# How to solve algebra problems
We will explore How to solve algebra problems can help students understand and learn algebra. We will also look at some example problems and how to approach them.
## How can we solve algebra problems
Read on for some helpful advice on How to solve algebra problems easily and effectively. The internet is a great place to start, as there are many websites that offer step-by-step solutions to common problems. In addition, most major textbook publishers offer online homework help services. These services typically provide access to a database of answers, as well as a variety of tools and resources that can help with the solution process. With a little bit of effort, it is usually possible to find the answer to any homework problem.
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the length of the hypotenuse. This theorem can be represented using the equation: a^2 + b^2 = c^2. In this equation, a and b represent the lengths of the two shorter sides, while c represents the length of the hypotenuse. To solve for a side, you simply need to plug in the known values and solve for the unknown variable. For example, if you know that the length of Side A is 3 and the length of Side B is 4, you can solve for Side C by plugging those values into the equation and solving for c. In this case, 3^2 + 4^2 = c^2, so 9 + 16 = c^2, 25 = c^2, and c = 5. Therefore, the length of Side C is 5.
In mathematics, "solving for x" refers to the process of finding the value of an unknown variable in an equation. In most equations, the variable is represented by the letter "x." Fractions can be used to solve for x in a number of ways. For example, if the equation is 2x + 1 = 7, one can isolated the x term by subtracting 1 from each side and then dividing each side by 2. This would leave x with a value of 3. In some cases, more than one step may be necessary to solve for x. For example, if the equation is 4x/3 + 5 = 11, one would first need to multiply both sides of the equation by 3 in order to cancel out the 4x/3 term. This would give 12x + 15 = 33. From there, one could subtract 15 from each side to find that x = 18/12, or 1.5. As these examples demonstrate, solving for x with fractions is a matter of careful algebraic manipulation. With a little practice, anyone can master this essential math skill.
This can be especially helpful when working with complex problems or when trying to learn a new concept. By seeing the step-by-step process that was used to solve the problem, students can better understand the material and develop their own problem-solving skills. In addition, a math solver with work can often be used to check answers that have been arrived at using other methods. This can help to ensure that the solution is correct and also help identify any mistakes that were made along the way. Whether you are a student who is struggling with math or a teacher who is looking for a way to check answers, a math solver with work can be an invaluable tool.
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Sometime back, I was searching in the Wide Web World for an easy-to-understand solution for Rubik’s cube. I found many, but none worked out for me. Finally my friends helped me out to solve it. As I got help from many, I started jotting down the moves and thought I will write those down for the benefit of others. That was the first version of this article.
Later, I discovered that people are still having problems with notations, and I introduced a new notation as well.
Read this completely once, before attempting to solve.
Most of us see Rubik’s cube as six faces. This will not help in solving the cube. From now onwards we should be able to see the cube, constituted of three layers. The method I discuss here is to solve the cube layer by layer.
Before getting into the solution, it is good to learn some terminology and notations, for that helps in writing smaller steps/instructions which will stay in memory. After a while, you will forget the instructions, and follow the patterns.
As we start with terminology, keep in mind that while solving a piece (or using a formula), you should not change the orientation of the cube. Thus said, let us name the faces. There are six faces for Rubik’s cube, and hold the cube in front of you. Let the face that is facing you is F (Front), and the opposite face is B (Back). The face on your left is L (Left) and on the right is R (Right). The upper side is U (Up) and the one that faces down is D (Down). The color of a face is the color of the center piece, as the center piece cannot be moved.
Consider that the Rubik’s cube is made up of 27 little cubes. One cube is in the core (in fact, there isn’t any) and it does not come in our way. Observe – the center pieces on all the faces are immovable (relatively). Now, that leaves 20 little cubes to play around. Let us name the small cubes with respect to the intersection of faces. We can even name each face of these little cubes by properly qualifying the names. The left hand top little cube is a corner piece and is the intersection of L, U and F faces; so we can call it LUF, LFU, ULF, UFL, FUL, or FLU. Let us stick to a convention that the first letter denotes the face of the little cube. So, if we are talking about the front face of the left hand top little cube we will call it FLU or FUL. If it is the upper face of the same we will call ULF or UFL. The edge pieces will have only two faces and their names will be like LF, RD, and DB etc.
We need to rotate these layers while solving the cube. For a layer, there can be two kinds of rotations – clockwise and anticlockwise. But, qualifying a rotation to be clockwise or anticlockwise depends on which way we are looking at. Let’s make it clear. A face’s rotation is said to be clockwise, if the face is rotating in the clockwise direction when you are looking at it. So, when you rotate the left face anticlockwise FLU goes away from you, and when you rotate the right face anticlockwise FRU becomes DRF. The rotations are then quantified as quarter, half, three-quarter and full turns. A full turn of a layer does not alter anything. A clockwise three-quarter turn is equal to an anticlockwise quarter turn and for the same reason we will not deal with three-quarter turns. Half turn of a layer is the same whether it is turned clockwise or anticlockwise.
The names of the faces are not needed often, so we will use the same names for clockwise rotations. Let’s call L as clockwise quarter turn of left face, R as clockwise quarter turn of right face and so on. The symbol’ (single quote) will denote anticlockwise rotation. For example, L’ will represent left face anticlockwise quarter turn and F’ will denote front face anticlockwise quarter turn. We will call half turns B2, F2 etc. Once you are familiar with the faces and the notation and the rotation, we can start solving the cube.
Solving the first layer
Solving the first face can be done by observing the movements of pieces (little cubes). Determine a U face (let’s take White as U face). Look for a white edge piece on the bottom layer, white not on’D face’. Determine the color of the bottom face of that little edge piece. (Say we found it to be Red). Now, that piece (White-Red) should be at the intersection of centerpieces colored with White and Red. White is already on U face, now turn the cube in such a way that Red is your F face. Now, bring the White-Red edge piece on the L face or R face, in such a way that White is LD or RD and Red is DL or DR. (Say the piece is on the right hand side and then White is RD). This White-Red edge piece is to occupy the UF position when the cube is solved. Let’s take it there. If you observe the possible rotations of UF position and RD, you can find that the circles of rotation intersect at FD. That’s the best location to swap these pieces. So turn the center layer of F face such that UF comes to FD. Now turn the bottom layer such that RD becomes FD (now FD is White-Red). Finally turn the FD layer back to UF position and the White-Red piece is correctly positioned. A hint that I can give you is that it is more like catching with a hook. The same exercise can be repeated for corner pieces and the pieces of the middle layer. But if the White face is on the D side, we have to first bring it to the bottom layer of R, L, F or B faces. While doing that, make sure that you are not disturbing the correctly set pieces on the U face. Also, if the piece is wrongly positioned on the U face, break it away first by bringing some unwanted piece in its position, and then start moving it to the correct place.
Solving the second layer
When the first layer is done it forms a T with the center pieces in the middle layer. That means we need to solve only four pieces to finish the second layer. The formulae start here. All the formulae can be mirrored. If you mirror the steps for one piece the formula will affect its opposite piece.
The following formula will move the cube RD to FL. First, find out which cube fits into FL and position it in RD. Make sure it is positioned as RD, not DR.
Formula 1 To move RD to FL :::: L D’ L’ D’ F’ D F
For the first timers, it is better to write down what is on the RD and DR, and do the steps. Then find out what happened to the piece. Remember, you should not change the orientation of the cube while doing the steps.
Since this is the first formula let’s write down the corollary. To move LD to FR :::: R’ D R D F D’ F’
This formula can be repeated to solve the second layer. If there is piece incorrectly positioned on the second layer, first take it out, and then position it correctly.
Solving the third layer
After two layers are completed it is advised to turn the cube upside down to get a better view as well as to reduce the complexity. After turning the cube upside down, now your top layer is scrambled and the bottom two layers are set. Our first goal is to form a cross (like a + symbol, not X) on the present U face (earlier it was D face). The best way to start is to find a “piece of cross”. Look for an L shape or a straight line on the U face, with centerpiece included. If you find a line you are lucky, you need to do the next formula only once. If you find an L shape then you have to do the formula twice. Else you have to keep doing the formula until you get a line or L shape on the U face. (Don’t worry the formula is simple and you will get it soon).
Once you find an L shape you need to convert it to a line. And the same formula will convert the line into a cross. Starting with the L shape, first position the shape such that it forms between UR, UF and the center piece.
Formula 2 To get a cross on the third layer :::: B L U L’ U’ B’
If you had the L shape now it is gone, and you got the line between UR – Centerpiece – UL. Repeat the formula to complete the cross.
(The L shape could be anywhere, but you have to make appropriate modifications to the formula)
Now position the cross such that the edge pieces are matching with the other layers. (Say, for Blue-Yellow edge piece, Yellow should be on the Yellow side, and so on). It does not happen easily. Sometimes two adjacent pieces will match, and sometimes two opposite pieces will match. If you find only one edge piece matching with the other layers, turn the U face and you will find some other two pieces are matching. Once you match two adjacent edge pieces with other layers and two others are not matching, we need to swap those two pieces which are not matching. If the two opposite pieces are matching we need to break that. Simply apply the next formula to break the matching and start with two adjacent matching pieces.
Let’s position the incorrect pieces on UF and UR for the third formula. Before doing the third formula, we need to be careful, because the third formula is presented in a reusable manner for memorizing. That necessitates to do an extra step before the third formula is applied. Once the UL and UB pieces are correct (i.e., UF and UR pieces are incorrect), quarter turn the U face clockwise, so that UF’looks’ correct. This is very important and often forgotten.
Formula 3 To swap the incorrect edge pieces :::: R U R’ U R U2 R’
By now all the edge pieces are correct, and the corner pieces needs to be solved. If you look carefully, one of the corner pieces may be in its right position, but not aligned correctly. We need it for the next step. If none of the corner pieces are in its position simply follow the next formula, and you can find one falling into position (the alignment may be wrong) and do the formula again to correct others. The formula cycles three corner pieces in clockwise direction. It means, UBR will go to UFR, UFR will move to UFL and UFL will go to UBR. Notice that UBL remains as it is. So the correctly positioned corner piece should be on UBL while other three pieces move in clockwise direction to take their places.
Formula 4 To cycle three corner pieces :::: R’ U L U’ R U L’ U’
After cycling enough, the corner pieces take their position, and may be misaligned. If two adjacent corner pieces are aligned properly, you are lucky. But if two opposite pieces are aligned properly, you have to break one of those. How sad!
The last and longest formula will solve the cube. This formula should be applied in such a way that when the little face RUF will become UFR (or rotates anticlockwise in its position) the piece will be solved. This is important. To reiterate, hold the cube such that the corner piece will be correctly aligned when RUF turns to UFR. It is also to be noted that this formula works with two corner pieces. So if you rotate RUF, RLF will also rotate. Now, if you do not find any piece that will satisfy the condition that when RUF becomes UFR the corner piece will be aligned, just do the next formula once, and you will find a piece satisfying the condition.
Formula 5 To rotate the corner piece RUF to UFR :::: R’ U2 R U R’ U R L U2 L’ U’ L U’ L’
Repeat the formula and rotate the pairs of unsolved corners such that when RUF becomes UFR the RUF (UFR) corner piece is correctly aligned, until THE CUBE IS SOLVED.
Notice that the first part of formula 5 is the reverse of formula 3 and the second part is mirrored moves of the first. That is why it was asked to do the extra step before the formula 3 is applied. Also, keep in mind that all these formulae can be mirrored to provide opposite/mirrored results. Practicing such mirrored formulae will improve the speed of solving.
There are more optimizations that can be done while solving, but it is better learned after familiarizing one solution. It would complicate this already complicated document, if I were to write down the optimal steps.
Last but not the least, do not get discouraged or disheartened if you cannot solve the cube overnight. It took me some sleepless nights to observe, learn and practice the cube, and now the cube just gets solved in my hands. I need not recollect these formulae or consciously apply them.
Wishing you all the best…
NOTE: The solution is also given as an image – rubiks-cube-solution.gif – if you need any visual assistance in solving. In case the image does not look good, change the background to white in your image viewer, as there are some transparent colors in the image.
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# Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities.
Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas: $$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$ $$\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$$
Any idea how to solve this problem?
Note that as $x_1 \le 2$, $x_n \le 2 \forall n$. This can be shown inductively.
Thus, we may write $x_n = 2 \cos \theta_n$ for some $\theta_n \in [0, \pi/2]$, with $\theta_1 = \pi/2$
Now, $2 \cos \theta_{n+1} = \sqrt{2(1+ \cos \theta_n)} = 2 \cos \frac{\theta_n}2 \implies \theta_{n+1} = \frac{\theta_n}{2} = \frac{\theta_1}{2^n} = \frac{\pi}{2^{n+1}}$.
This implies that $2^n\sqrt{2 - x_n} = 2^{n+1} \sin \frac{\pi}{2^{n+1}}$. Finish off by using $\frac{\sin x}{x} \to 1$ as $x \to 0$.
By induction and the formula from the hint, we can easily get $x_n=2\cos(\frac{\pi}{2^n})$. Then $$\lim_{n\rightarrow\infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow\infty}2^n\sqrt{2}\sqrt{1-\cos(\frac{\pi}{2^n})}=\lim_{n\rightarrow\infty}2^{n+1}\sin\left(\frac{\pi}{2^{n+1}}\right)=\pi$$
By mathematical induction it can be proved that $x_n=2\cos(\frac{\pi}{2^n})$. for base of induction $x_0=0=2\cos(\frac{\pi}{2})$. Also we have $x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(\frac{\pi}{2^n})}=2\sqrt{\frac{1+\cos(\frac{\pi}{2^n})}{2}}=2\cos(\frac{\pi}{2^{n+1}})$. Therefore $\sqrt{2-x_n}=2\sqrt{\frac{1-\cos(\frac{\pi}{2^n})}{2}}=2\sin(\frac{\pi}{2^{n+1}})$, so $\lim_{n\rightarrow \infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow \infty}2^{n+1}\sin(\frac{\pi}{2^{n+1}})=\lim_{n\rightarrow \infty}\pi( \frac{\sin(\frac{\pi}{2^{n+1}})}{\frac{\pi}{2^{n+1}}})=\pi$
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# Find Io in the network in Fig. P3.8
Find Io in the network in Fig. P3.8 using nodal analysis.
Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.
#### Solution:
Let us first “bend” the wires in the circuit diagram to see how the nodes connect.
Notice how we simply moved the resistor connection to the node as it was an “empty” wire. This is the same as the original circuit diagram, however, now, it is much easier for us to see the two main nodes.
Let us label the nodes and the currents as follows:
In the following equations, $k=10^3$ and $m=10^{-3}$
We can now write a KCL equation for the orange node, $V_1$.
$I_1+I_2+4m=2m$
Expressing these currents in terms of voltage and resistance using $I=\dfrac{V}{R}$ gives us:
$\dfrac{V_1}{1k}+\dfrac{V_1-V_2}{2k}+4m=2m\,\,\,\color{orange} {\text{(eq.1)}}$
Now, we will switch our attention to the purple node, $V_2$ and write a KCL equation.
$I_3+I_4=4m+6m$
Again, expressing these currents in terms of voltage and resistance gives us:
$\dfrac{V_2-V_1}{2k}+\dfrac{V_2}{1k}=10m\,\,\,\color{purple} {\text{(eq.2)}}$
Solving equations 1 and 2 simultaneously, we get :
$V_1=1$ v
$V_2=7$ v
From our diagram, we know that $I_0=I_4$.
$I_0=\dfrac{V_2}{1k}$
$I_0=\dfrac{7}{1k}$
$I_0=7$ mA
$I_0=7$ mA
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# The Pythagorean Triple Essay
Pythagoras a Greek philosopher and mathematician is very famous for its Pythagorean Theorem. This theorem states that if a, b and c are sides of a right triangle then a2 + b2 = c2 (Morris, 1997).
The study of the Pythagorean triples started long before Pythagoras knew how to solve it. There were evidences that Babylonians have lists of the triples written in a tablet. This would only mean that Babylonians may have known a method on how to produce such triples (Silverman, 16 – 26). Pythagorean triple is a set of number consisting of three natural numbers that can suit the Pythagorean equation a2 + b2 = c2. Some of the known triples are 3, 4, 5 and 5, 12, 13 (Bogomolny, 1996). How can we derive such triples?
If we multiple the Pythagorean formula by 2 then we generate another formula 2a2 + 2b2 = 2c2. This only means that if we multiply 2 to the Pythagorean triple 3, 4, 5 and 5, 12, 13 then we can get another set of Pythagorean triple. The answer to that is triple 6, 8, 10 and 10, 24, 26. To check whether the said triple are Pythagorean triple, we can substitute it to the original formula a2 + b2 = c2.
Check: is 6, 8, 10 Pythagorean triple?
62 + 82 = 102
36 + 64 = 100
100 = 100
Thus 6, 8 and 10 satisfy the Pythagorean equation.
Δ 6, 8, 10 is a Pythagorean triple.
Check: is 10, 24, 26 satisfy the Pythagorean equation?
102 + 242 = 262
100 + 576 = 676
676 = 676
Thus 10, 24, 26 satisfy the Pythagorean equation.
Δ 10, 24, 26 is a Pythagorean triple.
If we multiply the Pythagorean equation by 3 and using the first 2 Pythagorean triple mentioned above, we can yield another set of Pythagorean triple. Thus we can formulate a general formula that can produce different sets of Pythagorean triple. We can generate an infinite number of Pythagorean triple by using the Pythagorean triple 3, 4, 5. If we multiple d, where k is an integer, to that triple we will yield different sets of Pythagorean triple all the time.
d*(3, 4, 5) where d is an integer.
Check: if k is equal to 4 we get a triple 12, 16, and 20. Is this a Pythagorean triple?
By substitution,
122 + 162 = 202
144 + 256 = 400
400 = 400
Thus 12, 16, 20 satisfy the Pythagorean equation.
Δ 12, 16, 20 is a Pythagorean triple.
Check: if k is equal to 5 we get a triple 15, 20, 25. Is this a Pythagorean triple?
By substitution,
152 + 202 = 252
225 + 400 = 625
625 = 625
Thus 15, 20, 25 satisfy the Pythagorean equation.
Δ 15, 20, 25 is a Pythagorean triple.
But the formula given above is just a formula for getting the multiples of the Pythagorean triple. But is there a general formula in getting these triples? There are formulas that can solve each and every Pythagorean triple that one can ever imagine. One formula that can give us the triples is a = st, b = (s2 + t2)/2 and c = (s2 – t2)/2 (. A simple derivation of these formula will come from the main formula a2 + b2 = c2 (Silverman, 16 – 26). This is a shorten way to derive the formula from theorem 2.1(Pythagorean triples).
a2 + b2 = c2 with a is odd, b is even and a, b and c have no common factors.
a2 = c2 – b2 by additive property
a2 = (c – b)(c + b) by factoring (difference of two squares)
by checking
32 = (5 – 4)(5 + 4) = 1*9
52 = (13 – 12)(13 + 12) = 1*25
72 = (25 – 24)(25 + 24) = 1*49
Essay Topics:
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How Do You Do Multiplication? – tntips.com
# How Do You Do Multiplication?
## How Do You Do Multiplication?
These are the steps to do long multiplication by hand:
1. Arrange the numbers one on top of the other and line up the place values in columns. …
2. Starting with the ones digit of the bottom number, the multiplier, multiply it by the last digit in the top number.
3. Write the answer below the equals line.
## How do you multiply step by step?
These are the steps to do long multiplication by hand:
1. Arrange the numbers one on top of the other and line up the place values in columns. …
2. Starting with the ones digit of the bottom number, the multiplier, multiply it by the last digit in the top number.
3. Write the answer below the equals line.
## How do you do multiplication in math?
Multiplication is often described as a sort of shorthand for repeated addition. For example, when you multiply 4 by 3, you are adding 4 to itself 3 times: 4 + 4 + 4 = 12.
## How do you do 2 digit multiplication?
Multiplying 2-digit numbers by 2-digit numbers
1. You can follow the steps below to multiply 2-digit numbers. Try it with. × …
2. Then, multiply the top number by the ones digit of the bottom number. In this example, multiply. × …
3. Next, multiply the top number by the tens digit of the bottom number. In this example, multiply. …
4. So, ×
## What is short multiplication?
Short multiplication is usually applied when multiplying a two, three or four-digit number by a one-digit number. To do short multiplication, the layout of numbers is key. … To use short multiplication, you need to set the calculation out correctly by writing the first number and then writing the other one underneath it.
## What is basic multiplication?
Multiplication is when you take one number and add it together a number of times. Example: 5 multiplied by 4 = 5 + 5 + 5 + 5 = 20. We took the number 5 and added it together 4 times. This is why multiplication is sometimes called “times”.
## What is the easiest way to teach multiplication?
The Best Way to Teach Multiplication | 5 Simple Steps
2. Step two: introduce skip counting. …
3. Step three: highlight the commutative property. …
4. Step four: drill and practice multiplication facts. …
5. Step five: work with words.
## What is column multiplication?
Long multiplication (or column multiplication) is a written method of multiplying numbers (usually a two- or three-digit number by another large number). … The benefit of the grid method is that it encourages children to think about place value and multiply multiples of ten and one hundred.
## How can I learn my times tables?
10 fun tips for teaching times tables effectively
1. Use times tables chanting. …
2. Make times tables fun with songs and multiplication games. …
3. Make use of times tables grids. …
4. Use concrete resources. …
5. Get active outside the classroom. …
6. Use pupil’s interests to engage them with times tables.
## How do you calculate maths quickly?
For instance, consider 5 x 3.
1. Step 1: Subtract one from the number being multiplied by 5, in this instance the number 3 becomes the number 2.
2. Step 2: Now halve the number 2, which makes it the number 1. Make 5 the last digit. The number produced is 15, which is the answer.
## How do you multiply things Wikihow?
For example, for 5 × 3, add 5 three times: 5 + 5 + 5 =15. To multiply bigger numbers, place the larger number on top of the smaller number. Then, multiply the last digit in the bottom number by each individual digit in the top number.
## What is a multiply in math?
In math, multiplication is the method of calculating the product between two or more numbers. It is a primary arithmetic operation. We use multiplication more often in real life. When few groups of equal sizes are combined we perform multiplication there.
The answer to a multiplication problem is called the product. A product is the result of numbers, known as factors, being multiplied together, such as…
## What is the shortcut key for multiplication?
Dealing with spans of numbers and symbols related to numbers
Term Looks like How to get it (keyboard)
Multiplication sign × Alt+0215 *
Division sign ÷ Alt+0247 *
Plus/minus sign ± Alt+0177 *
Superscript number m3 Ctrl+Shift+=
## How do you teach multiplication in a fun way?
35 Fun, Hands-on Ways to Teach Multiplication
1. Draw Waldorf multiplication flowers. …
2. Play multiplication war. …
3. Put a mathematical twist on an old favorite. …
4. Repurpose an egg carton as a multiplication problem generator. …
5. Teach multiplication facts with a simple wheel. …
6. Make a deck of fact family triangles. …
7. Use LEGO bricks.
## What’s a multiplication pattern?
Patterns in Multiplying by 2’s
All multiples of 2 have a pattern of 2, 4, 6, 8, or 0 in the ones place. When multiplying ANY number by 2, the result is EVEN. … even numbers/even columns alternate with odd numbers/odd columns. There are 5 even numbers and 5 odd numbers in each range of 10 numbers (which is ½ or 50/50).
## Math Antics – Multi-Digit Multiplication Pt 2
Related Searches
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how to do multiplication without a calculator
how to do multiplication 2 digit
how to do multiplication with decimals
how to do multiplication by hand
how to multiply fast
how to do short multiplication
how to do long multiplication with 3 digits
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# What is 34/2 as a decimal?
## Solution and how to convert 34 / 2 into a decimal
34 / 2 = 17
34/2 or 17 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 34/2 is 34 divided by 2
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 34 is being divided into 2. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 34 divided by 2. We use this as our equation: numerator(34) / denominator (2) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 34/2 as our equation:
### Numerator: 34
• Numerators are the portion of total parts, showed at the top of the fraction. Overall, 34 is a big number which means you'll have a significant number of parts to your equation. The good news is that 34 is an even number which can simplify equations (sometimes). Values closer to one-hundred make converting to fractions more complex. Time to evaluate 2 at the bottom of our fraction.
### Denominator: 2
• Denominators represent the total parts, located at the bottom of the fraction. With 2 being a small number, you won't have as large of a sum. And it is nice having an even denominator like 2. It simplifies some equations for us. Overall, a small denominator like 2 could make our equation a bit simpler. Now let's dive into how we convert into decimal format.
## Converting 34/2 to 17
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 2 \enclose{longdiv}{ 34 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Solve for how many whole groups you can divide 2 into 34
$$\require{enclose} 00.17 \\ 2 \enclose{longdiv}{ 34.0 }$$
How many whole groups of 2 can you pull from 340? 34 Multiply this number by 2, the denominator to get the first part of your answer!
### Step 3: Subtract the remainder
$$\require{enclose} 00.17 \\ 2 \enclose{longdiv}{ 34.0 } \\ \underline{ 34 \phantom{00} } \\ 306 \phantom{0}$$
If there is no remainder, you’re done! If you have a remainder over 2, go back. Your solution will need a bit of adjustment. If you have a number less than 2, continue!
### Step 4: Repeat step 3 until you have no remainder
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 34/2 into a decimal
Dining - We don't give a tip of 34/2 of the bill (technically we do, but that sounds weird doesn't it?). We give a 1700% tip or 17 of the entire bill.
### When to convert 17 to 34/2 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 34/2 = 17 what would it be as a percentage?
• What is 1 + 34/2 in decimal form?
• What is 1 - 34/2 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 17 + 1/2?
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# What is the ZC?
Finding z Critical Values (zc) Critical values are frequently required. A critical value is frequently used to represent a rejection region cut-off value for a hypothesis test, also known as a confidence interval zc value. Furthermore, how do you locate ZC in C?
The cumulative z value for the given CI percentage (from any z table) is called “zc”. 95% is required to obtain zc. 1 –
95 = . 05 (as we have). 05 in both tails) 05/2 =
1 – 025 (in each tail) 025 = 975. Look up.
On any z table, 975. For z, the z value is. 975 is a value of 1.96.
As a result, the 95% confidence interval is 1.96. What is the z score of 95%, for example? Z=1.96 is the Z value for 95% confidence. As a result, what is the 98 ZC?
Confidence Level z*- value 90% 1.64 95% 1.96 98% 2.33 99% 2.58 According to Statistics For Dummies, the z score is 90%. 90% 0.4500 1.645 95% 0.4750 1.960 98% 0.4900 2.326 99% 0.4950 2.576 Area in Tails Confidence Level
## In statistics, what does ZC mean?
Finding z Critical Values (zc) Critical values are frequently required. A critical value is frequently used to represent a rejection region cut-off value for a hypothesis test, also known as a confidence interval zc value.
## From a table, how do you find the z value?
Go to the row that represents your z-value’s decimal point, as well as the first digit after it (the tenths digit). Go to the column that follows the decimal point (the hundredths digit) of your z-value’s second digit.
Steps 1 and 2 divide the row and column into two sections.
## What is the 90-degree confidence interval for Z?
Calculating the Confidence Interval Confidence Interval Z 90% 1.645 95% 1.960 99% 2.576 99.5% 2.807 Calculating the Confidence Interval Confidence Interval Z 90% 1.645 95% 1.960 99% 2.576 99.5%
## What is the percentile of the Z score?
Z-scores are used to define the scale by measuring how outstanding an individual is in relation to the mean of a population. The standard deviation for that population is used to determine how outstanding an individual is. It’s worth noting that percentiles use the median (50th percentile) as an average, whereas z-scores use the mean (z-score of zero).
## For 5%, what is the z score?
When we look at the table, we notice that the value 0.90 isn’t exactly there; however, the values 0.8997 and 0.9015 are present, respectively, and correspond to Z values of 1.28 and 1.29 (i.e., 89.97% of the surface area under the standard normal curve is below 1.28). Percentages of Computing When we look at the table, we notice that the value 0.90 isn’t exactly there; however, the values 0.8997 and 0.9015 are present and correspond to Z values of 1.28 and 1.29 (i.e., 89.97% of the surface area under the standard normal curve is below 1.28), respectively.
Percentages of Computing Percentile Z 2.5th -1.645 5th -1.645 10th -1.282 25th -0.675 Percentile Z 2.5th -1.960
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## Precalculus (6th Edition) Blitzer
The partial fraction is, $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$
$\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{{{\left( x-1 \right)}^{2}}}$ Now, multiply both sides by \begin{align} & {{x}^{2}}+2x+7=A{{\left( x-1 \right)}^{2}}+Bx\left( x-1 \right)+Cx \\ & =A\left( {{x}^{2}}+1-2x \right)+B{{x}^{2}}-Bx+Cx \\ & =A{{x}^{2}}+A-2Ax+B{{x}^{2}}-Bx+Cx \\ & =\left( A+B \right){{x}^{2}}+\left( -2A-B+C \right)x+A \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=1$ …… (1) $-2A-B+C=2$ …… (2) $A=7$ …… (3) And put the value of A into equation (1): \begin{align} & 7+B=1 \\ & B=-6 \end{align} Also, put the value of A and B into equation (2): \begin{align} & -2\left( 7 \right)+6+C=2 \\ & C=2+8 \\ & =10 \end{align} Therefore, $\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$ Thus, the partial fraction of the provided expression is $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$.
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# Question Video: Finding the Size of an Angle Using the Properties of Tangents to a Circle
In the figure below, π΄π΅ is a tangent to the circle π at π΅, πΆπ· is a diameter, πβ π΅π΄π = π₯, and πβ ππ·π΅ = 2π₯ β 55. Find the value of π₯ in degrees.
04:27
### Video Transcript
In the figure below, π΄π΅ is a tangent to the circle π at π΅, πΆπ· is a diameter, the measure of angle π΅π΄π equals π₯, and the measure of angle ππ·π΅ equals two π₯ minus 55. Find the value of π₯ in degrees.
Whenever you attempt this kind of question, I always say mark on diagram the angles that you already know or that you can work out first. Well, the first angle we know is that the angle π·π΅π is gonna be equal to two π₯ minus 55. And we know that it equals two π₯ minus 55 because itβs gonna be the same as angle ππ·π΅ because itβs an isosceles triangle. Well, how do we know that itβs an isosceles triangle? Well, itβs gotta be an isosceles triangle because π·π will have to equal ππ΅ because theyβre both radii of the circle.
Okay, great, so now what else do we know? Well, the next angle we know is angle π΄π΅π. We know that angle π΄π΅π is equal to 90 degrees. So itβs a right angle and this is because itβs the angle between a tangent and a radius and the angle between tangent and radius is always 90 degrees. Okay, so now, weβve actually written all the angles that we know.
So now, itβs actually time to actually calculate some additional angles. The first one weβre gonna start with is angle π·ππ΅. Well, we know that angle π·ππ΅ is gonna be equal to 180 minus two π₯ minus 55 minus two π₯ minus 55, which is gonna be equal to 180 plus 110 minus four π₯. Now, Iβll point out here a common mistake. The common mistake is actually to have had minus 110. And the reason you might have that is because obviously we could see that it was minus 55 minus 55. However, as itβs minus a minus, that makes it positive. So we get add 110. And we can say that angle π·ππ΅ is equal to 290 minus four π₯. And this is because the angles in a triangle add up to 180.
So now, we can move on to angle π΅ππ΄, which is going to be equal to 180 minus then weβve got 290 minus four π₯. And thatβs because that was the angle π·ππ΅. And this gives us four π₯ minus 110. Again, be very careful here of the negative numbers cause we have got minus a negative, again which gives us positive four π₯. And again, we include our reasoning. And the reason for this is because the angles on a straight line equal 180 degrees. So therefore, angle π·ππ΅ and angle π΅ππ΄ must be equal to 180 degrees.
Okay, great, so weβve now found that angle. So now, we can actually move on and find the angle weβre looking for, which is π΅π΄π. So therefore, we could say that the angle π΅π΄π is equal to 180 minus 90 minus four π₯ minus 110. Well, therefore, we can say that π₯ is equal to 200 minus four π₯. And we can say π₯ because we can see that angle π΅π΄π is equal to π₯. So now, weβve got π₯ is equal to 200 minus four π₯ and we can solve to find π₯. So then if we add four π₯ to each side, we get five π₯ equals 200. And then we divide both sides by five. We get π₯ equals 40. So therefore, we can say that the measure of angle π΅π΄π is equal to 40 degrees. And therefore, π₯ is equal to 40 degrees.
Okay, great, so weβve got to our final answer. But what we want to do here is actually quickly double check it. Iβm going to check it by using the first triangle, which is triangle π΅π·π. So we have all three angles which are two π₯ minus 55 plus two π₯ minus 55 plus 290 minus four π₯. So now, if I actually substitute in our value for π₯ which is 40, weβre gonna get 80 minus 55 because two multiplied by 40 gives us 80 plus another 80 minus 55 minus 290 minus 160 because again four multiplied by 40 equals 160.
Okay, great, so letβs calculate this. Well, this gives us 180. Well, we know that this is correct because the angles in a triangle add up to 180. So therefore, yes, we can say with confidence that π₯ is equal to 40 degrees.
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Mechanical Engineering - Linkage System Home : www.sharetechnote.com Linkage System : Relative Motion Analysis : Velocity All of linkage system problems involve relative-motion analysis because the motion of a link is affected directly by the motion of its adjacent link Must be comfortable with setting up vectors as well as executing vector addition, subtraction, and multiplication by hand Scalar method/approach can also be used as an alternative method for the vector method and is favored over the vector method as it is less time consuming Let's consider a simple linkage system as shown below Velocity of the link at point B is calculated by using the following equation if vector method is to be used Putting a short description on this equation as follows would help you with understanding this equation. The equation states that the velocity at point B of the rotating link AB is dependent on the sum of the velocity at point A and velocity of point B with respect to point A Example 1 > The slider block C moves at 8 m/s down the inclined groove. Determine the angular velocities of links AB and BC, at the instant shown. Step 1 > Situational analysis – set up the vectors For any linkage velocity problems, you must know the direction of the velocity vectors at each pivot points, which are points A,B, and C this case In this case, the velocity at point A is 0 because it is stationary To allow point C to move in such direction, link BC must rotate counter-clockwise and link AB must rotate clockwise Since link AB rotates in a clockwise direction about point A, the velocity vector at point B points downward can be represented as follows : Compute the distance vectors for link AB and link BC: Step 2 > Plug in the vectors into the relative velocity equations to obtain simplified expressions To calculate the angular velocity of link BC, first plug in all the vectors for the expression for velocity at point C Since point B rotates about point A in a clockwise direction, the velocity of point B can be expressed as following Plugging this into the first equation, you would get Example 2 > Scalar method of relative velocity analysis Scalar method of relative velocity analysisThe shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the angular velocity of the link CB at the instant shown, if the link AB is rotating at 4 rad/s Step 1 > locate the instantaneous center of zero velocity Start the question by drawing the velocity diagram of the linkage system Draw an orthogonal line for all the velocity vectors and locate the point of intersection for the lines The point of intersection is the instantaneous center(IC) of zero velocity Step 2 > Calculate the angular velocity of link CB Since point A is the stationary pivot point of link AB, velocity at point B can be calculated by using the length of link AB and the given angular velocity of the link: As velocity at point B is known, it can be used to calculate the angular velocity of link CB. And the angular velocity is calculated as following In the equation above, the magnitude of needs to be computed before calculating the angular velocity. The magnitude will be calculated through trigonometric analysis of the simplified instantaneous velocity diagram shown below. Plug in the distance between point B and IC into the angular velocity equation for link CB
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Adding and Subtracting Numbers to 20
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## How to Add and Subtract Numbers to 20 - Review
In second grade, you'll memorize the sums and differences up to 20.
Let’s review some of these! 🤗
### Adding Numbers Up to 20
This is an addition table up to 10.
To find the sum of any two numbers, find the number where they intersect on the table.
6 + 9 = ___
6 + 9 = 15
Great job! 👏
10 + 4 = _____
10 + 4 = 14
### Subtracting Numbers up to 20
Let’s look at a subtraction table now.
The minuends are at the top (👆) and the subtrahends are in the left column (👈).
Tip: The minuend is the bigger number that comes first in subtraction.
Let’s try to use this table to subtract:
15 - 3 = ?
👉 Find 15 on the top row and 3 on the left column.
What number do you find at the point where these numbers meet?
15 - 3 = 12
Great job! You now know how to use the addition and subtraction tables!
Study them.
Your goal is to have all of these addition and subtraction facts memorized.
Try the practice. By the time you're done, you'll have a lot more memorized.
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# How many ways can the letters of the word Arrange be arranged so that two As come together?
10Given, the word ARRANGE. It has 7 letters of which 2 letters (A, R) are repeating. The letter A is repeated twice, and the letter R is also repeated twice in the given word.To find: Number of ways the letters of word ARRANGE be arranged in such a way that not all R’s do come together.First, we find all arrangements of word ARRANGE, and then we minus all those arrangements of word ARRANGE in such a way that all R’s do come together, from it. This will exactly be the same as- all number of arrangements such that not all R’s do come together.Since we know, Permutation of n objects taking r at a time is nPr,and permutation of n objects taking all at a time is n!And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.A total number of arrangements of word ARRANGE: Total letters 7. Repeating letters A and R, the letter A repeating twice and letter R repeating twice. The total number of arrangementsNow we find a total number of arrangements such that all R’s do come together.A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here R, R) is assumed to be a single letter and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to nPr. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here R, R).Letters in word ARRANGE: 7 lettersLetters in a new word: A, RR, A, N, G, E: 6 letters (Letter A repeated twice).Total number of word arranging all the letters where second fraction 2! divided by 2! comes from arranging letters inside the group RR: arrangements of two letters where all the two letters are same = 2!/2! = 1 (Obviously! You can even think of it).Now, a Total number of arrangements where not all R’s do come together is equals to total arrangements of word ARRANGE minus the total number of arrangements in such a way that all R’s do come together.= 900Hence, the total number of arrangements of word ARRANGE in such a way that not all R’s come together is equals to 900. No worries! We‘ve got your back. Try BYJU‘S free classes today!Right on! Give the BNAT exam to get a 100% scholarship for BYJUS coursesNo worries! We‘ve got your back. Try BYJU‘S free classes today!No worries! We‘ve got your back. Try BYJU‘S free classes today!Open in App Suggest Corrections1 Text SolutionSolution : The letters of word ARRANGE can be rewritten as
A R N G E
A R
So we have 2 A's and 2 R's , and total 7 letters.
(i) Total number of words is `(7!)/(2!2!)=1260`.
The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g.,
The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g.,
(R R),A,A,N,G,E
Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is
`(7!)/(2!2!)-(6!)/(2!)=900`
(ii) The number of words in which both A's are together is `6!//2!=360`, e.g.,
(A A),R,R,N,G,E
The number of words in which both A's and both R's are together is 5!=120, e.g.,
(A A), (R R), N,G,E
Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240.
(iii) There are in all 900 words in each of which the two R's are never together. Consider any such word. Either the two A's are together or the two A's not together. But the number of all such arrangements in which the two A's are together is 240. Hence, the number of all such arrangements in which the two A's not together is 900-240=660.
|
### Lesson 1: Vectors - Fundamentals and Operations
```Vectors - Fundamentals and Operations
1.
2.
3.
4.
5.
6.
7.
8.
Vectors and Direction
Resultants
Vector Components
Vector Resolution
Relative Velocity and Riverboat Problems
Independence of Perpendicular Components of
Motion
Vector vs. scalar quantity
• Vector: a physical quantity that has both a
magnitude and a direction. We use an arrow
above the symbol to represent a vector.
A
• Scalar: a physical quantity that has only a
magnitude but no direction.
A
Representing Vectors
• Vectors on paper are simply arrows
– Direction represented by the way the ARROW POINTS
– Magnitude represented by the ARROW LENGTH
• Examples of Vectors
– Displacement
– Velocity
– Acceleration
– Force
Magnitude of a Vector
• The magnitude of a vector in a scaled vector
diagram is depicted by the length of the
arrow. The arrow is drawn a precise length in
accordance with a chosen scale.
Directions of Vector
Compass Point
The direction of a vector is often expressed as an angle of
rotation of the vector about its "tail" from east, west,
north, or south
20 meters at 10° south of west
34 meters at 42° east of north
N
W
0°
S
Directions of Vector
Reference Vector
Uses due EAST as the 0 degree reference,
all other angles are measured from that point
20 meters at 190°
34 meters at 48°
90°
0°
180°
270°
Changing Systems
• What is the reference vector angle for a vector
that points 50 degrees east of south?
270° + 50° = 320°
50°
• What is the reference vector angle for a vector
that points 20 degrees north of east?
20°
20°
What we can DO with vectors
– To produce a NEW VECTOR (RESULTANT)
• MULTIPLY/DIVIDE by a vector or a scalar
– To produce a NEW VECTOR or SCALAR
• Two vectors can be added together to determine the
result (or resultant). The resultant is the vector sum of
two or more vectors. It is the result of adding two or
more vectors together.
1. Graphical method: using
a scaled vector diagram
A
= ?
+ B
method (tip to tail)
• Parallelogram method
2. Mathematical method Pythagorean theorem and
trigonometric methods
• You are walking north 3 meters, then walking east 4
meters. What is your final displacement?
A
A
+
+
B
B
= ?
C A B
The resultant is from the first tail
Resultant
C A B
applying the head-to-tail method to determine the
sum of two or more vectors:
1. Choose a scale and indicate it on a sheet of paper. The best
choice of scale is one that will result in a diagram that is as
large as possible, yet fits on the sheet of paper.
2. Pick a starting location and draw the first vector to scale in the
indicated direction. Label the magnitude and direction of the
scale on the diagram (e.g., SCALE: 1 cm = 20 m).
3. Starting from where the head of the first vector ends, draw
the tail of second vector to scale in the indicated direction.
Label the magnitude and direction of this vector on the
diagram.
4. Repeat steps 2 and 3 for all vectors that are to be added
5. Draw the resultant from the tail of the first vector to the head
of the last vector. Label this vector as Resultant or simply R.
6. Using a ruler, measure the length of the resultant and
determine its magnitude by converting to real units using the
scale (4.4 cm x 20 m/1 cm = 88 m).
7. Measure the direction of the resultant using the
counterclockwise convention.
practice
method to determine the resultant, use a
ruler and a protractor.
1. 3 m east, and 4 m south.
2. 5 m north and 12 meters west.
3. 2 m east, 4 m north and 5 m west.
Graphical method 2: parallelogram (tail-tail)
• A cart is pushed in two directions, as the result, the
cart will move in the resultant direction
A
A
+
B
= ?
C A B
+
B
A
B
The resultant is diagonal of the parallelogram from the tail
of both vectors.
A
+
B
Parallelogram (tail-tail)
A
= ?
A
B
B
Parallelogram: tail and tail
touching, the resultant is the
diagonal.
the resultant is from first tail to last
example
• A model airplane heads due east at 1.50 meters per second,
while the wind blows due north at 0.70 meter per second.
1. Draw the resultant vector in the diagram
2. Determine the scale used in the diagram.
3. Determine, to the nearest degree, the angle between north
and the resultant velocity.
0.7 m/s
θ=?
R
1.5 m/s
θ = 65o
Vector properties
1. Vector can be moved parallel to themselves in a diagram.
A B
parallelogram
B
A
2. Vectors can be added in any order (commutative and
associative)
A B B A
B
A
A
B
3. To subtract a vector, add its opposite.
B
A
A B A (B)
4. Multiplying or dividing vectors by scalars results in vectors
with different size, but same direction.
Equilibrant
• The equilibrant vectors of the resultant of A
and B is the opposite of the resultant of
vectors A and B.
B
• Example:
A
Determine
equilibrant of A & B
B
A
R
A
B
R
Parallelogram
Commutative property of vectors
what is the title of this animation:?
Vector subtraction
A
-
=
B
= A
=
A-B
?
+
(-B)
+
Practice: determine resultant for the following diagrams
Mathematical Method - Use
Pythagorean Theorem to determine
magnitude
• The procedure is
restricted to the
that make right angles
to each other.
B
A
+
B
A2 + B2 = C2
= A
C
determine the magnitude of each resultant vector.
Mathematical Method - Using
Trigonometry to Determine a
Vector's Direction
Opp.
sin
Hyp.
cos
Hyp.
Opp.
tan
SOH
CAH
TOA
Example
• Example: Eric leaves the base camp and hikes 11 km, north
and then hikes 11 km east. Determine Eric's resulting
displacement.
θ
A2 + B2 = C2
(11 km)2 + (11 km) 2 = C2
C = 15.6 km
Opp. 11 km
tan
1
45o
Eric’s displacement is 15.6 km at 45o northeast
• Note: The measure of an angle as determined through use of
SOH CAH TOA is not always the direction of the vector.
Example: determine the magnitude and direction of
each resultant vector.
Vector Addition: 6 + 8 = ?
All that can be said for certain is that 8 + 6 can add up to a vector
with a maximum magnitude of 14 and a minimum magnitude of 2.
The maximum is obtained when the two vectors are directed in the
same direction. The minimum is obtained when the two vectors are
directed in the opposite direction. The sum of vectors 8 and vector 6
can be any number between 14 and 2.
Example
•
1.
2.
3.
4.
A 5.0-newton force and a 7.0-newton force
act concurrently on a point. As the angle
between the forces is increased from 0° to
180°, the magnitude of the resultant of the
two forces changes from
0.0 N to 12.0 N
2.0 N to 12.0 N
12.0 N to 2.0 N
12.0 N to 0.0 N
Example
•
1.
2.
3.
4.
A 3-newton force and a 4-newton force are
acting concurrently on a point. Which force
could not produce equilibrium with these two
forces?
1N
7N
9N
4N
Example
•
As the angle between two concurrent forces
decreases, the magnitude of the force
required to produce equilibrium
1. decreases
2. increases
3. remains the same
Example
• Two 20.-newton forces act concurrently on an
object. What angle between these forces will
produce a resultant force with the greatest
magnitude?
A. 0°
B. 45°
C. 90.°
D. 180.°
practices
1.
A person walks 150. meters due east and then walks 30. meters due west.
The entire trip takes the person 10. minutes. Determine the magnitude and
the direction of the person’s total displacement.
2. A dog walks 8.0 meters due north and then 6.0 meters due east.
a. Determine the magnitude of the dog’s total displacement.
b. Use appropriate scale to draw a vector diagram including the resultant
based on the information given.
3. A 20.-newton force due north and a 20.-newton force due east act
concurrently on an object. What is the additional force necessary to bring
the object into a state of equilibrium?
4. As the angle between two concurrent forces decreases, what happens to
the magnitude of the force required to produce equilibrium?
Vector Components
• In situations in which vectors are directed at angles to the
customary coordinate axes, we transform the vector into two
parts with each part being directed along the coordinate axes.
Each part is called a component of the vector. So any vector
can be transformed into two components.
Component 1
Component 1
Component 2
Component 2
• Any vector directed in two dimensions can be thought of as
having an influence in two different directions.
• Each part of a two-dimensional vector is known as a component.
• The components of a vector depict the influence of that vector in
a given direction.
• The combined influence of the two components is equivalent to
the influence of the single two-dimensional vector.
• The single two-dimensional vector could be replaced by the two
components.
Finding components –
Vector Resolution
•
To determine the magnitudes of the
components of a vector, we the
trigonometric method
Opp. Ay
sin
Hyp. A
Ay A sin
y
A
Ay
θ
Ax
x
cos
Hyp. A
Ax A cos
Trigonometric Method of Vector
Resolution
1. Construct a rough sketch (no scale needed) of the vector
in the indicated direction. Draw a rectangle about the
vector such that the vector is the diagonal of the
rectangle.
2. Label the components of the vectors with symbols to
indicate which component represents which side.
3. To determine the length of the side opposite the
indicated angle, use the sine function. Substitute the
magnitude of the vector for the length of the hypotenuse.
4. Repeat the above step using the cosine function to
determine the length of the side adjacent to the indicated
angle.
Example: Determine components for the vector 60 N at 40o
Example
• The vector diagram below represents the horizontal
component, FH, and the vertical component, FV, of a 24newton force acting at 35° above the horizontal.
• What are the magnitudes of the horizontal and vertical
components?
practices
1.
2.
3.
4.
5.
A baseball is thrown at an angle of 40.0° above the horizontal. The
horizontal component of the baseball’s initial velocity is 12.0 meters per
second. What is the magnitude of the ball’s initial velocity?
An airplane flies with a velocity of 750. kilometers per hour, 30.0° south
of east. What is the magnitude of the eastward component of the plane’s
velocity?
A vector makes an angle, θ, with the horizontal. What is θ if the
horizontal and vertical components of the vector are equal in magnitude?
A person exerting a 300.-newton force on the handle of a shovel that
makes an angle of 60.° with the horizontal ground. What is the
component of the 300.-newton force that acts perpendicular to the
ground?
As the angle between a force and level ground decreases from 60° to 30°,
what happens to the the vertical component of the force?
Component Method of
• In order to solve more complex vector addition problems, we
need to combine the concept of vector components and the
principles of vector resolution with the use of the
Pythagorean theorem.
Right Angle Vectors
components
components
R2 = (8.0 km)2 + (6.0 km)2
R2 = 64.0 km2+ 36.0 km2
R2 = 100.0 km2
R = SQRT (100.0 km2)
R = 10.0 km
Use Tangent Function (TOA) to find
Direction of Vectors
Tangent(Θ) = 8.0/6.0
Θ =53°
• Since the angle that the resultant makes with
east is the complement of the angle that it
makes with north, we could express the
direction as 53° CCW.
16 m
24 m
36 m
2m
X direction: Rx = (-16 m) + (-36 m)
Rx = -52 m
Y direction: Ry = (24 m) + (-2 m)
Ry = 22 m
R2 = Rx2 + Ry2 =(22.0 m)2 + (-52 m)2
R = 56.5 m
Θ
Tan(Θ)=52m/(22m)
Θ =67°
The resultant is the II Quadrant. The CCW direction is 157.1° CCW.
perpendicular
Add the components of the original displacement vectors to find
two components that form a right triangle with the resultant
vector.
1. Find the x and y components of all vectors.
Ax=AcosθA;
Ay=AsinθA;
Bx=BcosθB;
By=BsinθB
2. Find the x and y component of the resultant vector:
Rx=Ax + Bx;
Ry=Ay + By
3. Use the Pythagorean theorem to find the magnitude of the
resultant vector.
2
2
R Rx Ry
4. Use tan-1 function to find the angle the resultant vector
makes with the x-axis.
Ry
1
R tan ( )
Rx
Example
• A hiker walks 25.5 km from her base camp at 35o south of
east. On the second day, she walks 41.0 km in a direction 65o
north of east at which point she discovers a forest ranger’s
tower. Determine the magnitude and direction of her
resultant displacement between the base camp and the
ranger’s tower.
R
d2
35o
d1
65o
Example
A bus heads 6.00 km east, then 3.5 km north, then 1.50
km at 45o south of west. What is the total displacement?
A: 6.0 km, 0° CCW
B: 3.5 km, 90° CCW
C: 1.5 km, 225° CCW
+
A
+
B
C
Cx = Ccos225o = -1.06 km
Cy = Csin225o = - 1.06 km
Example
• Cameron and Baxter are on a hike. Starting from home base,
they make the following movements.
A: 2.65 km, 140° CCW
B: 4.77 km, 252° CCW
C: 3.18 km, 332° CCW
• Determine the magnitude and direction of their overall
displacement.
Vector
X Component (km)
Y Component
A
2.65 km
140° CCW
(2.65 km)•cos(140°) (2.65 km)•sin(140°)
= -2.030
= 1.703
B
4.77 km
252° CCW
(4.77 km)•cos(252°) (4.77 km)•sin(252°)
= -1.474
= -4.536
C
3.18 km
332° CCW
(3.18 km)•cos(332°) (3.18 km)•sin(332°)
= 2.808
= -1.493
Sum of
A+B+C
-0.696
R = 4.38 km
-4.326
Θ = 80.9°
Direction is at 260.9° (CCW)
Relative Velocity and Riverboat
Problems
When objects move within a medium that is also moving with respect
to an observer, the resultant velocity can be determined by vector
Analysis of a Riverboat's Motion
• The affect of the wind upon the plane is similar to the affect
of the river current upon the motorboat.
Example
Suppose that the river was moving with a velocity of 3 m/s,
North and the motorboat was moving with a velocity of 4 m/s,
East.
1. What is the resultant velocity (both magnitude and
direction) of the boat?
2. If the width of the river is 500 meters wide, then how
much time does it take the boat to travel shore to shore?
3. What distance downstream does the boat reach the
opposite shore?
To calculate resultant velocity
A2+B2 = R2
(4.0 m/s)2 + (3.0 m/s)2 = R2
16 m2/s2 + 9 m2/s2 = R2
25 m2/s2 = R2
SQRT (25 m2/s2) = R
5.0 m/s = R
tan(θ) = (3/4)
θ = tan-1 (3/4)
θ = 36.9o
To calculate time
Critical variable in multi dimensional problems is TIME.
We must consider each dimension SEPARATELY, using TIME as
the only crossover VARIABLE.
d
v
t
d
t
v
d 500 m
t
125 s
v 4 m/ s
d and v must be in the same dimension
To calculate distance downstream
Critical variable in multi dimensional problems is TIME.
We must consider each dimension SEPARATELY, using TIME as
the only crossover VARIABLE.
d
v
t
d v t
d and v must be in the same dimension
d (3 m / s)(125 s) 375 m
Example 1
•
A motorboat traveling 12 m/s, East
encounters a current traveling 5.0 m/s,
North.
1. What is the resultant velocity of the motorboat?
2. If the width of the river is 96 meters wide, then
how much time does it take the boat to travel
shore to shore?
3. What distance downstream does the boat reach
the opposite shore?
Example 2
•
1.
2.
3.
A motorboat traveling 4 m/s, East encounters a current
traveling 7.0 m/s, North.
What is the resultant velocity of the motorboat?
If the width of the river is 80 meters wide, then how much time
does it take the boat to travel shore to shore?
What distance downstream does the boat reach the opposite
shore?
Example 3
• A stream is 30. meters wide and its current flows southward
at 1.5 meters per second. A toy boat is launched with a
velocity of 2.0 meters per second eastward from the west
bank of the stream.
1. What is the magnitude of the boat’s resultant velocity as it
crosses the stream?
2. How much time does it take the toy boat to travel shore to
shore?
3. What distance downstream does the toy boat reach the
opposite shore?
Independence of Perpendicular
Components of Motion
• In the riverboat or airplane problems, the resultant
velocity is obtained by adding the perpendicular
components. These perpendicular components are
independent of each other, which means, as one
changes, the other is not affected at all. For
example, it does not matter with what speed I launch
the toy boat, the speed of the river remains the
same, even though the resultant velocity changes.
Extra credit question
•
Mia Ander exits the front door of her home and walks along
the path shown in the diagram at the right (not to scale). The
walk consists of four legs with the following magnitudes:
A = 46 m
B = 142 m
C = 78 m
D = 89 m
•
Determine the magnitude and direction of Mia's resultant
displacement. Consider using a table to organize your
calculations.
```
|
# What is a ratio analysis?
## What is a ratio analysis?
What is a ratio analysis? A ratio analysis is a way to compare the values of two or more variables. For example, if a product’s ratio is 1:1, its ratio can be measured, while the product’ (2:1) ratio is measured, and vice versa. Computing ratios In the past, ratios were used as a function of price. This became a goal of the game industry in the 1980s, and the ratio analysis was often used as a way to measure the price of a product. However, as the market was getting saturated, ratios were often used in other applications. In games, it is easy to use the ratio as a measure of a product price. A popular example is the ratio of a product versus a price (2:2). The product price is the ratio between the price of the product and the price of its competitors (3:1). The value of a product‘s price is the price of that product. A common way to measure price is to use the product‘ ratio. In the past, a product price was a function of both the price (2) and the ratio (3:2). Nowadays, people are working on a lot of different methods. In this post, I will come up with a simple algorithm to compute the ratio of two products. Here is a simple algorithm for computing the ratio of 2:2 and 3:2. The problem Let’s take a look at the algorithm below. The formula for computing the product ratio is as follows: The algorithm is as follows. 1. – (2:3) 2. – (1:1) 3. Since we are interested in the product ratio, we need to know the value of the product.
## We Do Your Homework For You
Let‘s take a closer look at the formula of the algorithm below: 1 2 3 2/3 I‘m going to use the formula: 2 / 3 3 / 2 2/(1:1/2) The ratio of a 3:2 product is defined as 1/3, which is the ratio when the product is produced by a 3:1 product. The price of a 3/2 product is 1/3. Therefore, the ratio of the 2:2 price is 1/2, which is 1/(1:2/3) = 1/3 = 1/2/3. So the ratio of 1/2 is 1/(2/3). The formula is as follows, which is as follows 1 / 2 2 / 2 3 / 3 2/(2/2) = 1/(2.3/2) / 3 = 1/(3/2). Now, what is the value of aWhat is a ratio analysis? A ratio analysis is a way of evaluating the relative proportions of the different groups of people in a population. The ratio can be used to describe the incidence of diseases in a population based on the ratios of the different populations. A standard ratio is a number, which is a percentage of the total population divided by the population size. The standard ratio can be divided into categories, such as 1. The proportion of people who are sickest in the population 2. The proportion who are sicker than the population If the proportion of people in the population is less than 1%, the population is a sicker population. In a number, the ratio is used to express the relative proportion of people with a certain type of disease. Examples of ratios in a population include 1-1/2 1/2.1 1 2-1/4 1,2 2 3-3/4 In a population, the ratios of 1 to 3 are the proportions of people with diseases that are most likely to be affected by a particular disease. For example, diseases that affect the body are more likely to be related to the body than are diseases that affect attention. By using a ratio, it is possible to measure the relative proportion that visit site a person with a certain disease, but the relative proportion is not a ratio. Example 1 The ratio of people in each age group is 1 = 1/2,2 = 1/4,3 = 1/6,4 = 1/12. This ratio is compared to the ratio of people who have the same disease, or are the same. After 10 years, the ratio of the population will be 1 – 10/12 = 1/1,2 – 10/2 = 1.
## Take My Test
If the population is larger or smaller than 10 years, it is necessary to take into account the age of the population, or the age of parents, which is more important than the population size at the time of the survey. For example, if the population was 40 years old, it would be necessary to take the age of a person of 40 years as a factor of the population size, or to divide the population into the following five groups of people: 1 – 40 = 1/3,4 = 4/3,5 = 5/3,6 = 6/3,7 = 7/3,8 = 8/3,9 = 9/3 2 – 42 = 2/3,3 = 4/4,5 = 6/4,8 = 6/6,9 = 7/6,10 = 7/7,11 = 8/7,12 = 8/8,13 = 9/8,14 = 9/9,15 = 9/10,16 = 10/10,17 = 10/11,18 = 10/12,19 = 10/13,20 = 10/14,21 = 10/15,22 = 10/16,23 = 10/17,24 = 10/18,25 = 10/19,26 = 10/20,27 = 10/21,28 = 10/22,29 = 10/23,30 = 10/24,31 = 10/25,32 = 10/26,33 = 10/27,34 = 10/28,35 = 10/29,36 = 10/30,37 = 10/31,38 = 10/32,39 = 10/33,40 = 10/34,41 = 10/35,42 = 10/36,43 = 10/37,44 = 10/38,45 = 10/39,46 = 10/40,47 = 10/41,48 = 10/42,49 = 10/43What is a ratio analysis? How does a ratio analysis relate to your job search? The ratio analysis is a method of comparing your work performance and the productivity of a company. It’s a method of looking at the number of people whose work performance is the same in the past and comparing it to the number of employees in the current company. It’s also a database that’s used to store your data, and it’s used to rank companies by their productivity. The number of people who have more work than are already in a company is called the productivity ratio, and it can be used to rank a company’s productivity. The ratio is a number that’s used by job search and vice versa, and it shows how many people are involved in the process of determining the productivity of the company. Are there any algorithms or rules that can be used in the ratio analysis to find out how many people work more or less, in a given company? I don’t think there’s a simple way to do this. If you’re looking for a way to rank your company’s productivity, you can use a number against the top of the table. This will tell you which companies are in the top, and which are not. What’s the difference between a ratio and a ranking? In the ratio analysis, it’s the number of men who are paid more than women, in terms of their productivity. Men who are paid at least half of their labor than women are in the company. In the ranking, the ranking is based on the number of women in the company, which is taken with the number of terms in the table. In this case, it’s a ratio, not a ranking. That doesn’t mean that the job is less or more productive, but it means that the work is more or less productive. You can have a ranked table without using a ratio, but you can do it with a ranking table. It’s also
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|
### Visual Models for the Division of Fractions
This summer I attended a workshop called Visual Models for the Division of Fractions. Moving up to 5th grade next year, there's a lot I need to re-teach myself, like dividing fractions! Eeekkk! I wanted to share a couple of examples of how to use a visual model to divide fractions. For each question, we had to draw a linear model or an area model. I chose a linear model (number line). Our district Math coach is all about number lines so I figured it would be best for me to try to "see" it that way.
Question #1:
Grandmother has three acres of land. When she divided it equally among all of her grandchildren, each got 1/2 acre. How many grandchildren does Grandmother have?
First, I drew a number line and recorded the wholes 0, 1, 2, and 3. I went up to 3 because there is 3 acres of land. Clearly, I free-handed this (which our presenter was adamant that we don't do...oops!). After my whole number were recorded, I divided them in half because the grandchildren received 1/2 acres. Then, I was able to see that there were 6 halves in 3. So, three wholes divided into halves equal six.
One thing our presenter said was to make students write the answer in word form as well as number form. Writing in word form helps you see that the child does understand the problem.
Question #2: Four students sitting at a table were given 1/3 of a pan of brownies to share. How much will each student get if they share the pan equally?
First, I drew a number line to the whole number 1 and split it into thirds since I knew I was dealing with thirds. Knowing from my problem that 1/3 was being divided into 4 parts, I used a red marker to show this. Then, I had to find out how much each small part equaled, which is 1/12. I colored in green to show what part of the number line represents my problem. Notice though that I still divided the other part of the number line into twelfths. So, if 4 students share 1/3 a pan of brownies, each student receives 1/12 piece.
Question #3:
We have 1/2 of a gallon of milk to pour equally among 3 pitchers. After pouring, how much will be in each pitcher?
First, I drew my number line to the whole number one and split into half since I have a half gallon of milk. Knowing the half gallon is split into 3 pitchers, I showed that on my number line as well. Again, note that I split it all the way to the whole. I didn't just stop at the half mark. When I did that, I determined that each spot equals 1/6. So, if we pour 1/2 gallon in 3 pitchers, each pitcher will hold 1/6 gallon.
When drawing visuals models, it really works best if you let students use colored pencils/crayons/pens. It really helped me to see it by using different colors.
Before going into this workshop, I couldn't even remember how to divide fractions. Then, I was shown the traditional version where you take one the second fraction and do the inverse and then multiply. So, here I changed the whole number 3 to 1/3 and then multiplied. Did you notice that this is the same problem as Question #3 above?
But, if I was trying to explain WHY 1/2 divided by 3 equals 1/6 and I did it the traditional way, I wouldn't be able to explain it. Therefore, I wouldn't truly understand it. Drawing visual models helps students (and teachers!) understand division so much more than using the traditional algorithm.
Now, give me some feedback! Do you teach division of fractions? If so, what are some ways you teach it? Would you like to see more sample problems other than the ones shown above?
1. Thank goodness for the sake of my kiddos I don't have to teach math this year. This look like a great idea though!! I had a colleague last year who would post things like this to her blog so parents could use the resource at home to help with HW. This would be great for that!
Lauren
The Sweetest Thing
2. Great tutorial! I'm not going to lie...5th grade math intimidates me! Ha!
Erin
Short and Sassy Teacher
1. Hey lady wanna you know
3. Great visuals I bet those will definitely help your students understand the concepts behind dividing fractions better.
Kaylee's Education Studio
4. That sounds like a great workshop! I pinned your post in the event I ever move up to fifth. What a great visual!
Meg
Third Grade in the First State
5. Please do more! I would love to make a PowerPoint presentation on this! More examples the better!
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# Substitution Rules
### A Lesson in Logic
Now that we’ve finished all of the rules of inference, it’s time to get into substitution rules. See, you can’t change a set of symbols without applying some kind of rule to do so, in order to make sure that the change actually follows from. Substitution rules are a different set of rules with a different function, and should be pretty straightforward.
Rules of inference, you’ll remember, let you draw conclusions from premises and, as long as the rules are followed and the premises are true, the conclusion has to be true. Substitution rules show that some logical statements can be exchanged for other ones without losing any of the information involved. This means that every time a premise is true, its substitution is true, every time it’s false, its substitution is false. They’re equivalent, alternate constructions of the same thing. We use them because sometimes we need one construction, and sometimes we need another. This argument will show why.
1. ~P → Q. If the pencil case isn’t red, then there is a quick brown fox on the mat
2. ~Q. There is not a quick brown fox on the mat.
3. Therefore ~~P. The pencil case isn’t not red.
This is a pretty straightforward use of Modus Tollens, where the negation of the antecedent (P) follows from the negation of the consequent (Q). But P was already negated, so the rule gives us ~~P. But we wouldn’t say “The pencil case isn’t not red”. It has the right meaning, but it’s awkward. So we apply a substitution rule, the rule of Double Negation. ~~P is equivalent to P, so we add a fourth line.
4. Therefore, P. Double Negation from 3.
Every time ~~P is true, P is true. And now, instead of saying “The pencil case isn’t not red”, we can show that it follows that “The pencil case is red.” You can have triple negations and quadruple negations as well, but the same rule applies. Removing two negations leaves you would the same meaning as if they were there. This is probably the most intuitive of the substitution rules, but there aren’t a lot, and we’ll get through them pretty quickly. Check out the whole series for more lessons in logic, and how we can use formal logic in our every day lives.
### Jim
1. I see my calculus stnetuds make these sorts of mistakes all the time. I also plug in numbers to show them that it doesn’t work, but they continually make the same mistake. The (a+b)^2 versus a^2+b^2 thing comes up at least once a month. I find that going into the theory helps a lot more, but there are still just some things that look so right. And so the kids (and the teachers sometimes!) just keep making silly mistakes.
2. Een mooi spel ja.Nu Z 4R biedt, kan N misschien alsnog 4H bieden. Na zijn passen op 3H kan dat vermoedelijk geen 4kaart zijn? Geen simpel 15 En een beetje achteraf-commentaar?Met een 4kaart H had W misschien wel 3S geboden..Maar misschien is het beter om na 3S nogmaals te doubleren. Noord zou kunnen passen en de 500 incasseren. Als 5R er in zit zal 3S vast wel (genoeg?) down gaan.Maar doublet met een singleton klaver?……Vandaar: een mooi spel!
3. How do I win my ex back who is now dating his ex?Okay, my boyfriend and I have been dating for 2 1/2 months. We’ve had problems with his ex girlfriend(which was his first love) for awhile. Anyway, last week he broke up with me because I supposedly “changed”. Which now I know was not why. He and his ex had been talking. And now they are dating. All I want is for him to be happy. But I also want to be with him. How do I win him back? I love him with all my heart.
4. Quisiera saber si el Partido Politico ESPERANZA ROSARINA sigue vigente.Desde ya agradezco su atención.Y respondió el 13 de julio de 2010 a las 16:25:Fernanda: No figura haber participado ningún partido de ese nombre ni en las elecciones de 2009 ni de 2007, lo que hace presumir que no existe
5. That’s a mold-breaker. Great thinking!
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# How to Calculate Angles
Co-authored by wikiHow Staff
Updated: December 11, 2019
In geometry, an angle is the space between 2 rays (or line segments) with the same endpoint (or vertex). The most common way to measure angles is in degrees, with a full circle measuring 360 degrees. You can calculate the measure of an angle in a polygon if you know the shape of the polygon and the measure of its other angles or, in the case of a right triangle, if you know the measures of two of its sides. Additionally, you can measure angles using a protractor or calculate an angle without a protractor using a graphing calculator.
### Method 1 of 2: Calculating Interior Angles in a Polygon
1. 1
Count the number of sides in the polygon. In order to calculate the interior angles of a polygon, you need to first determine how many sides the polygon has. Note that a polygon has the same number of sides as it has angles.[1]
• For instance, a triangle has 3 sides and 3 interior angles while a square has 4 sides and 4 interior angles.
2. 2
Find the total measure of all of the interior angles in the polygon. The formula for finding the total measure of all interior angles in a polygon is: (n – 2) x 180. In this case, n is the number of sides the polygon has. Some common polygon total angle measures are as follows:[2]
• The angles in a triangle (a 3-sided polygon) total 180 degrees.
• The angles in a quadrilateral (a 4-sided polygon) total 360 degrees.
• The angles in a pentagon (a 5-sided polygon) total 540 degrees.
• The angles in a hexagon (a 6-sided polygon) total 720 degrees.
• The angles in an octagon (an 8-sided polygon) total 1080 degrees.
3. 3
Divide the total measure of all of a regular polygon's angles by the number of its angles. A regular polygon is a polygon whose sides are all the same length and whose angles all have the same measure. For instance, the measure of each angle in an equilateral triangle is 180 ÷ 3, or 60 degrees, and the measure of each angle in a square is 360 ÷ 4, or 90 degrees.[3]
• Equilateral triangles and squares are examples of regular polygons, while the Pentagon in Washington, D.C. is an example of a regular pentagon and a stop sign is an example of a regular octagon.
4. 4
Subtract the sum of the known angles from the total measure of the angles for an irregular polygon. If your polygon doesn't have sides of the same length and angles of the same measure, all you need to do is add up all of the known angles in the polygon. Then, subtract that number from the total measure of all of the angles to find the missing angle.[4]
• For example, if you know that 4 of the angles in a pentagon measure 80, 100, 120, and 140 degrees, add the numbers together to get a sum of 440. Then, subtract this sum from the total angle measure for a pentagon, which is 540 degrees: 540 – 440 = 100 degrees. So, the missing angle is 100 degrees.
Tip: Some polygons offer “cheats” to help you figure out the measure of the unknown angle. An isosceles triangle is a triangle with 2 sides of equal length and 2 angles of equal measure. A parallelogram is a quadrilateral with opposite sides of equal lengths and angles diagonally opposite each other of equal measure.
### Method 2 of 2: Finding Angles in a Right Triangle
1. 1
Remember that every right triangle has one angle equal to 90 degrees. By definition, a right triangle will always have one angle that's 90 degrees, even if it's not labeled as such. So, you will always know at least one angle and can use trigonometry to find out the other 2 angles.[5]
2. 2
Measure the length of 2 of the triangle's sides. The longest side of a triangle is called the “hypotenuse.” The “adjacent” side is adjacent (or next to) to the angle you're trying to determine. The “opposite” side is opposite to the angle you're trying to determine. Measure 2 of the sides so you can determine the measure of the remaining angles in the triangle.[6]
Tip: You can use a graphing calculator to solve your equations or find a table online that lists the values for various sine, cosine, and tangent functions.
3. 3
Use the sine function if you know the length of the opposite side and the hypotenuse. Plug your values into the equation: sine (x) = opposite ÷ hypotenuse. Say that the length of the opposite side is 5 and the length of the hypotenuse is 10. Divide 5 by 10, which is equal to 0.5. Now you know that sine (x) = 0.5 which is the same as x = sine-1 (0.5).[7]
• If you have a graphing calculator, simply type 0.5 and press sine-1. If you don't have a graphing calculator, use an online chart to find the value. Both will show that x = 30 degrees.
4. 4
Use the cosine function if you know the length of the adjacent side and the hypotenuse. For this type of problem, use the equation: cosine (x) = adjacent ÷ hypotenuse. If the length of the adjacent side is 1.666 and the length of the hypotenuse is 2.0, divide 1.666 by 2, which is equal to 0.833. So, cosine (x) = 0.833 or x = cosine-1 (0.833).[8]
• Plug 0.833 into your graphing calculator and press cosine-1. Alternatively, look up the value in a cosine chart. The answer is 33.6 degrees.
5. 5
Use the tangent function if you know the length of the opposite side and the adjacent side. The equation for tangent functions is tangent (x) = opposite ÷ adjacent. Say you know the length of the opposite side is 75 and the length of the adjacent side is 100. Divide 75 by 100, which is 0.75. This means that tangent (x) = 0.75, which is the same as x = tangent-1 (0.75).[9]
• Find the value in a tangent chart or press 0.75 on your graphing calculator, then tangent-1. This is equal to 36.9 degrees.
## Community Q&A
Search
• Question
How do I create a 90 degree corner by swinging an arch?
Donagan
Pick a convenient point on a line to be the vertex of your 90° angle. Choose two points on the line, one on each side of the vertex and equidistant from the vertex. Use a compass to draw two arcs of the same diameter, each centered on one of those latter points. Draw a line connecting the vertex point with the intersecting point(s) of the arcs. That line describes a 90° angle with the first line.
• Question
How do I find the interior angles of a hexagon without base or height or anything?
Donagan
The sum of the six interior angles of a regular polygon is (n-2)(180°), where n is the number of sides. Therefore, in a hexagon the sum of the angles is (4)(180°) = 720°. All the angles are equal, so divide 720° by 6 to get 120°, the size of each interior angle.
• Question
How do I calculate the angle of a roof as opposed to the vertical wall it leans on?
Donagan
For a rough approximation, use a protractor to estimate the angle by holding the protractor in front of you as you view the side of the house. For the exact angle, measure the horizontal run of the roof and its vertical rise. Divide the horizontal measurement by the vertical measurement, which gives you the tangent of the angle you want. Use a trigonometry table to find the angle.
• Question
If I I have a pillow wedge that is 24" long and 12" tall, what is the degree of the wedge?
Donagan
A right triangle with legs of 24 and 12 has acute angles of 26.6° (opposite the 12 side)(the angle you're looking for), 63.4°(opposite the 24 side), and 90°.
• Question
How do I calculate if the angle is (n+11), the second angle is (4n-17), and the third angle is (5n+36)?
Donagan
If you are trying to calculate the three angles of a triangle, add together the three angles as expressed in terms of n. Set their sum equal to 180°, then solve for n. Thus, (n+11) + (4n-17) + (5n+36) = 10n + 30 = 180. So n = 15, making the angles equal to 26°, 43°, and 111°.
• Question
Translate 2 units down and 6 units to the right?
Donagan
If you're looking for the angle, use trigonometry: the angle's tangent is 6/2, or 3.
• Question
How do I calculate exterior angles?
Donagan
An exterior angle of a triangle is equal to the difference between 180° and the accompanying interior angle. Thus, if an angle of a triangle is 50°, the exterior angle at that vertex is 180° - 50° = 130°.
• Question
How can I find angles of a triangle based off of the 3 known side lengths?
Donagan
The easiest way is to construct the triangle and then use a protractor to measure the angles. If you can't use that method, you'll have to construct the triangle and do this with any angle: Drop an altitude to the opposite side, thus forming two new triangles. Measure the length of the altitude. For the two angles opposite the altitude, use the sine (opposite side divided by hypotenuse) to find the angles.
• Question
What would the angle be on a triangle that is 4" high and the base is 120" long?
Donagan
Assuming this is a right triangle and the angle you're looking for is the one opposite the 4" leg, the tangent of that angle is 0.0333. That means the angle is slightly less than 2° (about 1.9°).
• Question
I am building a solar panel to 20.5 degrees. The length of the longest side of a right-angle triangle is 3.4 meters. To achieve a 20.5 degree slope, what is the height of the triangle?
Donagan
Let x be the height of the panel. The tangent of 20.5° = 0.3739 = x / 3.4 m. x = (3.4)(.3739) = 1.27 m.
• How do I find the missing length for a right angle if I know the height and length?
• How do I calculate revolutionary angles?
• When calculating angles, when should a person subtract the angle from one-eighty or three-sixty?
200 characters left
## Tips
• Angles are given names according to how many degrees they measure. As noted above, a right angle measures 90 degrees. An angle measuring more than 0 but less than 90 degrees is an acute angle. An angle measuring more than 90 but less than 180 degrees is an obtuse angle. An angle measuring 180 degrees is a straight angle, while an angle measuring more than 180 degrees is a reflex angle.
Thanks!
• Two angles whose measures add up to 90 degrees are called complementary angles. (The two angles other than the right angle in a right triangle are complementary angles.) Two angles whose measures add up to 180 degrees are called supplementary angles.
Thanks!
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 9 references.
Co-authors: 10
Updated: December 11, 2019
Views: 211,480
Article SummaryX
To calculate angles in a polygon, first learn what your angles add up to when summed, like 180 degrees in a triangle or 360 degrees in a quadrilateral. Once you know what the angles add up to, add together the angles you know, then subtract the answer from the total measures of the angles for your shape. For example, add 60 and 80 to get 140 for 2 angles in a triangle, then deduct 140 from 180 to work out the third angle in the triangle, which will be 40 degrees. To find out how to calculate angle measure in a right triangle, read on!
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# A Problem in Pentagon with Supplementary Angles
### Solution
Much of the proof is based on angles chasing.
Join $D\,$ to $A,K,B.$
Since inscribed angles subtended by the same arc are equal,
\displaystyle\begin{align} &\angle BKC=\angle BDC=\frac{1}{2}(180^{\circ}-\angle BCD),\\ &\angle CKD=\angle CBD=\frac{1}{2}(180^{\circ}-\angle BCD),\\ &\angle AKE=\angle ADE=\frac{1}{2}(180^{\circ}-\angle AED),\\ &\angle DKE=\angle DAE=\frac{1}{2}(180^{\circ}-\angle AED),\\ \end{align}
\displaystyle\begin{align} \angle BKC+\angle CKD+\angle AKE+\angle DKE&=2\cdot 180^{\circ}-(\angle BCD+\angle AED)\\ &=2\cdot 180^{\circ}-180^{\circ}=180^{\circ}, \end{align}
implying that $AKB\,$ is a straight line. This answers #1. Along the way, we obtain
$\displaystyle\angle BKC+\angle AKE=180^{\circ}-\frac{1}{2}(\angle BCD+\angle AED)=\frac{180^{\circ}}{2},$
implying $\angle CKE=90^{\circ},\,$ i.e. #2.
To prove #3, we mimic Solution 2 from an earlier page. Consider a sequence of rotations: around $E\,$ through $\angle AED,\,$ around $C\,$ through $\angle DCB,\,$ through $180^{\circ}\,$ around $H.\,$ The result is either a translation or a rotation which lives $A\,$ fixed. It follows that the composition of the three rotations is the identity transform.
Let's trace $E:\,$ the first rotation leaves it stationary, the second moves it to some $E',\,$ the third takes $E'\,$ back to $E.\,$ It follows that $H\,$ is the midpoint of the base of the isosceles $\Delta ECE',\,$ such that $CH\perp EE',\,$ i.e., $\angle CHE=90^{\circ}.$
Lastly, $\angle CKF=\angle CKE=90^{\circ}.\,$ This makes $CF\,$ a diameter of $(BCD)\,$ and, since $B\,$ is the midpoint of one arc $\overset{\frown}{BD},\,$ $F\,$ is the midpoint of the other. Thus $CF\,$ is the bisector of $\angle BCD.$
Point $G\,$ is treated similarly. This proves #4. Join now $BE'.\,$ Since $H\,$ is the midpoint of both $EE'$ and $AB,\,$ $AEBE'\,$ is a parallelogram, so that $BE'=AE=DE.\,$
It follows that $\Delta BE'H=\Delta AEH\,$ and $\Delta BCE'=\Delta DCE.\,$ As we can see,
\begin{align}[\Delta ECH] &= [\Delta E'CH]\\ &=[\Delta CBK]+[\Delta BE'H]+[\Delta BCE']\\ &=[\Delta CBK]+[\Delta AEH]+[\Delta DCE], \end{align}
which implies #5.
### Acknowledgment
The problem due Thanos Kalogerakis is a generalization of an earlier one. The problem was posted at the Peru Geometrico and, as #1082, Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook groups.
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# Introduction to Linear Regression - UTEP MATHEMATICS
Introduction to Linear Regression You have seen how to find the equation of a line that connects two points. You have seen how to find the equation of a line that connects two points. Often, we have more than two data points, and usually the data points do not all lie on a single line.
You have seen how to find the equation of a line that connects two points. Often, we have more than two data points, and usually the data points do not all lie on a single line. It is possible to find the equation of a line that most closely fits a set of data points. Such a line is called a regression line or a linear regression equation. You have seen how to find the equation of a
line that connects two points. Often, we have more than two data points, and usually the data points do not all lie on a single line. It is possible to find the equation of a line that most closely fits a set of data points. Such a line is called a regression line or a linear regression equation. Our goal here is to learn what a regression line is. You can then watch the presentation on how to find the equation of a regression
Consider the following table that the average price of a two-bedroom apartment in downtown New York City from 1994 to 2004, where t=0 represents 1994. Consider the following table that the average price of a two-bedroom apartment in downtown New York City from 1994 to 2004, where t=0 represents 1994. Consider the following table that the average price of a two-bedroom apartment in downtown New York City from 1994 to 2004,
where t=0 represents 1994. We can plot each of these data points on a graph. Each point is of the form (t, p), so we have 6 points to plot. Consider the following table that the average price of a two-bedroom apartment in downtown New York City from 1994 to 2004, where t=0 represents 1994. We can plot each of these data points on a graph. Each point is of the form (t, p), so we have 6 points to plot.
They are (0, 0.38), (2, 0.40), (4, 0.60), (6, 0.95), (8, 1.20), and (10, 1.60). Just looking at them like this doesnt give much indication of a pattern, although we can see that the p- When we plot the points all together on a set of axes, we get the following scatter plot: 1.8 Price p in millions of \$ 1.6 1.4 1.2 1 0.8
0.6 0.4 0.2 0 0 2 4 6 8 Time t in years since 1994 10 12
When we plot the points all together on a set of axes, we get the following scatter plot: Price p in millions of \$ 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2
4 6 8 10 12 Time t in years since 1994 It seems that the data do follow a somewhat linear pattern. We can find the line the line that most closely fits the equation and graph it over the data points.
1.8 Price p in millions of \$ 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4
6 8 Time t in years since 1994 10 12 We can find the line the line that most closely fits the equation and graph it over the data points. 1.8 Price p in millions of \$
1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 Time t in years since 1994
10 12 We can find the line the line that most closely fits the equation and graph it over the data points. Price p in millions of \$ 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4
0.2 0 0 2 4 6 8 10 12 Time t in years since 1994
Notice that the line does not go through all of the data points. We can also find the equation of this line of best fit. 1.8 Price p in millions of \$ 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
0 2 4 6 8 Time t in years since 1994 10 12 We can also find the equation of this line of best fit. 1.8 Price p in millions of \$
1.6 1.4 f(x) = 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6
8 Time t in years since 1994 10 12 We can also find the equation of this line of best fit. We can also get whats called the correlation coefficient. 1.8 Price p in millions of \$
1.6 1.4 f(x) = 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
Time t in years since 1994 10 12 We can also find the equation of this line of best fit. We can also get whats called the correlation coefficient. 1.8 Price p in millions of \$ 1.6
1.4 R f(x)==0.95 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
Time t in years since 1994 10 12 1.8 Price p in millions of \$ We can also find the equation of this line of best fit. We can also get whats called the correlation coefficient. 1.6
1.4 R f(x)==0.95 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
10 12 Time t in years since 1994 You will be able to do all of this on Excel once you watch the instructional video and read the PDFs for this material. For now, we just want to get an idea of what the regression line is and what the correlation coefficient tells us What does the regression equation tell us about the relationship between time and sale price? 1.8 Price p in millions of \$
1.6 1.4 R f(x)==0.95 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4
6 8 Time t in years since 1994 10 12 What does the regression equation tell us about the relationship between time and sale price? Price p in millions of \$ 1.8 1.6
1.4 R f(x)==0.95 0.13 x + 0.22 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
10 12 Time t in years since 1994 The slope and the vertical intercept (usually the y-intercept, here the p-intercept) tell us different things. In this case, the p-intercept tells us what the sale price is predicted to be when t=0 (that is, in the year 1994).
In this case, the p-intercept tells us what the sale price is predicted to be when t=0 (that is, in the year 1994). The regression equation is p=0.1264t+0.2229. Recall that price is in millions of dollars. In this case, the p-intercept tells us what the sale price is predicted to be when t=0 (that is, in the year 1994). The regression equation is p=0.1264t+0.2229. Recall that price is in
millions of dollars. Thus, if t=0, the regression equation predicts a price of \$0.2229 million or \$222,900. In this case, the p-intercept tells us what the sale price is predicted to be when t=0 (that is, in the year 1994). The regression equation is p=0.1264t+0.2229. Recall that price is in millions of dollars. Thus, if t=0, the regression equation predicts a price of \$0.2229 million or \$222,900.
According to the table, the actual price was \$0.38 million or \$380,000. These values dont have to be the same however, since the regression equation cant match every point exactly. It is only a model that most closely fits the data points. What does the slope of the regression equation tell us? What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264.
What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264. We can always write a number x as x divided by 1, so we can write this slope as .
What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264. We can always write a number x as x divided by 1, so we can write this slope as . Recall that the definition of slope is .
What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264. We can always write a number x as x divided by 1, so we can write this slope as . Recall that the definition of slope is . In this case we are using p and t, so its .
What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264. We can always write a number x as x divided by 1, so we can write this slope as . Recall that the definition of slope is . In this case we are using p and t, so its
. So for our problem, we have . What does the slope of the regression equation tell us? The slope of our regression equation is 0.1264. We can always write a number x as x divided
by 1, so we can write this slope as . Recall that the definition of slope is . In this case we are using p and t, so its . So for our problem, we have . For this problem, t is measure in years and p is measured in millions of dollars.
For this problem, t is measure in years and p is measured in millions of dollars. So more specifically, the slope can be interpreted to mean that if t increases by 1 year, the model predicts that the average price p of a two-bedroom apartment will increase by about \$0.1264 million dollars, or \$126,400.
For this problem, t is measure in years and p is measured in millions of dollars. So more specifically, the slope can be interpreted to mean that if t increases by 1 year, the model predicts that the average price p of a two-bedroom apartment will increase by about \$0.1264 million dollars, or \$126,400. Even more plainly, we can say that the model predicts that the average price of a twobedroom apartment in New York City will increase by about \$126,400 per year.
For this problem, t is measure in years and p is measured in millions of dollars. So more specifically, the slope can be interpreted to mean that if t increases by 1 year, the model predicts that the average price p of a two-bedroom apartment will increase by about \$0.1264 million dollars, or \$126,400. Even more plainly, we can say that the model predicts that the average price of a twobedroom apartment in New York City will increase by about \$126,400 per year. We can now use the linear regression model to predict future prices. For example, if we wanted to predict what the price of an apartment was in 2008, we could plug in 14 Plugging in 14 for t into the regression equation gives
p=0.1264(14)+0.2229=1.9925. Plugging in 14 for t into the regression equation gives p=0.1264(14)+0.2229=1.9925. This means that if the trend continued, we can expect that the price of a two-bedroom apartment was around \$1,992,500 in 2008. Plugging in 14 for t into the regression
equation gives p=0.1264(14)+0.2229=1.9925. This means that if the trend continued, we can expect that the price of a two-bedroom apartment was around \$1,992,500 in 2008. You can also use the regression equation to check how closely the model matches the actual price in some years that were given on the table. For example, for 2000 the equation predicts a price of p=0.1264(6)+0.2229=0.9813, or \$981,300.
Plugging in 14 for t into the regression equation gives p=0.1264(14)+0.2229=1.9925. This means that if the trend continued, we can expect that the price of a two-bedroom apartment was around \$1,992,500 in 2008. You can also use the regression equation to check how closely the model matches the actual price in some years that were given on the table. For example, for 2000 the equation predicts a price of p=0.1264(6)+0.2229=0.9813, or \$981,300. According to the table, the actual price was \$950,000, so the regression equation is pretty close. It is important to remember that the regression equation is just a model, and it
wont give the exact values. It is important to remember that the regression equation is just a model, and it wont give the exact values. If the equation is a good fit to the data however, it will give a very good approximation, so it can be used to forecast what may happen in the future if the current trend continues.
It is important to remember that the regression equation is just a model, and it wont give the exact values. If the equation is a good fit to the data however, it will give a very good approximation, so it can be used to forecast what may happen in the future if the current trend continues. Next, lets take a quick look at how a regression equation is derived, and then take a look at what the correlation coefficient (or the r-squared value on Excel) tell us about the regression equation. Lets take another look at the data points and the regression line. 1.8 Price p in millions of \$
1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
Time t in years since 1994 10 12 Lets take another look at the data points and the regression line. Price p in millions of \$ 1.8 1.6 1.4 1.2 1 0.8 0.6
0.4 0.2 0 0 2 4 6 8 10 12 Time t in years since 1994
Why does this particular line give the best fit for the data? Why not some other line? It has to do with what is called a residual. 1.8 Price p in millions of \$ 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0
2 4 6 8 Time t in years since 1994 10 12 It has to do with what is called a residual. Price p in millions of \$
1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8
10 12 Time t in years since 1994 A residual is the difference between a particular data point and the regression line. If we zoom in on a particular data point, we can see what a residual is. 1.8 Price p in millions of \$ 1.6
1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 Time t in years since 1994
10 12 If we zoom in on a particular data point, we can see what a residual is. Lets zoom in on this particular data point. Zooming into this box:
Zooming into this box: We see the data point and the line. Zooming into this box: We see the data point and the line. The vertical distance between the line and the data point is the residual. Zooming into this box:
We see the data point and the line. The vertical distance between the line and the data point is the residual. Zooming into this box: We see the data point and the line. The vertical distance between the line and the data point is the residual.
Zooming into this box: We see the data point and the line. The vertical distance between the line and the data point is the residual. The idea behind linear regression is to keep the residuals as small as possible. There is a method that allows us to minimize the sum of all of the residuals.
There is a method that allows us to minimize the sum of all of the residuals. This is called the least-squares method. You can read about it in the PDF for linear regression. There is a method that allows us to minimize the sum of all of the residuals. This is called the least-squares method. You can read about it in the PDF for linear regression. Since these formulas can get fairly complicated, you will not be required to use them in the course.
There is a method that allows us to minimize the sum of all of the residuals. This is called the least-squares method. You can read about it in the PDF for linear regression. Since these formulas can get fairly complicated, you will not be required to use them in the course. You will only need to know how to find a regression line using Excel. You can watch the video on how to do this, or read through the
PDF, or both. There is a method that allows us to minimize the sum of all of the residuals. This is called the least-squares method. You can read about it in the PDF for linear regression. Since these formulas can get fairly complicated, you will not be required to use them in the course.
You will only need to know how to find a regression line using Excel. You can watch the video on how to do this, or read through the PDF, or both. Next, we look at what the correlation coefficient tells us about the regression equation. Recall that in our graph, a number was given, called the correlation coefficient, denoted by the letter r. Recall that in our graph, a number was given, called the correlation coefficient, denoted by
the letter r. The correlation coefficient tells us how closely the regression line fits the data points. Recall that in our graph, a number was given, called the correlation coefficient, denoted by the letter r. The correlation coefficient tells us how closely the regression line fits the data points. It has a value between -1 and 1. A value very close to 1 indicates a very good fit with a positive sloping linear function.
Recall that in our graph, a number was given, called the correlation coefficient, denoted by the letter r. The correlation coefficient tells us how closely the regression line fits the data points. It has a value between -1 and 1. A value very close to 1 indicates a very good fit with a positive sloping linear function. A value very close to -1 indicates a very good fit with a negative sloping linear function.
Recall that in our graph, a number was given, called the correlation coefficient, denoted by the letter r. The correlation coefficient tells us how closely the regression line fits the data points. It has a value between -1 and 1. A value very close to 1 indicates a very good fit with a positive sloping linear function. A value very close to -1 indicates a very good fit with a negative sloping linear function. A value very close to 0 indicates a very poor fit with the data, so there will be no linear
relationship between variables in this case. Excel will not give the value of r, instead it gives the value of r squared. Excel will not give the value of r, instead it gives the value of r squared. The r-squared value basically tells us the same thing, but it will only be between 0 and 1.
Excel will not give the value of r, instead it gives the value of r squared. The r-squared value basically tells us the same thing, but it will only be between 0 and 1. If the r-squared value is close to 1, there is a very good linear fit for the data points. Excel will not give the value of r, instead it
gives the value of r squared. The r-squared value basically tells us the same thing, but it will only be between 0 and 1. If the r-squared value is close to 1, there is a very good linear fit for the data points. If the r-squared value is close to 0, there is a very poor fit between the data points. Excel will not give the value of r, instead it
gives the value of r squared. The r-squared value basically tells us the same thing, but it will only be between 0 and 1. If the r-squared value is close to 1, there is a very good linear fit for the data points. If the r-squared value is close to 0, there is a very poor fit between the data points. We will now look at some examples of what it looks like with an r-squared value close to 1 and with an r-squared value close to 0. Consider the following set of data points. Consider the following set of data points. 8
7 6 5 4 3 2 1 0 0 2 4 6 8 10
12 Consider the following set of data points. 8 7 6 5 4 3 2 1 0 0 2
4 6 8 10 12 They follow a clear linear pattern, so we should expect the r-squared value to be close to 1. Consider the following set of data points. 8 7 6
f(x) = 0.51 x + 1.94 R = 0.99 5 4 3 2 1 0 0 2 4
6 8 10 12 They follow a clear linear pattern, so we should expect the r-squared value to be close to 1. And it is. Now consider the following set of data points. Now consider the following set of data points.
20 18 16 14 12 10 8 6 4 2 0 0 2 4 6 8
10 12 Now consider the following set of data points. 20 18 16 14 12 10 8 6 4 2 0 0
2 4 6 8 10 12 These points seem to be scattered everywhere and dont follow any linear pattern.
Now consider the following set of data points. 20 18 16 14 12 10 8 6 4 2 0 0 2
4 6 8 10 12 These points seem to be scattered everywhere and dont follow any linear pattern. We expect the r-squared value to be close to 0. Now consider the following set of data points.
20 18 16 14 12 10 8 6 f(x) = 0.18 x + 8.33 R = 0.01 4 2 0 0
2 4 6 8 10 12 These points seem to be scattered everywhere and dont follow any linear pattern. We expect the r-squared value to be close to 0.
And it is. So, to summarize, a linear regression equation is a line that most closely fits a given set of data points. So, to summarize, a linear regression equation is a line that most closely fits a given set of data points. The regression equation can be used to predict future values, or values that are outside of the given data range.
So, to summarize, a linear regression equation is a line that most closely fits a given set of data points. The regression equation can be used to predict future values, or values that are outside of the given data range. We can find regression equation for any set of data points, no matter how scattered the data look, but we can tell how closely the data follow a linear pattern by looking at the rsquared value.
So, to summarize, a linear regression equation is a line that most closely fits a given set of data points. The regression equation can be used to predict future values, or values that are outside of the given data range. We can find regression equation for any set of data points, no matter how scattered the data look, but we can tell how closely the data follow a linear pattern by looking at the rsquared value. An r-squared value close to 1 indicates a very good fit to the given data, and an r-squared value close to zero indicates a very poor fit to the data.
The topic of linear regression is very deep, and we have only given a very brief introduction to it here. The topic of linear regression is very deep, and we have only given a very brief introduction to it here. You can read more about it in the PDF given on the Assigned Reading for section 1.4.
The topic of linear regression is very deep, and we have only given a very brief introduction to it here. You can read more about it in the PDF given on the Assigned Reading for section 1.4. Be sure you also watch the video about how to find a linear regression on Excel! You can find the video link in the Assigned Reading for section 1.4.
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The exponent of a number says how many times to use the number in a multiplication.
You are watching: What is 4 to the 6th power
In 82 the "2" says to use 8 twice in a multiplication,so 82 = 8 × 8 = 64
In words: 82 could be called "8 to the power 2" or "8 to the second power", or simply "8 squared"
Exponents are also called Powers or Indices.
Some more examples:
### Example: 53 = 5 × 5 × 5 = 125
In words: 53 could be called "5 to the third power", "5 to the power 3" or simply "5 cubed"
### Example: 24 = 2 × 2 × 2 × 2 = 16
In words: 24 could be called "2 to the fourth power" or "2 to the power 4" or simply "2 to the 4th"
So in general:
an tells you to multiply a by itself,so there are n of those a"s:
## Another Way of Writing It
Sometimes people use the ^ symbol (above the 6 on your keyboard), as it is easy to type.
## Negative Exponents
Negative? What could be the opposite of multiplying? Dividing!
So we divide by the number each time, which is the same as multiplying by 1number
## Negative? Flip the Positive!
That last example showed an easier way to handle negative exponents: Calculate the positive exponent (an)
More Examples:
Negative Exponent Reciprocal ofPositive Exponent Answer
4-2 = 1 / 42 = 1/16 = 0.0625
10-3 = 1 / 103 = 1/1,000 = 0.001
(-2)-3 = 1 / (-2)3 = 1/(-8) = -0.125
## What if the Exponent is 1, or 0?
1 If the exponent is 1, then you just have the number itself (example 91 = 9) 0 If the exponent is 0, then you get 1 (example 90 = 1) But what about 00 ? It could be either 1 or 0, and so people say it is "indeterminate".
## It All Makes Sense
If you look at that table, you will see that positive, zero ornegative exponents are really part of the same (fairly simple) pattern:
Example: Powers of 5
.. etc..
See more: How To End A Program In C Library Function, End, Stop And Null Statements
52 5 × 5 25
51 5 5
50 1 1
5-1 15 0.2
5-2 15 × 15 0.04
.. etc..
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## Complementary and Supplementary Angles Definition
Complementary and supplementary angles are the two types of angles based on their measures. An angle that measures 900 is a complementary angle and whose measure is 1800 is a supplementary angle.
We have studied basic geometrical terms. We know that two lines having a common vertex form an angle. Various parts of an angle are vertex, arms, interior, and exterior of angles. When two lines intersect they form four angles at the point of intersection. An angle is denoted by the symbol ∠.
From the figure, ∠ABC is an angle. B is the point of intersection called the vertex and AB and BC are the sides of the angle. Angles are commonly measured in terms of degree.
Angles are Classified According to Their Sizes as Follows
• Acute angle
• Obtuse angle
• Right angle
• Reflex angle
There are more types of angles on the basis of the pairs of angles. They are as follows
• Complementary angles
• Supplementary angles
• Vertically opposite angles
• Linear Pairs
In this session, we will be learning complementary and supplementary angles in detail.
### Complementary Angles Definition
When the sum of the measure of two angles is 900, then the pair of angles is said to be complementary angles. In complementary angles one angle is a complement of the other making a sum of 900 or you can say forming a right angle.
From the figure, we can say that ∠ABC + ∠CBD = 50 + 40 = 900. This is an example of complementary angles example
But it is not necessary that the two complementary angles are always adjacent to each other. They can be different angles, only their sum should be 900
Figure given below is a complementary angle example 27 + 63 = 900.
### Supplementary Angles Definition
When the sum of the measure of two angles is 1800, then the pair of angles is said to be supplementary angles. Here the supplementary meaning is one angle is supplemented to another angle to make a sum of 1800.
Supplementary angles example
From the figure, we can say that ∠1 + ∠2 = 1800
But it is not necessary that the two supplementary angles are always adjacent to each other. They can be different angles, only their sum should be 1800
From the figure, ∠3 and ∠4 are not adjacent but they make a supplementary angle if their sum is 1800.
Supplementary angles example
### Properties of Supplementary Angles
• According to supplementary angles definition two angles are said to be a supplementary angle if the sum of their measures is 1800.
• It is not necessary that a supplementary angle will lie on the same line, they can be on different lines but should measure 1800.
• In the supplementary angles if one is an acute angle then the other is an obtuse angle.
• In a supplementary angle if one angle is 900 the other angle will also be 900.
### Properties of Complementary Angles
• According to supplementary angles definition two angles are said to be a complementary angle if the sum of their measures is 900.
• It is not necessary that a supplementary angle will lie on the same line, they can be on different lines but should measure 900.
How to Find Two Angles are Complementary?
If measures of two angles are given. Add them. If the sum of the measures of these two angles is 900, then the two angles are complementary angles.
If the two angles are given as complementary angles and if the measure of one angle is given we can find the other angle.
For example : ∠A + ∠B = 90° and ∠A = 300 , then ∠B = ?
∠B = 90 – ∠A
∠B = 90 – 30
∠B = 60
The formula for finding a complementary angle is 90 – x, where x is the measure of one of the angles.
How to Find Two Angles are Supplementary?
If measures of two angles are given. Add them. If the sum of the measures of these two angles is 1800, then the two angles are supplementary angles.
If the two angles are given as supplementary angles and if the measure of one angle is given we can find the other angle.
For example : ∠A + ∠B = 180° and ∠A = 800 , then ∠B = ?
∠B = 180 – ∠A
∠B = 180 – 80
∠B = 1000
The formula for finding a supplementary angle is 180 – x, where x is the measure of one of the angles.
### Fun Facts
• In a right-angled triangle, the two non-right angles are complementary angles to each other.
• Complementary comes from the Latin word ‘completum’ meaning “completed” because the right angle is thought of as being a complete angle.
• Supplement comes from Latin word ‘supplere’, to complete or “supply” what is needed.
### Solved Examples
Example 1: Two angles are supplementary. One of these two angles is 1100 and finds the other angle.
Solution: One angle is given 1100
Let the other angle be x
Given that the two angles are supplementary we have,
The sum of the measures of these two angles is 1800
x + 110 = 1800
x = 180 – 110
x = 700
So the other angle is 700
Example 2: The measure of an angle is 72°. What is the measure of a complementary angle?
Solution: Let x be the measure of the complementary angle.
Because x and 62° are complementary angles, we have
x + 62° = 90°
x = 90 – 72
x = 18
So, the measure of the complementary angle is 18°.
Example 3: Find the Supplement of the angle 1/3 of 240°.
Solution:
Convert 1/3 of 240°
That is, 1/3 x 240° = 80°
The sum of the measures of these two angles is 1800
x + 80 = 1800
x = 180 – 80
x = 1000
Therefore, Supplement of the angle 1/3 of 240° is 100°
### Quiz Time
Supplementary Angles Examples
1. The measure of an angle is 100°. What is the measure of a supplementary angle?
2. .Find the value of x
Complementary Angles Examples
1. The measure of an angle is 22°. What is the measure of a complementary angle?
2. Find the value of x
What is the difference between Complementary and Supplementary angles?
Here are some differences between difference between Complementary
and Supplementary angles
• According to supplementary angle definition If the sum of the measure of two angles is 1800 then the angles are said to be supplementary angles, whereas according to complementary angles definition if the sum of measures of two angles is 900 then the angles are called complementary angles.
• Supplementary angles together form a straight line whereas complementary angles together form a right angle.
• Two angles ∠A and ∠B are supplementary angles if ∠A + ∠B = 1800, while two angles ∠A and ∠B are complementary if ∠A + ∠B = 900
• Example of supplementary angle
∠1 + ∠2 = 1800
Example of Complementary angle
∠1 + ∠2 = 900
What is another name for supplementary angles?
Another name for supplementary angle is straight angle. Because according to supplementary angle definition two angles are supplementary if their sum adds up to 1800. Similarly the straight angle measures 1800. They both make a complete half turn.
What is zero angle?
As the name implies, an angle that measures 0o is called a zero angle. You can assume two sides of an angle overlaps each other.
|
# Factors of 50
Factors of 50 are basically the numbers that divide it evenly or exactly without leaving any remainder i.e if a number divides 50 with a remainder of zero, then that number is called a factor.
• All Factors: 1, 2, 5, 10, 25, and 50
• Factors in pairs : (1,50), (2,25), and (5,10)
• Prime factors : 2, 5
Now, let us see the prime factors and prime factorization of the number.
Index
## Prime Factorization of 50
Prime factorization is a method of “expressing” or finding the given number as the product of prime numbers. If a number occurs more than once in prime factorization, it is usually expressed in exponential form to make it more compact.
Here Prime factorization comes out to be 2 x 5 x 5
### Prime Factorization of 50 by Upside-Down Division Method
Upside-Down Division is one of the techniques used in the Prime factorization method to find factors of numbers.
In this method, you will divide a given “composite” number evenly by the several prime numbers(starting from the smallest) till it gets a prime number.
It is called Upside-Down Division because the symbol is flipped upside down.
Here, 50 is an even number. So it is undoubtedly divisible by 2 with no remainder.
Thus we do 50÷ 2 = 25. Now find the prime factors of the obtained quotient.
Repeat Step 1 and Step 2 until we get a result of prime number as the quotient.
Here, 25 is the quotient. Now find the prime factor of 25
25÷ 5 = 5. Here, 5 is the prime number.
So we can stop the process.
So, now the prime factorization of 50 with the upside-down division method is 2 × 5 × 5.
## Prime Factorization of 50 by Factor Tree Method
The Factor tree method is another technique for producing the prime factorization and all factors of a given number.
To use this method for a number x,
Firstly consider two factors say a,b of x such that a*b is equal to x and at least one of them (a, b) is a prime factor say a.
Then consider two factors of b say c, d such that again at least one of them is a prime factor. This process is repeated until both the factors are prime i.e if we get both the factors as prime at any step, we stop the process there.
Following is the factor tree of the given number.
So here also we get the prime factorisation as 2 * 5 * 5 and the two prime factors are 2, 5
## FAQs
What are the factors of 50?
All the factors are: 1, 2, 5, 10, 25, and 50
What are the prime factors of 50?
Prime factors are: 2 and 5
What is the sum of factors of 50?
The Sum of factors of 50 is 93 which is found by (2(1+1) – 1)/(2-1) * (5(2+1) – 1)/(5-1) = 3 * 124/4 = 93
More factors:
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# Number Properties You Must Know
Now that you have taken the first step in learning about the properties of numbers, you might be curious about the properties of numbers and number properties and the different numbers that exist. These are called number properties. These are all the distributive, associative and commutative properties.
Factoring Trinomials. Factoring trinoms, also known as factoring with trigonometry, are used to multiply a number and another number together. In factoring, each side of the equation is multiplied by the other side. Finding the average of both sides.
Addition by Multiplication. The answer of an add-and-subtract is that it is an equation which can be solved. It is also an equation that uses one number, or its addition, and another number, or its addition and subtraction. This can be combined with the previous properties to find the average. For example, a number, A, is multiplied by another number, B.
Factoring with Trinomials. To learn how factoring works, you must first know that factoring with trigonometry involves addition by multiplication. That is, two numbers are added together and their difference is subtracted from the sum. Once you know this you can go ahead and do factoring.
Division by Multiple Factors. The division of a number, say A, by another number, say B, is the division of A by B. Once you know this you can figure out the average by taking the difference and dividing by A.
Common addition. The common addition of two numbers is that they are added together. You can multiply both sides together by the common factor. This can be used to find the mean of both sides.
Common subtraction. The common subtraction of two numbers is that they are subtracted from a number. You can take the difference of both sides and divide by the difference. This can be used to find the difference between any two numbers. It can also be used to find the mean of the difference between two numbers and another number.
Fraction. Fraction is a way of dividing an equation. Divide a number by a second number. Divide the number by its denominator. Divide both sides together by the numerator.
Integral number. This is a number that is both constant and varies over time. In other words, it cannot be changed at all. If you multiply a constant number by a constant number, then it will always remain constant. A variation in the value of an integral number will cause the number to change as well.
Numerator. This is the number of times the numerator equals the denominator. The denominator is the number of times the denominator equals the numerator. When these two numbers are equal, this is a zero.
Denominator. This is the number of times the denominator is equal to the numerator. When the numerator and denominator are equal, this is one. When the numerator is equal to the denominator, it is also called a minus sign.
Multiplication and Division. When you multiply both a numerator and denominator together, then you get the product of both. In addition, by division you are multiplying both a numerator and denominator together.
Subdivision by Zero. Subdivision by zero is the ability to divide a number into parts without losing any information about its parts. In addition by subtraction you can divide a number without losing any information about its parts. By subtraction, you can divide a number by a smaller number.
Addition by Zero. Adding to two negative numbers can get you an addition with the zero sign.
Subtraction by Zero. Subtracting from a positive number can get you a subtraction with the zero sign. Addition by subtraction can get you a positive addition.
Division by Zero. Division by zero is the ability to divide a number in half without losing any information about its parts. This can be used to get a division of an unknown number into parts.
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# Practice English Speaking&Listening with: Math Antics - Percents Missing Total
Normal
(0)
Difficulty: 0
Hi! Welcome to Math Antics. In our last lesson about percents,
we learned that therere three main types of percent problems because theres three different numbers that could be missing.
Those three numbers are thepart’, thetotal’, and thepercent’.
Theyre just the three variable (or changeable numbers) in the percentage equation.
The forth number is always 100, since thats what percent means: per 100.
In the last two videos, we learned how to solve problems where thepartwas unknown and where thepercentwas unknown.
In this video, were gonna learn how to solve problems where thetotalis unknown or missing.
With this type of problem, youll be told what the percent is,
and youll be told what part of the total you have.
But youll need to figure out what that total itself is.
Heres an example of a problem like that:
Your friend has a bag of marbles, and he tells you that 20% of the marbles are red.
If theres 7 red marbles,
how many marbles does he have altogether?
Okay, so how do you know that its the total thats missing in this problem?
Well, the wordaltogetheris a big clue because it means almost the same thing astotal”.
So, if the question has words like, “altogether”, orin all”, ortotal”, orwhole”, orentire
those can help you know that you need to find the total.
And another way that we can tell is by the numbers that we ARE given.
In this problem, we know that the percent is 20, and were also told that PART of the marbles are red,
so we know that the PART is 7.
So that means that it must be the total thats missing!
Alright then, so how do we figure out what the total is?
Well, using a little algebra (which you dont need to know how to do right here)
we can re-arrange our percent equation like this:
What this new form of the equation tells us is that,
if we take thepartand multiply it by 100, and then we divide that by thepercent’, well get thetotal’.
That seems simple enough. Its just two steps!
Lets try it out on our word problem about the marbles.
We know that thepart’ (that are red) is 7,
so step one is to just multiply that part by 100.
7 × 100 is 700.
And in step two, we take that 700 and divide it by thepercent’, which were told is 20.
Okay, 700 divided by 20…. hmmm….
Well, we could use a calculator to divide, but this doesnt seem too hard,
so Ill just do the division the long way.
20 is to big to divide into the first digit (7) so well need to include the digit next to it as well.
Now we ask, “How many20s does it take to make 70 or almost 70”.
That would be 3 because 3 × 20 is 60.
70 minus 60 leaves 10 as the remainder.
And then we bring down the zero and then we ask, “How many20s will divide into 100?”
Ah-ha5, because 5 × 20 is 100, so that leaves no remainder.
So 700 divided by 20 is 35.
And that means that the total number of marbles is 35.
And in a problem like this, you can always check your answer
by making sure that the fraction of thepartover thetotalwould give you the correct percent.
For example, in this case, you could make sure that the fraction 7 over 35 would really be 20%.
Now that wasnt so tough, was it?
Lets see one more example to make sure youve got the procedure down before you try some on your own.
The next problem says:
A high school marching band has 12 flute players.
[frantic flute music]
If 8% of the band members play the flute,
then how many members are in the entire band?
Okay, so the smallerpartin this problem is 12 since theres 12 flute players.
And were told that they make up 8 percent of the band, so thepercentis 8.
Again, its thetotalthats missing,
and to find it, we just need to follow our 2-step procedure.
For step one, we multiply thepartby 100:
12 × 100 = 1,200
For step two, we divide that 1,200 by the percent, which is 8.
(This time I think Ill use a calculator to divide.)
1,200 divided by 8 equals 150.
Greatthat means that the total number of band members is 150.
And again, you can always check your answer the way we did in the last example.
Alrightthat does it for this lesson.
Rememberthe key to getting really good at math is to do it yourself.
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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10
# SSLC (English Medium) Class 10 - Tamil Nadu Board of Secondary Education Question Bank Solutions for Mathematics
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Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A
[0.01] Relations and Functions
Chapter: [0.01] Relations and Functions
Concept: Ordered Pair
If B × A = {(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} find A and B
[0.01] Relations and Functions
Chapter: [0.01] Relations and Functions
Concept: Ordered Pair
Multiple Choice Question :
If the ordered pairs (a + 2, 4) and (5, 2a + b) are equal then (a, b) is
[0.01] Relations and Functions
Chapter: [0.01] Relations and Functions
Concept: Ordered Pair
If the ordered pairs (x2 – 3x, y2 + 4y) and (– 2, 5) are equal, then find x and y
[0.01] Relations and Functions
Chapter: [0.01] Relations and Functions
Concept: Ordered Pair
Find all positive integers, when divided by 3 leaves remainder 2
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Prove that the product of two consecutive positive integers is divisible by 2
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Prove that two consecutive positive integers are always co-prime
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Prove that n2 – n divisible by 2 for every positive integer n
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
When the positive integers a, b and c are divided by 13 the respective remainders is 9, 7 and 10. Find the remainder when a b + + 2 3c is divided by 13
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Show that 107 is of the form 4q +3 for any integer q
[0.02] Numbers and Sequences
Chapter: [0.02] Numbers and Sequences
Concept: Euclid’s Division Lemma
Solve the following system of linear equations in three variables
x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16
[0.03] Algebra
Chapter: [0.03] Algebra
Concept: Simultaneous Linear Equations in Three Variables
Solve the following system of linear equations in three variables
1/x - 2/y + 4 = 0; 1/y - 1/z + 1 = 0; 2/z + 3/x = 14
[0.03] Algebra
Chapter: [0.03] Algebra
Concept: Simultaneous Linear Equations in Three Variables
Solve the following system of linear equations in three variables
x + 20 = (3y)/2 + 10 = 2z + 5 = 110 – (y + z)
[0.03] Algebra
Chapter: [0.03] Algebra
Concept: Simultaneous Linear Equations in Three Variables
Discuss the nature of solutions of the following system of equations
x + 2y – z = 6; – 3x – 2y + 5z = – 12; x – 2z = 3
[0.03] Algebra
Chapter: [0.03] Algebra
Concept: Simultaneous Linear Equations in Three Variables
< prev 1 to 20 of 921 next >
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# How do you divide (4x^3-5x^2-4x-12)/(3x-4) ?
Nov 24, 2017
$\frac{4}{3} {x}^{2} + \frac{1}{9} x - \frac{32}{27} \text{ Remainder: } \frac{- 196}{27 \left(3 x - 4\right)}$
#### Explanation:
This is best solved using long division. The question is, what do I have to multiply by $3 x - 4$ to get $4 {x}^{3} - 5 {x}^{2} - 4 x - 12$?
$\text{ "4/3 x^2 + 1/9x - 32/27 " Remainder: } \frac{- 196}{27 \left(3 x - 4\right)}$
$3 x - 4 | \overline{4 {x}^{3} - 5 {x}^{2} - 4 x - 12}$
$\text{ "-ul((4x^3-16/3 x^2)) downarrow " } \downarrow$
$\text{ } \frac{1}{3} {x}^{2} - 4 x$
$\text{ } - \underline{\left(\frac{1}{3} {x}^{2} - \frac{4}{9} x\right)}$
$\text{ } - \frac{32}{9} x - 12$
$\text{ } - \underline{\left(- \frac{32}{9} x + \frac{128}{27}\right)}$
$\text{ } - \frac{196}{27}$
Nov 24, 2017
color(magenta)(1.3x^2+0.1x-1.2 and remainder of color(magenta)(-6
#### Explanation:
$\frac{4 {x}^{3} - 5 {x}^{2} - 4 x - 12}{3 x - 4}$
color(white)(................)color(magenta)(1.3x^2+0.1x-1.2
$\textcolor{w h i t e}{a} 3 x - 4$$|$$\overline{4 {x}^{3} - 5 {x}^{2} - 4 x - 12}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} \underline{4 {x}^{3} - 5.3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 0.3 {x}^{2} - 4 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{0.3 {x}^{2} - 0.4 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 3.6 x - 12$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- 3.6 x + 4.8}$
color(white)(...........................................)color(magenta)(-6
color(magenta)((4x^3-5x^2-4x-12)/(3x-4)=1.3x^2+0.1x-1.2and remainder of color(magenta)(-6
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# ${\text{Write an equation of the horizontal parabola with the given vertex and passing through the given point}}{\text{.}} \\ {\text{Vertex at }}\left( { - 8,2} \right){\text{ and passing through }}\left( {2, - 3} \right). \\$
${\text{As you know that general equation of a horizontal parabola is:}} \\ {\left( {y - k} \right)^2} = 4a\left( {x - h} \right){\text{ }} \\ {\text{As Vertex is given to us as }}\left( { - 8,2} \right){\text{, therefore we have}} \\ {\left( {y - 2} \right)^2} = 4a\left( {x - \left( { - 8} \right)} \right){\text{ }} \\ \Rightarrow {\left( {y - 2} \right)^2} = 4a\left( {x + 8} \right){\text{ }} - Equation(1){\text{ }} \\ {\text{Now, putting }}\left( {2, - 3} \right){\text{ in Equation(1) we get}} \\ {\left( { - 3 - 2} \right)^2}{\text{ }} = {\text{ }}4a\left( {2 + 8} \right) \\ \Rightarrow {\left( { - 5} \right)^2}{\text{ }} = {\text{ }}4a\left( {10} \right){\text{ }} \\ \Rightarrow 25 = 40a \\ \Rightarrow a = \frac{5}{8}{\text{ }} - Equation(2) \\ {\text{Using Equation(2) in Equation(1)}} \\ \Rightarrow {\left( {y - 2} \right)^2} = 4\left( {\frac{5}{8}} \right)\left( {x + 8} \right) \\ \Rightarrow {\left( {y - 2} \right)^2} = \frac{5}{2}\left( {x + 8} \right){\text{ is the required Equation}} \\ {\text{Note: For these kinds of questions we must remember the general equation of a horizontal }} \\ {\text{parabola and then put vertex's coordinates in it to get the equation in only one variable 'a'}}{\text{. }} \\ {\text{Now, Put the coordinates if the point given to be on the parabola and solve to get value of 'a'}}{\text{. }} \\ {\text{Put the value of 'a' in the equation to get desired equation}}{\text{.}} \\$
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# How do you find the critical points for f(x) = (x^2-10x)^4 and the local max and min?
##### 1 Answer
Jul 3, 2016
three critical points: $x = 0 , 5 , 10$
Relative minimum at $x = 0 , 10$
Relative maximum at $x = 5$
#### Explanation:
First, we need to find $f ' \left(x\right)$.
Using rules for taking derivative, we get following:
$f ' \left(x\right) = 4 \cdot {\left({x}^{2} - 10 x\right)}^{3} \cdot \left(2 x - 10\right)$
Critical points occur when the slope is zero meaning $f ' \left(x\right) = 0$.
$f ' \left(x\right) = 4 \cdot {\left({x}^{2} - 10 x\right)}^{3} \cdot \left(2 x - 10\right) = 0$
$\left(2 x - 10\right) = 0$ => $x = 5$
$\left({x}^{2} - 10 x\right) = 0$ => $x = 0 , 10$
So, three critical points would be $x = 0 , 5 , 10$.
To find relative maximum and minimum, we can do a sign test.
For $x \in$(-∞, 0) => $f ' \left(x\right)$ is negative.
For $x \in$(0, 5) => $f ' \left(x\right)$ is positive.
For $x \in$(5, 10) => $f ' \left(x\right)$ is negative.
For $x \in$(10, ∞) => $f ' \left(x\right)$ is positive.
At $x = 0$, function has relative minimum because $f ' \left(x\right)$ changes signs from negative to positive.
At $x = 5$, function has relative maximum because $f ' \left(x\right)$ changes signs from positive to negative.
At $x = 10$, function has relative minimum because $f ' \left(x\right)$ changes signs from negative to positive.
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# Improper Fractions and Mixed Numbers
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1 This assignment includes practice problems covering a variety of mathematical concepts. Do NOT use a calculator in this assignment. The assignment will be collected on the first full day of class. All work must be shown in order to earn full credit. The assignment will be graded for completion. REMEMBER: There are lots of ways of solving math problems. The suggested methods in this packet are just one way of solving the problem. If you have learned another method through your elementary math program that it is fine.
2 Improper Fractions and Mixed Numbers An improper fraction is a fraction with the numerator that is greater than the denominator. A mixed number is a whole number and a fraction. and 7 5 are improper fractions is a mixed number. An improper fraction can be written as a whole or mixed number. A mixed number can be written as a improper fraction. Write as whole Write 7 5 as a Write numbers. mixed numbers. as an improper fraction. Multiply the whole number by the denominator. Add this Divide the Divide the numerator by the product to the numerator. Then write the sum over the numerator by the denominator. Write the denominator. denominator. remainder as a fraction by writing the remainder over the 4 divisor = = = = 11 6 = = 11 6 Write as a whole number. 1) 16 4 = 2) 15 8 = ) 28 4 = 4) 48 6 = Write as a mixed number. 1) 15 4 = 2) 21 5 = ) 2 6 = 4) 7 2 = Write as an improper fraction. 1) 2 7 = 2) 8 1 = ) = 4) 2 5 =
3 Adding Mixed Numbers, Whole Numbers, and Fractions To add mixed numbers, whole numbers, and fractions, first check for unlike denominators. Write mixed numbers and fractions as equivalent fractions with like denominators. Add he fractions. Then add the whole numbers and simplify. Find: Write the fractions with Add the Add the whole Simplify. like denominators. fractions. numbers. 2 1 = = = = = = = 9 2 Add and simplify. 1) ) 1 4 ) 1 6 4) ) 1 6) ) 1 5 8)
4 Subtraction of Mixed Numbers with Regrouping To subtract mixed numbers, it may be necessary to regroup first. Write the whole number part as a mixed number. Add the mixed number and the fraction. Then subtract and simplify. Find: Write the fractions with like denominators. Compare the numerators. 5 7 = = is greater than 7. You cant subtract the fractions. To regroup, write 5 as the following mixed number. 5 = 4 Add the mixed number and the fraction. 5 7 = = 4 19 Now you can subtract and simplify. 5 7 = = 1 9 = 5 6 Subtract and Simplify. 1) 6 8 2) 7 4 ) ) ) )
5 Multiplication of Fractions To multiply fractions, multiply the numerators and multiply the denominators. Simplify the product. Find: 4 x 5 6 Multiply the numerators. Multiply the denominators. Simplify = 5 = = = = 5 8 Multiply and simplify. 1) 4 1 = 5 2) 1 = 5 ) 4 = 7 5 4) = 5) = 6) 1 11 = Multiplication of Whole Numbers by Fractions To multiply a whole number by a fraction, first write the whole number as an improper fraction. Multiply the numerators and the denominators. Simplify. Find: Write the whole number as an improper fraction = Multiply. Simplify. 6 5 = 0 = 1 14 = Multiply and simplify. 1) 2 6 2) 25 ) 25 5
6 Multiplication of Mixed Numbers by Whole Numbers To multiply a mixed number by a whole number, first write the mixed number and the whole number as improper fractions. Multiply the numerators and denominators. Simplify the answer. Find: Write the whole number and the mixed number as improper fractions = Multiply the numerators and denominators. Simplify. 9 4 = 6 = 4 4 = Multiply and Simplify. 1) = 2) 2 1 = ) = Multiplication of Mixed Numbers by Fractions To multiply a mixed number by a fraction, first write the mixed number as an improper fraction. Multiply the numerators and denominators. Simplify. Find: Write the mixed number as an improper fraction = Multiply the numerators and denominators. Simplify = 9 6 = 1 6 = Multiply and Simplify. 1) = 2) = ) =
7 Multiplication of Mixed Numbers by Mixed Numbers To multiply a mixed number by a mixed number, write both mixed numbers as improper fractions. then multiply the numerators and denominators. Simplify. Find: Write the mixed numbers as improper fractions. Multiply the numerators and denominators. Simplify = = 209 = 17 5 Multiply and Simplify. 1) = 2) = ) = 4) = 5) = 6) =
8 Division of Fractions by Fractions To divide a fraction by a fraction, multiply by the reciprocal of the second fraction. Simplify the answer if needed. Remember, only the second fraction is inverted. Find: Multiply by the reciprocal of the second fraction = Multiply the numerators and denominators. Simplify. 5 4 = 20 = Divide and Simplify. 1) 4 1 = 5 2) 5 1 = 4 ) 1 1 = ) = 5) = 6) = Division of Mixed Numbers by Fractions To divide a mixed number by a fraction, write the mixed number as an improper fraction. Multiply by the reciprocal of the second fraction. Simplify the answer. Find: Write the mixed number Multiply by the reciprocal Multiply and simplify. as an improper fraction. of the second fraction = = 40 = 4 = 1 4 Divide and Simplify. 1) = 2) = ) =
9 Fraction and Decimal Equivalents Sometimes you will need to either change a decimal to a fraction or fraction to a decimal. To write a decimal as a fraction, identify the value of the last place in the decimal. Use this place value to write the denominator. To write a fraction that has a denominator of, 0, or 1,000 as a decimal, write the digits for the numerator. Then write the decimal point. Decimal Fraction or Mixed Number Fraction or Mixed Number Decimal 0.7 = 7 9 = = 0.09 = 1.72 = ,000 or , or = 0.14 = 0.48 = 9.06 Write each decimal as a fraction. 1) 0.5 2) 0.06 ) ) 6.54 Write each decimal as a mixed number. 5) 2. 6) ) ) 7.5 Write each fraction as a decimal. 6) 9 7) 7 0 8) 9) 1 00 ) ) ) 17 1) 9 00
10 Addition and Subtraction of Decimals To add or subtract decimals, line up the decimal points. Write zeros as needed. Then add or subtract the same way as whole numbers. Examples: Find: Write a zero. Add. Write a decimal point in the sum. Find: Write a zero. Regroup. Subtract. Write a decimal point in the difference Add or subtract. Write zeros as needed
11 Multiplying Decimals by Decimals To multiply decimals by decimals, multiply the same way as whole numbers. Place the decimal point in the product by counting the number of decimal places in each factor. The product will have the same number of decimal places. Remember, sometimes you might need to write a zero in the product in order to place the decimal point correctly. Examples: Find: Multiply. Write the decimal point in the product. Find: Multiply. Write the decimal point in the product places + 1 place places Write a zero. 2 places + 2 places 4 places Multiply. Write zeros as needed
12 Dividing Decimals by Decimals To divide decimal by decimal, change the divisor to a whole number by moving the decimal point. Move the decimal point in the dividend the same number of places. Then divide. Remember, write a decimal point in the quotient directly above the position of the new decimal point in the dividend. Examples: Find: Move each decimal Divide. Find: Move each decimal Divide. point 1 place. point 2 places ) ) ) ) Divide ) ) ) ) ) )
13 Multiplying by Powers of To multiply decimals by powers of ten, move the decimal point in the product to the right as many places as there are zeros in the multiplier. Remember, sometimes you might need to write zeros in the product in order to move the decimal point the correct number of places. Study these examples. 0.2 = = 20 1, = = = 5.1 1, = 51 Multiply. Write zeros as needed = = ,000 = Dividing by Powers of To divide a decimal by a power of ten, move the decimal point in the dividend to the left as many places as there are zeros in the divisor. Remember, sometimes you might need to write zeros in the quotient in order to correctly insert the decimal point. Study these examples. 0.2 = = ,000 = = = ,000 = Divide. Write zeros as needed = = ,000 =
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# How to Use Distance Formula to Find the Length of a Line
You can measure the length of a vertical or horizontal line on a coordinate plane by simply counting coordinates; however, measuring the length of a diagonal line is trickier. You can use the Distance Formula to find the length of such a line. This formula is basically the Pythagorean Theorem, which you can see if you imagine the given line segment as the hypotenuse of a right triangle.[1] By using a basic geometric formula, measuring lines on a coordinate path becomes a relatively easy task.
### Part 1 of 2: Setting up the Formula
1. 1
Set up the Distance Formula. The formula states that ${\displaystyle d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}$, where ${\displaystyle d}$ equals the distance of the line, ${\displaystyle (x_{1},y_{1})}$ equal the coordinates of the first endpoint of the line segment, and ${\displaystyle (x_{2},y_{2})}$ equal the coordinates of the second endpoint of the line segment.[2]
2. 2
Find the coordinates of the line segment’s endpoints. These might already be given. If not, count along the x-axis and y-axis to find the coordinates.[3]
• The x-axis is the horizontal axis; the y-axis is the vertical axis.
• The coordinates of a point are written as ${\displaystyle (x,y)}$.
• For example, a line segment might have an endpoint at ${\displaystyle (2,1)}$ and another at ${\displaystyle (6,4)}$.
3. 3
Plug the coordinates into the Distance Formula. Be careful to substitute the values for the correct variables. The two ${\displaystyle x}$ coordinates should be inside the first set of parentheses, and the two ${\displaystyle y}$ coordinates should be inside the second set of parentheses.[4]
• For example, for points ${\displaystyle (2,1)}$ and ${\displaystyle (6,4)}$, your formula would look like this: ${\displaystyle d={\sqrt {(6-2)^{2}+(4-1)^{2}}}}$
### Part 2 of 2: Calculating the Distance
1. 1
Calculate the subtraction in parentheses. By using the order of operations, any calculations in parentheses must be completed first.[5]
• For example:
${\displaystyle d={\sqrt {(6-2)^{2}+(4-1)^{2}}}}$
${\displaystyle d={\sqrt {(4)^{2}+(3)^{2}}}}$
2. 2
Square the value in parentheses. The order of operations states that exponents should be addressed next.[6]
• For example:
${\displaystyle d={\sqrt {(4)^{2}+(3)^{2}}}}$
${\displaystyle d={\sqrt {16+9}}}$
3. 3
Add the numbers under the radical sign. You do this calculation as if you were working with whole numbers.
• For example:
${\displaystyle d={\sqrt {16+9}}}$
${\displaystyle d={\sqrt {25}}}$
4. 4
Solve for . To reach your final answer, find the square root of the sum under the radical sign.
• Since you are finding a square root, you may have to round your answer.
• Since you are working on a coordinate plane, your answer will be in generic “units,” not in centimeters, meters, or another metric unit.
• For example:
${\displaystyle d={\sqrt {25}}}$
${\displaystyle d=5}$ units
## Community Q&A
Search
• Question
What do we call points that are on same line?
Donagan
Points on the same straight line are said to be "collinear" or "colinear."
• Question
How do I simplify the square root?
Donagan
Look for a perfect square inside the radical sign, find its square root, and put that square root out in front of the radical sign, indicating that it's to be multiplied by the radical. For example: √50 = √(2 x 25) = 5√2.
• Question
The rise is 4 inches. The angle is 90 degrees from the height to the base. What is the slope?
Donagan
To find the slope, you have to know the horizontal distance over which the rise occurs. The 90° angle is not relevant.
• Question
How do I find the distance of a line if I don't know the end points?
Donagan
Assuming you don't have a measuring tool (tape measure, ruler, etc.), you cannot use the distance formula outlined above without knowing the end points. However, if the line is given as part of a geometrical figure, there are various geometrical and trigonometrical methods to find the length (too numerous and complicated to treat here).
• Question
How would I find the length of each segment?
For a line segment, you should create a right triangle by making two more lines. If this line segment is on a dot grid, every line between two dots is one unit. If the line segment is on a coordinate plane, then you can do the same or use the distance formula for a line.
• Question
A square table has an area of 16 sq cm. How do I find the length of its side?
Donagan
√16 = 4 cm.
• Question
On a number line, the length of line segment joining 3 and -3 is what?
Donagan
On a number line 3 and -3 are each 3 units from zero (in opposite directions). So the total distance between them is 3 + 3 = 6.
• Question
Would line AB be the same as line BA?
Donagan
Yes.
• Question
I don’t understand the distance formula. Can you help?
Donagan
It's simply the Pythagorean theorem. Find the vertical change (between endpoints of the line). Square it. Find the horizontal change. Square that, too. Add together the two squares. Then find the square root of that sum. This gives you the line length you're looking for.
• Question
If I know the length of the line and one of the endpoints, how do I find the other endpoint?
Donagan
You can't do it unless you also know the slope of the line.
200 characters left
## Tips
• Do not confuse this formula with others, like the Midpoint Formula, Slope Formula, Equation of a Line or Line Formula.
Thanks!
• Remember the order of operations when calculating your answer. Subtract first, then square the differences, then add, and then find the square root.
Thanks!
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This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 368,291 times.
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Updated: September 23, 2020
Views: 368,291
Categories: Coordinate Geometry
Article SummaryX
To use the distance formula to find the length of a line, start by finding the coordinates of the line segment's endpoints. Then, plug the coordinates into the distance formula. Next, subtract the numbers in parenthesis and then square the differences. Once you've done that, just add the numbers that are under the radical sign and solve for d. To learn how to set up the distance formula, keep reading!
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# 2009 USAMO Problems/Problem 6
## Problem
Let $s_1, s_2, s_3, \ldots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \cdots.$ Suppose that $t_1, t_2, t_3, \ldots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.
## Solution
Suppose the $s_i$ can be represented as $\frac{a_i}{b_i}$ for every $i$, and suppose $t_i$ can be represented as $\frac{c_i}{d_i}$. Let's start with only the first two terms in the two sequences, $s_1$ and $s_2$ for sequence $s$ and $t_1$ and $t_2$ for sequence $t$. Then by the conditions of the problem, we have $(s_2 - s_1)(t_2 - t_1)$ is an integer, or $(\frac{a_2}{b_2} - \frac{a_1}{b_1})(\frac{c_2}{d_2} - \frac{c_1}{d_1})$ is an integer. Now we can set $r = \frac{b_1 b_2}{d_1 d_2}$, because the least common denominator of $s_2 - s_1$ is $b_1 b_2$ and of $t_2 - t_1$ is $d_1 d_2$, and multiplying or dividing appropriately by $\frac{b_1 b_2}{d_1 d_2}$ will always give an integer.
Now suppose we kept adding $s_i$ and $t_i$ until we get to $s_m = \frac{a_m}{b_m}$ in sequence $s$ and $t_m = \frac{c_m}{d_m}$ in sequence $t$ so that $(t_m - t_i)(s_m - s_i)$ is an integer for all $i$ with $1 \le i < m$, where $m$ is a positive integer. At this point, we will have $r$ = $\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}$, because these are the least common denominators of the two sequences up to $m$. As we keep adding $s_i$ and $t_i$, $r$ will always have value $\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}$, and we are done.
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# How do you solve x^2+8x+1=0 by completing the square?
Jun 6, 2016
$x = - 4 - \sqrt{15}$ or
$x = - 4 + \sqrt{15}$
#### Explanation:
In ${x}^{2} + 8 x + 1 = 0$, we observe that coefficient of $x$ is $8$ and hence, we can complete the square by adding square of half of $8$ i.e. ${4}^{2} = 16$.
Hence ${x}^{2} + 8 x + 1 = 0$ can be written as ${x}^{2} + 8 x + 16 - 16 + 1 = 0$ or
${\left(x + 4\right)}^{2} - 15 = 0$ which can be written as
${\left(x + 4\right)}^{2} - {\left(\sqrt{15}\right)}^{2} = 0$
Now using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this can be written as
$\left(x + 4 + \sqrt{15}\right) \left(x + 4 - \sqrt{15}\right) = 0$
Hence either $x + 4 + \sqrt{15} = 0$ i.e. $x = - 4 - \sqrt{15}$
or $x + 4 - \sqrt{15} = 0$ i.e. $x = - 4 + \sqrt{15}$
|
## College Algebra (11th Edition)
$6a^6 (4a+5b)(2a-3b)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $48a^8-12a^7b-90a^6b^2 ,$ factor first the $GCF.$ Then find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 48,-12,-90 \}$ is $6 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ a^8,a^7,a^6 \}$ is $a^6 .$ Hence, the entire expression has $GCF= 6a^6 .$ Factoring the $GCF= 6a^6 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6a^6 \left( 8a^2-2ab-15b^2 \right) .\end{array} In the trinomial expression above the value of $ac$ is $8(-15)=-120$ and the value of $b$ is $-2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{ 10,-12 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6a^6 \left( 8a^2+10ab-12ab-15b^2 \right) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 6a^6 [( 8a^2+10ab)-(12ab+15b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 6a^6 [2a( 4a+5b)-3b(4a+5b)] .\end{array} Factoring the $GCF= (4a+5b)$ of the entire expression above results to \begin{array}{l}\require{cancel} 6a^6 [(4a+5b)(2a-3b)] \\\\= 6a^6 (4a+5b)(2a-3b) .\end{array}
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# The population of Greenville is approximately 75 times the population of Fairview. There are 2.50 times 10^4 people living in Greenville. Approximately how many people are living in Fairview?
##### 1 Answer
Nov 14, 2017
See a solution process below:
#### Explanation:
Let's call the population of Fairview: $f$
And, let's call the population of Greenville: $g$
Then we can write an equation:
$g = 75 f$ or $\frac{g}{75} = f$
Substituting $2.50 \times {10}^{4}$ for $g$ gives:
$\frac{2.50 \times {10}^{4}}{75} = f$
$\frac{2.50}{75} \times {10}^{4} = f$
$0.0 \overline{3} \times {10}^{4} = f$
We can move the decimal point two places to the left and subtract from the exponent for the 10s term giving:
$3. \overline{3} \times {10}^{2} = f$
The population of Fairview is approximately $3.33 \times {10}^{2}$
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GeoGebra Classroom
# Primitive Notions and Propositions
## What is an axiom?
Axioms (or postulates) are unquestionably universally valid truths, often used as principles in the construction of a theory or as the basis for an argument. That is, an axiom is a proposion which is so clear, that is assumed as true without a demonstration or proof. An axiomatic system is the set of axioms that define a given theory and that constitute the simplest truths from which the new results of that theory are demonstrated. What this means is that for every theorem in math, there exists an axiomatic system that contains all the axioms needed to prove that theorem. The axiom contains evidence in itself and therefore does not need not be proved.
## Incidence axioms
Axiom 1- Whatever the line, there are points that belong to this line, and points not belonging to it.
Axiom 2- Given two distinct points, there is a single line that contains them. Note: when two lines have a point in common, they are said to intersect or cut at that point.
## Activity 1
In the previous Geogebra applet, select the Line option (window 3) and draw a line that passes through points B and D. Draw another line that passes through D and C. Is this last line in the same plane of the other two lines?
## Proposition 1
Two distinct lines either do not intersect or intersect at a single point.
## Activity 2
Which pairs of lines intersect? Do lines g and h intersect?
## Proof of Proposition 1
Hypothesis: Consider m and n as two distinct lines. Thesis: m and do not intersect or intersect at a single point. The intersection of these two lines cannot contain two or more points, otherwise, when looking at the truth provided by axiom 2 (Given two distinct points there is a single line that contains them), they would coincide. Therefore the intersection of m and n either happens at only one point or it doesn't happen.
## Activity 3
Did you understand the proof? Can you explain it in another way? If you didn't understand, write what you didn't understand.
## Axiom 3
Given three distinct points on a line, one and only one of them is located between the other two.
## Definition 1: Segment
The set consisting of two points A and B and all points between A and B is called line segment AB. Points A and B are called line segment endpoints.
## Activity 4
In the previous construction, select the Segment tool (window 3) and create the line segments BC, CE and AC. If you didn't understand, write what you didn't understand.
## Definition 2: Ray
Consider A and B as two distinct points. The set consisting of points of segment AB and of all points C so that B is between A and C, is called a ray starting at A passing through B.
## Note 1
In reference books, in general, the ray is represented only with an "Vector".
## Note 2
Note that two points A and B determine two rays: SAB and SBA.
## Axiom 4
Consider two points A and B of a line. There is always a point C between A and B , and a point D so that B is between A and D.
## Note 3
The result in this axiom is that, between any two points on a line, there is an infinity of points.
## Definition 3: Half-plane
Consider m as a line and A a point that does not belong to line m. The set consisting of m and all points B, such that A and B are on the same side of line m, is called a half-plane determined by m containing A.
## Axiom 5
A line m outlines two distinct half-planes whose intersection is line m.
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# Inverse functions
## Example: an inverse function in a population model
• Recall the exponential model for the population of the Earth:
• $$f(t)=(1436.53)\cdot(1.01395)^t$$, where $$t$$ is the number of years since $$1900$$, and $$P=f(t)$$ is the population measured in millions of people.
• We saw that it can be used to predict the population at eg. 2020: $$f(t=120)\approx7573.549$$.
• Note that $$f(t)$$ assigns years $$t$$ to population values $$P$$.
• Suppose you want to predict when the population will exceed 8 billion.
• This means that you want to reverse the assignment, and assign to population values $$P$$ the years $$t$$.
• This can be done via the inverse function $$t=f^{-1}(P)$$ of $$f(t)$$.
• The prediction will be $$f^{-1}(P=8000)\approx123.954$$, that is the model predicts that population will exceed 8 billion in 2023.
## One-to-one functions
• Recall that a function $$y=f(x)$$ assigns precisely one value $$f(x)$$ to every $$x$$ in its domain.
• This means that in order for the inverse $$x=f^{-1}(y)$$ to be a function, for each $$y$$, there can be at most one $$x$$ such that $$f(x)=y$$.
• Definition. A function $$f(x)$$ is called one-to-one, if it never takes on the same value twice: $$f(x_1)\ne f(x_2)$$ whenever $$x_1\ne x_2$$.
• For example, the function $$f(x)=x^2$$ is not one-to-one, because we have $$f(x=-1)=(-1)^2=1=1^2=f(x=1)$$.
• On the other hand, the function $$f(x)=x^3$$ is one-to-one.
• In the next week, we will learn about continuous functions. In particular, we will learn the Intermediate Value Theorem, which will imply that if the domain $$D$$ of a continuous function $$f(x)$$ is one connected interval, with its endpoints possibly $$\pm\infty$$, then it's one-to-one precisely when it's strictly monotonous.
• Horizontal line test. A function is one-to-one if and only if no horizontal line intersects its graph more than once.
• Exercises: 1.5: 10,12
## Definition of inverse functions
• Definition. Let $$y=f(x)$$ be a function with domain $$D$$ and range $$R$$. Then its inverse function $$x=f^{-1}(y)$$ has domain $$R$$ and range $$D$$, and it is defined by $$$\tag{*} f^{-1}(y)=x\text{ if and only if }y=f(x)\text{ for any y in R}.$$$
• Caution. Do not confuse the inverse function $$f^{-1}(x)$$ with the reciprocal $$f(x)^{-1}=\frac{1}{f(x)}$$. Note that in the inverse function, the $${}^{-1}$$ is after the function name, and in the reciprocal, it is after the argument.
• Cancellation equations. By substituting the first equation in the definition (*) to the second and vice versa, we get the following. \begin{align*} y=f(f^{-1}(y))&\text{ for any y in R}\\ f^{-1}(f(x))=x&\text{ for any x in D} \end{align*}
## Finding inverse functions
• Let $$f(x)$$ be a one-to-one function with range $$R$$.
• To find its inverse function $$f^{-1}(y)$$, you need to solve the equation $$y=f(x)$$ for $$x$$ for every $$y$$ in $$R$$.
• Afterwards, you can interchange $$x$$ and $$y$$ to get the inverse in the form $$f^{-1}(x)$$, if you want to.
• Exercises: Find the inverses of $$2x-3$$, $$x^3+2$$, and $$x^2-x$$, the latter with domain $$x\ge\frac12$$.
• Click here for the graphs of the functions and their inverses.
• Note that the graph of the inverse $$y=f^{-1}(x)$$ is obtained by mirroring the graph of the original function $$y=f(x)$$ with respect to the line $$y=x$$.
• Exercises: 1.5.22,20
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Trying to find out how to convert 24/5 into a mixed number or fraction? Have I got the answer for you! In this guide, we"ll walk you through the step-by-step process of converting an improper fraction, in this case 24/5, to a mixed number. Read on!
Want to quickly learn or show students how to convert 24/5 to a mixed number? Play this very quick and fun video now!
Before we begin, let"s revisit some basic fraction terms so you understand exactly what we"re dealing with here:
Numerator. This is the number above the fraction line. For 24/5, the numerator is 24.Denominator. This is the number below the fraction line. For 24/5, the denominator is 5.Improper fraction. This is a fraction where the numerator is greater than the denominator.Mixed number. This is a way of expressing an improper fraction by simplifying it to whole units and a smaller overall fraction. It"s an integer (whole number) and a proper fraction.
You are watching: 24 5 as a mixed number
Now let"s go through the steps needed to convert 24/5 to a mixed number.
## Step 1: Find the whole number
We first want to find the whole number, and to do this we divide the numerator by the denominator. Since we are only interested in whole numbers, we ignore any numbers to the right of the decimal point.
24/5= 4.8 = 4
Now that we have our whole number for the mixed fraction, we need to find our new numerator for the fraction part of the mixed number.
## Step 2: Get the new numerator
To work this out we"ll use the whole number we calculated in step one (4) and multiply it by the original denominator (5). The result of that multiplication is then subtracted from the original numerator:
24 - (5 x 4) = 4
## Step 3: Our mixed fraction
We"ve now simplified 24/5 to a mixed number. To see it, we just need to put the whole number together with our new numerator and original denominator:
4 4/5
## Step 4: Simplifying our fraction
In this case, our fraction (4/5) can be simplified down further. In order to do that, we need to calculate the GCF (greatest common factor) of those two numbers. You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 4 and 5 is 1.
See more: Used 1999 Dodge Ram 1500 5.9 Specs & Features, 1999 Dodge Ram Specifications
We can now divide both the new numerator and the denominator by 1 to simplify this fraction down to its lowest terms.
4/1 = 4
5/1 = 5
When we put that together, we can see that our complete answer is:
4 4/5
Hopefully this tutorial has helped you to understand how to convert any improper fraction you have into a mixed fraction, complete with a whole number and a proper fraction. You"re free to use our calculator below to work out more, but do try and learn how to do it yourself. It"s more fun than it seems, I promise!
## Improper Fraction to Mixed Number
Enter an improper fraction numerator and denominator
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# Definition of magnitude in Math
May 30, 2016
#### Definition of a vector
A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.
The vector addition is the way forces and velocities combine. For example, if a car is travelling due north at 20 miles per hour and a child in the back seat behind the driver throws an object at 20 miles per hour toward his sibling who is sitting due east of him, then the velocity of the object (relative to the ground!) will be in a north-easterly direction. The velocity vectors form a right triangle, where the total velocity is the hypotenuse. Therefore, the total speed of the object (i.e., the magnitude of the velocity vector) is $\sqrt{20^2+20^2}=20\sqrt{2}$ miles per hour relative to the ground.
Addition of vectors satisfies two important properties.
1. The commutative law, which states the order of addition doesn't matter: $$\vc{a}+\vc{b}=\vc{b}+\vc{a}.$$ This law is also called the parallelogram law, as illustrated in the below image. Two of the edges of the parallelogram define $\vc{a}+\vc{b}$, and the other pair of edges define $\vc{b}+\vc{a}$. But, both sums are equal to the same diagonal of the parallelogram.
2. The associative law, which states that the sum of three vectors does not depend on which pair of vectors is added first: $$(\vc{a}+\vc{b})+\vc{c} = \vc{a} + (\vc{b}+\vc{c}).$$
You can explore the properties of vector addition with the following applet. (This applet also shows the coordinates of the vectors, which you can read about in another page.)
The sum of two vectors. The sum $\vc{a}+\vc{b}$ of the vector $\vc{a}$ (blue arrow) and the vector $\vc{b}$ (red arrow) is shown by the green arrow. As vectors are independent of their starting position, both blue arrows represent the same vector $\vc{a}$ and both red arrows represent the same vector $\vc{b}$. The sum $\vc{a}+\vc{b}$ can be formed by placing the tail of the vector $\vc{b}$ at the head of the vector $\vc{a}$. Equivalently, it can be formed by placing the tail of the vector $\vc{a}$ at the head of the vector $\vc{b}$. Both constructions together form a parallelogram, with the sum $\vc{a}+\vc{b}$ being a diagonal. (For this reason, the commutative law $\vc{a}+\vc{b}=\vc{b}+\vc{a}$ is sometimes called the parallelogram law.) You can change $\vc{a}$ and $\vc{b}$ by dragging the yellow points.
##### Vector subtraction
Before we define subtraction, we define the vector $-\vc{a}$, which is the opposite of $\vc{a}$. The vector $-\vc{a}$ is the vector with the same magnitude as $\vc{a}$ but that is pointed in the opposite direction.
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Operations with powers with the same and different base. Exercises resolved
# Operations with powers with the same and different base. Exercises resolved
I’m going to explain how to operate with powers with the same and different base. You will learn how to multiply and divide powers of different base, both of variables and with numbers.
## Multiplication of powers with the same base
When we have two powers multiplying, it is not a question of applying the property of the multiplication of powers with the same base and that’s it, but it is necessary to finish simplifying the operation with other properties.
Let’s see it with an example:
The first step is to check if they have the same base, that they have it.
Therefore, when we have multiplications with the same base, the power multiplication property is applied with the same base:
Keep the base and add the exponents.
In this case, we have a negative exponent, but it doesn’t matter, because we add a negative number and that’s it:
We are left with a negative base power (the exponent affects the minus sign because it is enclosed in parentheses), elevated to a negative exponent.
The next step is to apply the negative exponent property:
We pass that exponent to positive and then resolve the power in the denominator, which is negative because the exponent is odd:
As you can see we have applied two properties until we have simplified the operation. After adding or subtracting the exponents, always pass the exponent to positive.
Power properties must be applied until the operation is completely simplified.
## Division of powers with the same base
With the division of powers with the same base, the same thing happens as with multiplication. It is not enough to apply only the property of power division with the same base.
For example:
We have two powers that are dividing and have the same base, therefore, the first thing we must do is apply the property of the division of powers with the same base:
We maintain the base and subtract the exponents:
Nos has left a power with negative exponent, which we have to pass to positive exponent with this property:
That’s why we pass the power to the denominator with the positive exponent:
Summarizing, when we have multiplications or divisions of powers with the same base, we add or subtract exponents, that can be positive or negative and then we pass the exponent to positive.
## Multiplications and divisions with powers with the same base
In the same operation, we can have multiplications and divisions of powers with the same base. In other words, we would have a fraction with more than one power
In this case, we must apply the multiplication property, separately, in the numerator and in the denominator, then apply the division property and finally, pass the exponent to positive, if we have been negative.
Let’s take a slower look at an example:
We have an operation where several powers with the same base are multiplying and dividing.
We apply the multiplication property to the numerator and denominator. We maintain the base and add the exponents:
We are left with a fraction that has 2 particularities:
1 – We get a 2 raised to 0 in the numerator and we already know from the first property that any number raised to 0 is 1:
2 – We have a negative exponent in the denominator. We convert the exponent to positive by passing the power to the numerator. It is the same property as that of a power with negative exponent:
Continuing with our operation, we have the following:
Once we have passed the exponent to positive, the power can be resolved.
## Multiplications and divisions of powers with different base
In one operation we can find powers of different base, which are multiplying and dividing. Keep in mind that we can only multiply and divide powers when they have the same base.
If we have a multiplication of two powers that have different bases, such as this one:
We cannot operate with them because we cannot apply any property of the powers. It would stay as it is.
Remember that the properties of multiplication and division of powers are applied when we have the same base.
Therefore, the first thing we have to do is to look for the powers that have the same base, to multiply or divide them separately.
Let’s look at this concept with another example:
We have two bases: x e y.
With base x, we have two powers that are multiplying, so we can add the exponents. With base y, we can’t do anything and it stays as it is:
Do you see what the procedure is? You always have to look for powers of the same base to be able to apply the properties of the corresponding powers.
Let’s see another example:
We again have two bases: x and y.
We cannot multiply powers in the numerator and denominator, since we have powers of different base.
On the other hand, we do have power divisions with base x and with base y.
We divide separately with each of the bases. We treat them as if they were two fractions that are multiplying:
On the one hand, for the base x, the exponents are subtracted and on the other hand for the base y the exponents are also subtracted:
For each of the bases, we have a negative exponent left, which we turn positive by passing the power to the denominator:
Let’s see another example where we also have numbers, besides variables:
In this case we have on the one hand a fraction of numbers, on the other hand a division of powers with base x and on the other hand a division of base powers y.
With numbers we simplify the fraction, whose result is an integer:
With the bases x and y, we maintain the base and subtract the exponents. So we have our equation:
In the base y, we have the exponent equal to 0. We know by its corresponding property, that any variable or number a raised to 0 is 1, so we have:
And this would simplify the expression.
As you can see, it is always the same to solve separately powers with the same base, which are multiplied in the final result.
## Operations with powers of numbers with different base
When we work only with numbers and we have powers of different bases, we must look for the powers to have the same base, that is to say, we must express all the powers with the same base or if it is not possible to express all the powers with a single base, with the minimum possible number of bases.
And how do we express the number in another base? Then by breaking the number down into factors.
Let’s look at it with a very simple example:
In this multiplication of powers, in principle we can’t do anything, because we have a multiplication of powers of different base and we can’t add their exponents.
But we can decompose the 4:
Therefore, in the operation we are solving, we substitute 4 with its decomposition and in this way we have a multiplication of powers with the same base:
Before multiplying the powers, it is necessary to solve the parenthesis, multiplying the exponents:
Now we can multiply. We maintain the base and add the exponents
At the end, we can also solve the power.
Let’s see another example:
In principle, we have four bases: 2, 3, 4 and 9.
We want all powers to have the same base or the minimum number of bases possible. To do this, we must break down into prime factors the numbers that can be expressed in this way in the equation.
In this case we can break down 4 and 9, which we indicate in the equation as 2² and 3²:
We are left with two bases: 2 and 3.
The next step is to remove parentheses, multiplying the outer exponents by the inner exponents:
In the numerator we have two powers with base 2 multiplied, so we keep the base and add the exponents. We do the same in the denominator with two base powers 3:
Nos has remained a division of powers of base 2 and another of base 3. For each one we maintain the base and subtract the exponents:
Y with this we have finished simplifying the expression, since we do not have any negative exponent.
## Operations with high powers at other powers
Let us now see the steps to follow when we have multiplications or divisions with powers, which in turn are elevated to another power, such as:
We begin by multiplying the powers within the parenthesis:
Nos has been raised to another power. So now we multiply exponents:
We have made the negative exponent positive by passing it to the denominator.
We continue with a high power division at a negative exponent:
We begin operand within the parenthesis, subtracting the exponents:
We are left with one power raised to another power, so we multiply the exponents:
Let’s see a last example, in which we have all the operations with powers that we have seen until now:
First, we apply the property of power multiplication in the numerator and denominator. We maintain the base and add the exponents:
We are left with a division of powers. We maintain the base and subtract the exponents:
We are left with one power raised to another. Maintain the base and multiply the exponents:
At the end we have a power with negative exponent, which we turn positive by passing it to the denominator. Once we have the positive exponent, we can solve the power:
## Operations with powers from different base elevated to other powers
We are going to see the steps to follow when you have to simplify an operation in which you have multiplications and divisions of different base, which are also part of another power, as for example:
In the first place we simplify as much as possible inside the parenthesis.
Same as before, on the one hand we simplify the numbers and on the other hand, with each base x and y, we maintain the bases and subtract the exponents:
We can no longer operate within the parenthesis, so we proceed to resolve the parenthesis.
To solve the parenthesis, you have to multiply the exponent from outside by each of the exponents inside, according to this property:
Multiplying exponents leaves us:
Finally, we have to express the solution with all positive exponents.
We have negative exponents in the numerator and denominator.
I remind you that the powers with negative exponent that are in the numerator pass to the denominator with positive exponent and vice versa, according to this property:
Applied to our equation we have:
We finish the operation by solving the base power 2.
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1. ## solving equations
Pls help...pls show me the steps to solve the following probs...thanks lots..
Qn1: Limin bought 2.5 metres of cloth. Each dress used
metres. If she made 3 dresses, how much cloth
was left?
Qn2: The ratio of men to women in a food court is 4:3. The ratio
of women to children is 5:2. If there are 24 children, how
many men are there?
Qn3: of Limin's money can buy 40 books. How many books
can she buy with of her money?
2. Originally Posted by Bryan
Qn1: Limin bought 2.5 metres of cloth. Each dress used $\displaystyle \frac34$ metres. If she made 3 dresses, how much cloth
was left?
I'll do this first one completely:
One dress uses $\displaystyle \frac34$ meters of cloth, so three dresses will use $\displaystyle 3\cdot\frac34 = \frac94$ meters. So if she started with $\displaystyle 2.5 = \frac52$ meters of cloth, she will have
$\displaystyle \frac52 - \frac94 = \frac{10}4 - \frac94 = \frac14$
meters of cloth remaining.
Originally Posted by Bryan
Qn2: The ratio of men to women in a food court is 4:3. The ratio
of women to children is 5:2. If there are 24 children, how
many men are there?
Let the number of men, women, and children, be $\displaystyle m, w,\text{and }c,$ respectively.
Then we know that $\displaystyle \frac mw = \frac43$ and that $\displaystyle \frac wc = \frac52$.
But you also know that $\displaystyle c = 24$. Substitute this into the second relation, and you can solve for the number of women. Then substitute that back into the first equation and you will be able to solve for the number of men.
Originally Posted by Bryan
Qn3: $\displaystyle \frac25$ of Limin's money can buy 40 books. How many books
can she buy with $\displaystyle \frac34$ of her money?
We will have to assume that these books are all the same price. Let that price be $\displaystyle c$.
Suppose Limin has $\displaystyle x$ units of money (dollars, pounds, euros, whatever--makes no difference here). Then we know that two-fifths of this amount will total to the cost of 40 books: $\displaystyle \frac{2x}5 = 40c$.
Solving for $\displaystyle x$: $\displaystyle x = 40c\cdot\frac52 = 100c$. Now, $\displaystyle \frac34$ of her money will buy $\displaystyle n$ books: $\displaystyle \frac{3x}4 = nc$. Substitute for $\displaystyle x$ and solve for $\displaystyle n$.
(Of course, if you wanted to be picky, the initial relation should actually be $\displaystyle 40c\le\frac{2x}5$; the answer should work out the same)
,
,
### the ratio of men to women in a food court is 4:3. the ratio of women to children is 5:2. if there are 24 children, how many men are there? anwers
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Resource Lesson Vertical Circles and Non-Uniform Circular Motion
Unlike horizontal circular motion, in vertical circular motion the speed, as well as the direction of the object, is constantly changing. Gravity is constantly either speeding up the object as it falls, or slowing the object down as it rises.
We will begin by looking at two special positions which are usually analyzed in problems: the very top of a vertical circle and the very bottom of the circle.
Suppose a block is being whirled on the end of a string in a vertical circle. Let's look at the freebody diagrams of the forces acting on the block at the top of the circle and at the bottom of the circle. Remember when drawing freebody diagrams for objects moving in circular motion that the net force on the object is ALWAYS directed towards the center of the circle, no matter where the object is located in its circular path.
topbottom
Top of the circle:
Let's calculate the tension in the string at the top of the circle. Notice that both of the forces T and mg are directed downwards towards the center. Since our block is in circular motion, we know that the NET FORCE must act towards the center of the circle.
That is, the NET FORCE towards the center, or the centripetal force, is the resultant or SUM of these two REAL forces. You should never label Fc in a freebody diagram!
net force to the center = T + mg
Fc = T + mg
m(v2/r) = T + mg
T = m(v2/r) - mg
If we wanted to calculate the minimum or critical velocity needed for the block to just be able to pass through the top of the circle without the rope sagging then we would start by letting the tension in the rope approaches zero.
0 = m(v2/r) - mg
m(v2/r) = mg
v2/r = g
v2 = rg
v = √(rg)
Refer to the following information for the next three questions.
Consider a 1-kg brick being whirled in a vertical circle at the end of a 1-meter rope.
What critical velocity must the brick achieve in order to pass safely through the top of its circular path?
What is the tension in the rope as it is passing through the top of its circular path?
How would be the critical velocity of the brick if it were to be whirled on the moon where the acceleration due to gravity is 1/6th that on earth?
Bottom of the circle:
Now let's calculate the tension in the string at the bottom of the circle. Since the block is maintaining a circular path, we take the direction towards the center as positive. The NET FORCE acting towards the center, Fc, is the resultant force or the difference between T and mg since they now point in opposite directions.
net force to the center = T - mg
Fc = T - mg
m(v2/r) = T - mg
T = m(v2/r) + mg
This formula will be used frequently to calculate the tension in the string in a simple pendulum as the pendulum bob swung through its lowest position - the equilibrium position, the point of greatest KE.
Refer to the following information for the next question.
Consider a 1-kg brick being whirled in a vertical circle at the end of a 1-meter rope.
If the brick is being whirled so that it just passes through the top of its circular path without allowing the rope to sag, then what tension will be in the rope as the brick passes through the lowest point in its path?
Notice that the tension in the string is GREATEST as the block passes through the bottom of the circle and LEAST while it passes through the top of the circle.
If you were asked to calculate the tension at any intermediate point in a pendulum's swing, then the net force to the center would equal T - mg cos θ.
net Fc = T - mg cos θ
mv2/r = T - mg cos θ
T = mv2/r + mg cos θ
Notice that components of the weight are taken, not components of the tension. It is a component of the weight vector that accelerates the bob towards equilibrium.
In order to solve for tension, you would have to use conservation of energy techniques to first solve for the velocity at the requested intermediate position.
Refer to the following information for the next two questions.
Suppose a 1-meter pendulum is released from a horizontal position as shown below.
What would be the speed of the bob just as it reaches an arc equal to 37º?
What would be the tension in the string just as the bob reaches an arc equal to 37º?
Now, consider an example of a person riding a roller coaster through a circular section of the track, a "loop-the-loop."
Let's look at the formulas needed to calculate the normal force, N, exerted on a object traveling on the inside surface of a vertical circle as it passes through the bottom and through the top of the ride.
At the top:
net force to the center = N + mg
N + mg = m(v2/r)
N = m(v2/r) - mg
While at the bottom:
net force to the center = N - mg
N - mg = m(v2/r)
N = m(v2/r) + mg
If we let the value of normal approach zero in the formula for the top of the roller coaster we would get the same value for the critical velocity that we got when solving for the tension in the string in our previous discussion, v = √(rg). This principle of critical velocity is used in many places. When you watch clothes drying in a dryer, they are being rotated in a vertical circle. But the rate of rotation does not allow the clothes to achieve this critical value as they pass through the top of the circle. Therefore the clothes fall away from the drum and are "fluffed" as they spin. In roller coasters, this critical velocity is a safety threshold. The coasters MUST exceed this minimum value in order to be certified. Obviously, no one would want the cars to fall away from the rails as the participants experience a thrill passing through a "loop-the-loop" section of the track!
The value of the normal at the bottom of the ride is equivalent to questions asking about the apparent weight of a pilot as he pulls out of a vertical drive. The expression a = v2/r + g is often called the "g's" a pilot is experiencing.
Note that the normal, N, appears to play the same role as the tension, T, in our equations for vertical circular motion.
Refer to the following information for the next question.
While driving to work you pass over a "crest" in the road that has a radius of 30 meters.
How fast would you need to be traveling to experience apparent "weightlessness" while passing over the crest?
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# 3.3: Least Squares Solution
Let $$x$$ be a particular solution of (1a). Denote by $$x_{A}$$ its unique projection onto the range of $$A$$ (i.e. onto the space spanned by the vectors $$a_{i}$$) and let $$x_{A^{\perp}}$$ denote the projection onto the space orthogonal to this. Following the same development as in the proof of the orthogonality principle in Lecture 2, we find
$x_{A}=A \prec A, A \succ^{-1} \prec A, x \succ \ \tag{3a}$
with $$x_{A^{\perp}}= x- x_{A}$$. Now (1a) allows us to make the substitution $$y= \prec A, x \succ$$ in (3a), so
$x_{A}=A \prec A, A \succ^{-1} y\ \tag{3b}$
which is exactly the expression we had for the solution $$\check{x}$$ that we determined earlier by inspection, see (2b).
Now note from (3b) that $$x_{A}$$ is the same for all solutions $$x$$, because it is determined entirely by $$A$$ and $$y$$. Hence it is only $$x_{A^{\perp}}$$ that is varied by varying $$x$$. The orthogonality of $$x_{A}$$ and $$x_{A^{\perp}}$$ allows us to write
$<x, x>=<x_{A}, x_{A}>+<x_{A^{\perp}}, x_{A^{\perp}}>\nonumber$
so the best we can do as far as minimizing $$<x,x>$$ is concerned is to make $$x_{A^{\perp}}=0$$. In other words, the optimum solution is $$x = x_{A} = \check{x}$$.
Example 3.3
For the FIR filter mentioned in Example 3.1, and considering all input sequences $$x[k]$$ that result in $$y[0] = 7$$, find the sequence for which $$\sum_{i=-N}^{N} x^{2}[i]$$ is minimized. (Work out this example for yourself!)
Example 3.4
Consider a unit mass moving in a straight line under the action of a force $$x(t)$$, with position at time $$t$$ given by $$p(t)$$. Assume $$p(0)=0, \dot{p}(0)=0$$, and suppose we wish to have $$p(T ) = y$$ (with no constraint on $$\dot{p}(T )$$). Then
$y=p(T)=\int_{0}^{T}(T-t) x(t) d t=<a(t), x(t)> \ \tag{4}$
This is a typical underconstrained problem, with many choices of $$x(t)$$ for $$0 \leq t \leq T$$ that will result in $$p(T ) = y$$. Let us find the solution $$x(t)$$ for which
$\int_{0}^{T} x^{2}(t) d t=<x(t), x(t)> \ \tag{5}$
is minimized. Evaluating the expression in (2a), we find
$\check{x}(t)=(T-t) y /\left(T^{3} / 3\right) \ \tag{6}$
How does your solution change if there is the additional constraint that the mass should be brought to rest at time $$T$$, so that $$\dot{p}(T ) = 0$$?
We leave you to consider how weighted norms can be minimized.
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Divisibility tests
Lesson
"Divisible by" means "when you divide one number by another the result is a whole number." For example $77$77is divisible by $7$7 because $77\div7=11$77÷7=11. We could also say that $7$7 and $11$11 are factors of $77$77 (see Breaking Down Numbers for more information on factors).
Remember, we can use our knowledge of multiplication and division to complete sets of related number facts.
Example
$13\times4=52$13×4=52
$4\times13=52$4×13=52
$52\div4=13$52÷4=13
$52\div13=4$52÷13=4
Quick checks of divisibility
Divisible by 2
All even numbers are divisible by $2$2.
Example
$28$28 is even, so it is divisible by $2$2.
$28\div2=14$28÷2=14
Divisibility by 3
A number is divisible by $3$3 if the sum of the digits is divisible by $3$3.
Example
$234$234 is divisible by $3$3 since $2+3+4=9$2+3+4=9.
$234\div3=78$234÷3=78
Divisibility by 5
All numbers ending with a $5$5 or a $0$0 are divisible by $5$5.
Example
$85$85 ends with a $5$5, so it is divisible by $5$5.
$85\div5=17$85÷5=17
Divisibility by 9
A number is divisible by $9$9 if the sum of the digits is divisible by $9$9.
Example
$324$324 is divisible by $9$9 since $3+2+4=9$3+2+4=9
$324\div9=36$324÷9=36
Divisibility by 10
All numbers ending in a 0 are divisible by 10.
Example
$5200$5200 ends in a $0$0, so it is divisible by $10$10.
$5200\div10=520$5200÷10=520
Worked Examples
Question 1
Are the following numbers divisible by $4$4?
1. $58784$58784
Yes
A
No
B
2. $372938$372938
Yes
A
No
B
3. $58832$58832
Yes
A
No
B
Question 2
Which of the following numbers are exactly divisible by 3?
1. $73500$73500
A
$73495$73495
B
2. $545824$545824
A
$599772$599772
B
Question 3
Are the following numbers divisible by $8$8?
1. $195992$195992
Yes
A
No
B
2. $1669488$1669488
Yes
A
No
B
3. $1669374$1669374
Yes
A
No
B
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# Graphing Rational Functions: Definition, Examples
Mar 8, 2022
Do you know how graphing simple rational functions can be done with ease? You all must be acquainted with fractions. The top portion of a fraction is known as the numerator, and the bottom is known as the denominator. The numerator of a rational function reveals the x-intercept of the graph. On the other hand, the denominator reveals the vertical asymptotes of the graph. The factors of the numerator have integer powers greater than one in polynomials
## Graphing Rational Functions – 1/x
Can you guess which of the following rational functions is graphed below? The vertical asymptotes linked to the denominator’s factors will mirror one of the two reciprocal functions. You must know that when the degree of the factor in the denominator is odd, the graph heads positive infinity upwards on one side of the vertical asymptote. In contrast, the other side is towards negative infinity.
## Graphing Rational Functions – 1/x2
Here’s another example. Can you guess one more time which of the following rational functions is graphed below? You must know when the degree of the denominator is even; the characteristic graph either heads toward the positive infinity on both sides of the vertical asymptote or towards the negative infinity on both sides.
Let us understand the above graphing rational function by an example. Suppose, the fraction is f (x) = (x + 1)2 (x – 3)/(x + 3)2 (x – 2), then, the graph of the function will be:
This graph is drawn by putting 1, 2,…, and so on to get the values for the function to plot a graph. Here a key points that can be drawn from the graph above:
You must have noticed that at the x-intercept x = -1 corresponding to the (x + 1)2 factor present in the numerator, the graph consistently bounces with the quadratic nature of the factor.
Another thing to notice is that at the x-intercept x = 3 corresponding to the (x-3) factor of the numerator, the graph passes through the axis as in the case of a linear factor.
Also, at the vertical asymptote, x = -3 corresponding to the (x + 3)2 factor of the denominator, the graph is consistent with the behavior of the function f (x) = 1/x2. Due to this, the graph heads up on both sides of the asymptote.
Lastly, at the vertical asymptote x = 2, corresponding to the (x – 2) factor in the denominator, consistent behavior of the function f (x) = 1/x is followed. The graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side.
### How to Graph Rational Functions? – Key Tips to Follow
If you have complexities making graphs of rational functions, then it’s time to follow these key tips listed below:
1. You must evaluate the function in the first place. Evaluate it at 0 to find the y-intercept.
2. Next, you need to factor in the numerator and denominator.
3. After factoring, for factors in the numerator not common to the denominator, you need to determine where each factor is zero to find the x-intercepts.
4. After performing step 3, find the multiplicities of the x-intercepts. You are doing this step to determine the behavior of the graph at those points.
5. You need to note the multiplicities of the zeroes to determine the local behavior for the factors in the denominator. It would help to find the vertical asymptotes for the uncommon factors to the numerator. To do this, you need to set those factors equal to zero and then solve them.
6. After getting zeroes in step 5, you need to find the removable discontinuities by setting those factors equal to 0. This must be done for the factors in the denominator common to the factors in the numerator.
7. Now, compare the numerator and denominator degrees to determine the slant or horizontal asymptotes.
8. Finally, you can sketch the graph you wanted to.
Let us learn graphing simple rational functions via an example.
Example: Sketch a graph for the function, f (x) = (x + 2) (x – 3)/(x + 1)2 (x -2) .
Solution: You can follow the steps to sketch the graph for the following function:
Step 1: The first step to sketch the graph is to factor the function. Since the function is already factored, it saves your first step.
Step 2: You need to find the intercepts. Evaluate the function at zero to get the y-intercept as shown,
f (0) = (0 + 2) (0 – 3)/(0 + 1)2 (0 – 2)= 3.
Step 3: After getting the y-intercept, to find the x-intercept, determine the value when the numerator of the function is zero. Set each factor equal to zero, then find the x-intercepts at x = -2 and x = 3. You will see the behavior of the function is linear (multiplicity of 1) along with the graph through the intercept.
Step 4: Now, you have x and y-intercepts as (-2, 0) and (3, 0), and (0, 3), respectively.
Step 5: Next, to get the vertical asymptotes, you need to keep the value of the denominator equal to zero. That will occur when x + 1 = 0 and x – 2 = 0, giving the value for vertical asymptotes at x = -1 and x = 2. Since there are no common factors in the numerator and denominator, there will be no removable discontinuities.
Step 6: At last, the degree of the denominator is larger than the degree of the numerator, which tells that the graph has a horizontal asymptote at y = 0.
Step 7: To sketch the graph, you need to plot the three intercepts. Since you know that the y-intercept is positive and there is no x-intercept between the vertical asymptotes, you get to know that the function must remain positive between the asymptotes. It will let you fill the middle portion of the graph.
Step 8: The factor associated with the vertical asymptote at x = -1 is squared, so you know the behavior of such a function will be the same on both sides of the asymptote. The graph directs towards positive infinity as the inputs approach the asymptote on the right. This way, the graph will also head towards the positive infinity on the left.
Step 9: The factor is not squared in the vertical asymptote at x = 2. Hence, the graph will have opposite behavior on either side of the asymptote. After the graph passes the x-intercepts, you will see that the graph will then level off toward an output of zero. This was, however, indicated by the horizontal asymptote.
Try it yourself!
### After Graphing Rational Functions, How to Write Them?
If you are given a graph of a rational function and need to write the function, you must check the intercepts in the first place. If a rational function has x-intercepts at x = x1, x2, x3…, xn, vertical asymptotes at x = v1, v2, …vm, and no xi = any vj, then the function can be written in the form:
f (x) = a(x – x1)p1 (x – x2)p2… (x – xn)pn(x – v1)q1(x – v2)q2… (x – vm)qn
Where the behavior of the graph can find the powers of pi or qi on each factor at the corresponding intercept or asymptote, you can also find the stretch factor ‘a’ from the given value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero.
### After Graphing Simple Rational Functions, Write the Function
The steps to write the function after graphing simple rational functions are illustrated below:
Step 1: You need to determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to find the zeros and their multiplicities. This step is easy to find the simplest function with small multiplicities, such as 1 or 3. It gets complicated when larger numbers, such as 8 or 9, come into the picture.
Step 2: Next, you need to determine the factors of the denominator. Examine the behavior of both sides of each vertical asymptote to determine the factors and their powers.
Step 3: After performing steps 1 and 2, the last step is to use any clear point on the graph to find the stretch factor.
It is crucial to learn graphing rational functions as they can be useful tools for representing real-life situations where you need to find the solutions to complex problems. The equations representing inverse, direct or joint variation are rational formulas that can model many real-life situations. Hence, it would help if you practiced graphing rational functions.
### 1. What is the Definition of a Rational Function?
Ans. A rational function is a function of the form f(x) = p(x)/q(x), where p(x) and q(x) are polynomials. Rational functions are important because they can be expressed as the quotient of two polynomials, which makes them easy to factor.
### 2. How Do You Know If a Function is Rational?
Ans. If you want to know if a function is rational, you can either look at its graph or at its equation. If the graph of a function contains only vertical and horizontal lines, then it is rational. If the equation contains only integers, then it is rational.
### 3. How to Graph a Rational Function?
Ans. You can graph rational functions by using a vertical line test. The vertical line test is a way to determine whether a rational function is defined at a given point. It’s based on the fact that if you graph two different points on the function, then draw a vertical line connecting them, you’ll get a straight line if the function is defined at those points.
### 4. How to Find Asymptotes of Rational Functions?
Ans. Asymptotes of rational functions are the roots of the denominator, and they are always an integer.To find the asymptotes of a rational function, first, isolate the denominator by taking it outside of brackets. Then test all the roots of that denominator for being an integer. If you find an integer root, then you have found your asymptote.
### 5. What are the Applications of Rational Function?
Ans. Rational functions are the most common type of polynomial functions, and they can be applied in many different ways. One application of rational functions is in solving systems of linear equations. Rational functions can also be used to represent the relationship between two variables on a graph, known as a phase line.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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Dividing Polynomials
# Dividing Polynomials
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## Dividing Polynomials
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##### Presentation Transcript
1. Dividing Polynomials Section 2.4
2. Objectives • Divide two polynomials using either long division or synthetic division. • Use the Factor Theorem to show that x-c is a factor of a polynomial. • Use the Remainder Theorem to evaluate a polynomial at a given value.
3. Vocabulary • quotient • remainder • dividend • divisor
4. Division Algorithm If f (x) and d (x) are polynomials, with d(x) ≠ 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exists unique polynomials q(x) and r(x) such that f(x) = d(x) · q(x) + r(x) The remainder, r(x), equals 0 or it is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly into f(x) and that d(x) and q(x) are factors of f(x).
5. Find the quotient and remainder using long division.
6. Find the quotient and remainder using long division.
7. Find the quotient and remainder using synthetic division.
8. Remainder Theorem If the polynomial f (x) is divided by x – c, then the remainder f (c).
9. Use synthetic division to show that x = 6 is a root of
10. Factor Theorem • Let f (x) be a polynomial • If f(c) = 0, then x – c is a factor of f (x). • If x – c,is a factor of f(x), then f(c) = 0.
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## FANDOM
1,140 Pages
A fraction is a number expressed as the quotient of two mathematical expressions, often integers. Fractions come in the form $\frac{a}{b}$ , where $a$ is called the numerator and $b$ is called the denominator.
Sometimes the word "fraction" means the quotient of two integers. Under this definition, $\frac34$ is a fraction, but $\frac1{\pi+\sqrt2}$ is not.
## Types of fractions
If the numerator is smaller than the denominator, it is called a proper fraction, and is less than one. An example of this is $\frac35$ .
If the numerator is larger than the denominator, it is called an improper fraction, and is greater than one. An example of an improper fraction is $\frac{51}{2}$ , which is 25 wholes and a half; [1] $25\frac12$ as a mixed number.
In addition, $\frac{a}{b}=a\div b$ . For example, $2\frac63$ would be $2\times (6\div3)$ .
## Manipulating fractions
Multiplying fractions is very easy:
$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$
For $a\ne0$ , $\frac{a}{a}=1$ . This gives us a useful identity:
$\frac{a}{b}=\frac{a}{b}\cdot\frac{c}{c}=\frac{a\cdot c}{b\cdot c}$
allowing us to "force" values of the numerator or denominator. Some call this multiplying by one creatively.
Adding fractions is somewhat harder. If both the summands have the same denominator, we can use the distributive property:
$\frac{a}{b}+\frac{c}{b}=a\cdot\frac1b+c\cdot\frac1b=(a+c)\cdot\frac1b=\frac{a+c}{b}$
But if they're different, we can multiply by one creatively to force their denominators to line up:
$\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+bc}{bd}$
This formula is not terribly useful to memorize; knowing the technique is much more helpful. If we're working with integers, instead of $bd$ in the denominator, we can use the least common multiple of $b,d$.
### Example
$\frac29+\frac56=\frac{12}{54}+\frac{45}{54}=\frac{57}{54}=\frac{19}{18}$
Note that we could have taken a shortcut using the least common multiple:
$\frac29+\frac56=\frac4{18}+\frac{15}{18}=\frac{19}{18}$
We multiplied the first fraction by 2/2 and the second by 3/3.
## Conversions, mixed numbers and improper fractions
Explain in more detail what mixed numbers and improper fractions are.
You can convert a mixed number (a, surprisingly, mixed fraction — $5\frac45$ is a mixed number) into an improper fraction (a fraction in which the numerator is larger than the denominator) by doing three things:
1. Multiply the whole number (the $5$) by the denominator (the other $5$) of the fractional part. In the example above, to change $5\frac45$ to a mixed number, first multiply $5\times5=25$ .
2. Add the numerator of the fractional part to that product. In the example, $4+25=29$ .
3. The resulting sum is the numerator of the new (improper) fraction, with the denominator remaining the same in both the mixed number and the improper fraction. In the example, $5\frac45=\frac{29}{5}$ .
Likewise, you can convert an improper fraction into a mixed number by following another 3 rules:
1. Divide the numerator by the denominator. Let's say, for example, $\frac{32}{7}$. $32\div7=4$ with 4 left over (the remainder)
2. The quotient (without the remainder) becomes the whole number part of the mixed number. The remainder becomes the numerator of the fractional part. In the example, 4 is the whole number part and 4 is the numerator of the fractional part.
3. The new denominator is the same as the denominator of the improper fraction. In the example, the denominator is 7. Thus $\frac{32}{7}=4\frac47$ .
## Rationalizing the denominator
In elementary algebra, we are sometimes asked to rationalize the denominator of an expression like this one:
$\frac{1+\sqrt5}{\sqrt{11}-\sqrt7}$
To rationalize the denominator is to create a fraction equal to this one with an integer in the denominator.
As with adding fractions, we multiply by one creatively. The trick here is to use the difference of squares factorization $(a-b)(a+b)=a^2-b^2$ . We can turn the $\sqrt{11}-\sqrt7$ into a whole number by multiplying it by $\sqrt{11}+\sqrt7$ :
$(\sqrt{11}-\sqrt7)(\sqrt{11}+\sqrt7)=\sqrt{11}^2-\sqrt7^2=11-7=4$
So by multiplying our fraction by $\frac{\sqrt{11}+\sqrt7}{\sqrt{11}+\sqrt7}$ , we get a nice, round 4 in the denominator:
$\frac{1+\sqrt5}{\sqrt{11}-\sqrt7}=\frac{1+\sqrt5}{\sqrt{11}-\sqrt7}\cdot\frac{\sqrt{11}+\sqrt7}{\sqrt{11}+\sqrt7}=\frac{(1+\sqrt5)(\sqrt{11}+\sqrt7)}{(\sqrt{11}-\sqrt7)\cdot(\sqrt{11}+\sqrt7)}=\frac{\sqrt{11}+\sqrt7+\sqrt{55}+\sqrt{35}}{4}$
## Fractal fractions
Fractal fractions are fractions on which the numerator and/or the denominator consist of infinite other fractions. A simple fraction or number can be expressed as an infinite fraction in this way:
1. Express a number as a fraction with its numerator or denominator including the number. All square roots have this property, for example:
$\sqrt2=\frac2{\sqrt2}$
2. Now, replace the part of the fraction with the chosen number with the fraction itself. This step has to be repeated infinitely on the new fractions:
$\sqrt2=\frac2{\frac2{\sqrt2}}$
$\sqrt2=\frac2{\frac2{\frac2{\frac2{\ldots}}}}$
Then, the fractions can be split into a sum of two different fractions.
### Fractal fraction simplification
To convert an infinite fraction to a simpler one or a number equations must be used since taking a part of the infinite part of the fraction doesn't lead to the result it corresponds to:
$\frac22=1\ne\sqrt2$
$\frac2{\frac22}=\frac21=2\ne\sqrt2$
The equation in this example is:
$x=\frac2{\frac2{\frac2{\ldots}}}\rArr x=\frac2x\rArr x^2=2\rArr x=\sqrt2$
## References
1. Wikipedia:Fraction (mathematics)#Mixed numbers.
1. Divide the numerator by the denominator. In the example, $\tfrac{11}{4}$ , divide 11 by 4. $11\div 4=2$ with remainder 3.
2. The quotient (without the remainder) becomes the whole number part of the mixed number. The remainder becomes the numerator of the fractional part. In the example, 2 is the whole number part and 3 is the numerator of the fractional part.
3. The new denominator is the same as the denominator of the improper fraction. In the example, they are both 4. Thus $\tfrac{11}{4}=2\tfrac34$"
'''''''''''This article has excessive material copied from Wikpedia. It would be better if we could write our own articles, so help reword this article if you can. The talk page may contain suggestions.
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# A triangle has corners at (3 ,1 ), (7 ,-5 ), and (-4 ,-2 ). If the triangle is dilated by a factor of 5 about point (7 ,-6 ), how far will its centroid move?
Feb 5, 2018
Distance moved by centroid
$\vec{G G '} \approx \textcolor{g r e e n}{26.54}$ rounded to two decimals
#### Explanation:
Given : A (3,1), B (7,-5), C(-4,-2)
Dilated about D(7,-6) and dilation factor 5
To find distance the centroid has moved
$C e n t r o i d$G = (3+7+(-4))/3, (1-5-2)/3 = color(brown)((2,-2)
$\vec{A ' D} = 5 \cdot \vec{A D}$
$a ' - d = 5 \left(a - d\right)$ or $a ' = 5 a - 4 d$
$\implies 5 \left(\begin{matrix}3 \\ 1\end{matrix}\right) - 4 \left(\begin{matrix}- 4 \\ - 2\end{matrix}\right) = \left(\begin{matrix}15 \\ 5\end{matrix}\right) - \left(\begin{matrix}- 16 \\ - 8\end{matrix}\right) = \left(\begin{matrix}31 \\ 13\end{matrix}\right)$
color(blue)(A' (31, 13)
Similarly,
$\vec{B ' D} = 5 \cdot \vec{B D}$
$b ' - d = 5 \left(b - d\right)$ or $b ' = 5 b - 4 d$
$\implies 5 \left(\begin{matrix}7 \\ - 5\end{matrix}\right) - 4 \left(\begin{matrix}- 4 \\ - 2\end{matrix}\right) = \left(\begin{matrix}35 \\ - 25\end{matrix}\right) - \left(\begin{matrix}- 16 \\ - 8\end{matrix}\right) = \left(\begin{matrix}51 \\ 17\end{matrix}\right)$
color(blue)(B' (51, 17)
$\vec{C ' D} = 5 \cdot \vec{C D}$
$c ' - D = 5 \left(c - d\right)$ or $c ' = 5 c - 4 d$
$\implies 5 \left(\begin{matrix}- 4 \\ - 2\end{matrix}\right) - 4 \left(\begin{matrix}- 4 \\ - 2\end{matrix}\right) = \left(\begin{matrix}- 20 \\ - 10\end{matrix}\right) - \left(\begin{matrix}- 16 \\ - 8\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 2\end{matrix}\right)$
color(blue)(C' (-4, -2)
New centroid G' = (31 + 51-4)/3, (13+17-2)/3 = color(brown)((26, 28/3)
Distance moved by centroid is
$\vec{G G '} = \sqrt{{\left(2 - 26\right)}^{2} + {\left(- 2 - \left(\frac{28}{3}\right)\right)}^{2}} \approx \textcolor{g r e e n}{26.54}$ rounded to two decimals
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# Quadratics in the xb Plane
### Sean Johnston
When looking at quadratic equations, we often apply the quadratic formula to the standard form, a+ bx + c = 0, to find the roots. In this assignment, we will explore the roots of quadratics with particular values of a and c. We will first look at the case when a = 1 and c = 1, giving us the equation + bx + 1 = 0. We can graph this in the xb plane to get the following hyperbola:
The hyperbola actually tells us the values of the roots of the equation + bx + 1 = 0 for particular values of b. We can do this by looking at the horizontal line segments of the graph and associating the horizontal line segment with its b value. When b = 2 or b = -2, we get exactly 1 root. When -2 < b < 2, there are no real roots. When b < 2, we have 2 positive roots. When b > 2, we have 2 negative roots. we can see that the hyperbola is asymptotic to the lines x = 0 and y = -x. This tells us that we will never have a negative root less than -b or a positive root greater than b.
Next, we will look at similar hyperbolas for different values of c:
The red hyperbolas are the graphs of + bx + 1 = 0, + bx + 2 = 0, and + bx + 3 = 0. + bx + 1 = 0 is the red hyperbola closest to the origin, and + bx + 3 = 0 is the red hyperbola furthest from the origin.
The blue hyperbolas are the graphs of + bx - 1 = 0, + bx - 2 = 0, and + bx - 3 = 0. + bx - 1 = 0 is the blue hyperbola closest to the origin, and + bx - 3 = 0 is the blue hyperbola furthest from the origin.
The graph of + bx + 0 = 0, or + bx = 0 is actually the union of the lines x = 0, and b = -x (both in black). Interestingly enough, all of the hyperbolas formed from changing the values of c are asymptotic to + bx + 0 = 0.
First, we will focus on the hyperbolas. In the case of the red hyperbolas, we see that as we increase the value of c, we also increase the amount of space where no real roots occur. We also see that when b is positive, we always have either two positive roots or two negative roots, or no real roots.
In the case of the blue hyperbolas (which are the hyperbolas with a negative value for c), there is always one positive root and one negative root. We also see that as we decrease the value of c, we increase the distance between the hyperbolas, which increases the distance between the positive root and the negative root.
Now we will look at the meaning of the line 2x + b = 0 or b = -2x (shown in purple). This line represents the arithmetic mean of the two roots for any values of b and c where real roots exist. This is important because the x coordinate of the vertex of a parabola always lies on the arithmetic mean of the two roots. For example, we can use the graph to see that when b = 4 and c = 3, the top of the red hyperbolas intersect the line b = 4 at -1 and -3, which means the roots of + 4x - 3 are -1 and -3. This tells us that the x coordinate of the vertex is -2. A quicker way to do this would have been to simple use the equation b = -2x and plug in 4 for b to find that x = -2 is the location of the vertex of the parabola.
Return
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# Introduction To Trigonometry Exercise 8.4
### CLICK HERE to watch second part
Q.1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Sol. Consider a $\Delta ABC$, in which $\angle B = 90^\circ$.
For $\angle A$, we have
Base = AB, Perp = BC
and Hyp = AC
Therefore, $\cot A = {{Base} \over {Perp}} = {{AB} \over {BC}}$
$\Rightarrow {{AB} \over {BC}} = \cot A = {{\cot A} \over 1}$
Let AB = k cotA and BC = k.
By Pythagoras Theorem,
$AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{k^2}{{\cot }^2}A + {k^2}}$
$= k\sqrt {1 + {{\cot }^2}A}$
Therefore, $\sin A = {{Perp} \over {Base}} = {{BC} \over {AC}} = {k \over {k\sqrt {1 + {{\cot }^2}A} }}$
$= {1 \over {\sqrt {1 + {{\cot }^2}A} }}$
$\sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}} = {{k\sqrt {1 + {{\cot }^2}A} } \over {k\cot A}}$
$= \sqrt {{{1 + {{\cot }^2}A} \over {\cot A}}}$
and, $\tan A = {{Perp} \over {Base}} = {{BC} \over {AB}} = {k \over {k\cot A}} = {1 \over {\cot A}}$
Q.2 Write the other trigonometric ratios of A in terms of secA.
Sol. Consider a $\Delta ABC$, in which $\angle B = 90^\circ$.
For $\angle A,$ we have
Base = AB, Perp = BC
and Hyp = AC
Therefore, $\sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}}$
$\Rightarrow {{AC} \over {AB}} = \sec A = {{\sec A} \over 1}$
$\Rightarrow {{AB} \over {AC}} = {1 \over {\sec A}} = {{{1 \over {\sec A}}} \over 1}$ (Note this step)
Let $AB = k\left( {{1 \over {\sec A}}} \right),AC = k(1)$
By phythagoras Theorem
$BC = \sqrt {A{C^2} + A{B^2}} = \sqrt {{k^2} - {k^2}\left( {{1 \over {{{\sec }^2}A}}} \right)}$
$= k\sqrt {{{{{\sec }^2}A - 1} \over {{{\sec }^2}A}}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec }^2}A}}$
Therefore, $\sin A = {{BC} \over {AC}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec A} \over k}}} = {{\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}$
$\cos A = {{AB} \over {AC}}{{k\left( {{1 \over {\sec A}}} \right)} \over k} = {1 \over {\sec A}}$
$\tan A = {{BC} \over {AB}} = {{{{k\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}} \over {k\left( {{1 \over {\sec A}}} \right)}} = \sqrt {{{\sec }^2}A - 1}$
$\cot A = {1 \over {\tan A}} = {1 \over {\sqrt {{{\sec }^2}A - 1} }}$
$\cos ecA = {1 \over {\sin A}} = {{\sec A} \over {\sqrt {{{\sec }^2}A - 1} }}$
Q.3 Evaluate :
(i) ${{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ } \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ }}$
(ii) sin 25° cos 65° + cos25° sin65° $[Since,sin\left( {90^\circ - \theta } \right) = \cos \theta ]$
Sol. (i) Here, sin63° = sin(90° – 27°) = cos27°
and cos17° = cos(90° – 73°) = sin 73° $[Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]$
Therefore, ${{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ} \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ}} = {{{{\cos }^2}27^\circ + {{\sin }^2}27^\circ }\over {{{\sin }^2}73^\circ + {{\cos }^2}73^\circ}}$
$= {1 \over 1} = 1$
$[Since,{\cos ^2}A + {\sin ^2}A = 1]$
(ii) sin25°cos65°+cos25°sin65°
= sin(90°– 65°). cos65° + cos(90°– 65°) sin65°
= cos65° cos65° + sin 65° sin65°
$[Since,\sin \left( {90^\circ - \theta } \right) = \cos \theta ]$
= ${\cos ^2}65^\circ + {\sin ^2}65^\circ = 1$ $[Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]$
Q.4 Choose the correct option. Justify your choice :
(i) $9{\sec ^2}A - 9{\tan ^2}A =$
(A) 1 (B) 9
(C) 8 (D) 0
(ii) (1+ tan$\theta$ + sec$\theta$)(1 + cos $\theta$ – cosec $\theta$) =
(A) 0 (B) 1
(C) 2 (D) None of these
(iii) (secA + tanA)(1 – sinA) =
(A) secA (B) sinA
(C) cosecA (D) cosA
(iv) ${{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}} =$
(A) ${{{\sec }^2}A}$ (B) –1
(C) ${{{\cot }^2}A}$ (D) none of these
Sol. (i) (B), because
$9{\sec ^2}A - 9{\tan ^2}A = 9\left( {{{\sec }^2}A - {{\tan }^2}A} \right)$
= 9 × 1 = 9 $\left[ {Since,1 + {{\tan }^2}A = {{\sec }^2}A} \right]$
(ii) (C), because
$\left( {1 + \tan \theta + \sec \theta } \right)\left( {1 + \cot \theta - \cos ec\theta } \right)$
$= \left( {1 + {{\sin \theta } \over {\cos \theta }} + {1 \over {\cos \theta }}} \right)\left( {1 + {{\cos \theta } \over {\sin \theta }} - {1 \over {\sin \theta }}} \right)$
$= \left( {{{\cos \theta + \sin \theta + 1} \over {\cos \theta }}} \right)\left( {{{\sin \theta + \cos \theta - 1} \over {\sin \theta }}} \right)$
= ${{{{\left( {\cos \theta + \sin \theta} \right)}^2} - 1} \over {\sin \theta \cos \theta }}$ $[Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]$
= ${{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta} \right) - 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }}$ $[Since,{\sin ^2}\theta + {\cos ^2}\theta = 1]$
= ${{1 + 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }} = {{2\cos \theta \sin \theta } \over {\sin \theta \cos \theta }} = 2$
(iii) (D), because
(secA + tanA) (1 – sinA)
= $\left( {{1 \over {\cos A}} + {{\sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right)$
= $\left( {{{1 + \sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right)$ $[Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]$
$= {{1 - {{\sin }^2}A} \over {\cos A}}{{{{\cos }^2}A} \over {\cos A}} = \cos A[Since,{\sin ^2}A + {\cos ^2}A = 1]$
(iv) (D), because
${{1 + {{\tan }^2}A} \over {1 - {{\cot }^2}A}} = {{1 + {{\tan }^2}A} \over {1 + {1 \over {{{\tan }^2}A}}}} = {{1 + {{\tan }^2}A} \over {{{{{\tan }^2}A + 1} \over {{{\tan }^2}A}}}}$
$= \left( {1 + {{\tan }^2}A} \right) \times {{{{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {\tan ^2}A$
Q.5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined :
(i) ${\left( {\cos ec\theta - \cot \theta } \right)^2} = {{1 - \cos \theta } \over {1 + \cos \theta }}$
(ii) ${{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}} = 2\sec A$
(iii) ${{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta$
(iv) ${{1 + \sec A} \over {\sec A}} = {{{{\sin }^2}A} \over {1 - \cos A}}$
(v) ${{\cos A - \sin A+ 1} \over {\cos A + \sin A - 1}} = \cos ecA + \cot A,$ using the identity $\cos e{c^2}A = 1 + {\cot ^2}A$
(vi) $\sqrt {{{1 + \sin A} \over {1 - \sin A}} = } \sec A + \tan A$
(vii) ${{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = \tan \theta$
(viii) ${{\rm{(sinA + cosecA)}}^2} + {{\rm{(cosA + secA)}}^2} = 7 + {\tan ^2}A + {\cot ^2}A$
(ix) (cosecA – sinA)(secA – cosA) = ${1 \over {\tan A + \cot A}}$
(x) $\left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {\left( {{{1 - \tan A} \over {1 - {{\cot }^2}A}}} \right)^2} = {\tan ^2}A$
Sol. (i) We have,
L.H.S. = ${{\rm{(cosec}}\theta - \cot \theta {\rm{ )}}^2}$
$= {\left( {{1 \over {\sin \theta }} - {{\cos \theta } \over {\sin \theta }}} \right)^2} = {\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)^2}$
$= {{{{\left( {1 - \cos \theta } \right)}^2}} \over {{{\sin }^2}\theta }} = {{{{\left( {1 - \cos \theta } \right)}^2}} \over {1 - {{\cos }^2}\theta }}$ $[Since,{\sin ^2}\theta = 1 - {\cos ^2}\theta ]$
$= {{{{\left( {1 - \cos \theta } \right)}^2}} \over {\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} = {{1 - \cos \theta } \over {1 + \cos \theta }}$
R.H.S. $[Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$
(ii) We have,
L.H.S. = ${{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}}$
= ${{{{\cos }^2}A + {{\left( {1 + \sin A} \right)}^2}} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{{{\cos }^2}A + 1 + 2\sin A + {{\sin }^2}A} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{\left( {{{\cos }^2}A + {{\sin }^2}A} \right) + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{1 + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}[Since {\sin ^2}A + {\cos ^2}A = 1]$
= ${{2 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}} = {{2\left( {1 + \sin A} \right)} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${2 \over {\cos A}} = 2\sec A = R.H.S.$
(iii) We have,
L.H.S. = ${{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }}$
= ${{\tan \theta } \over {1 - {1 \over {\tan \theta }}}} + {{{1 \over {\tan \theta }}} \over {1 - \tan \theta }}$
$= {{\tan \theta } \over {{{\tan \theta - 1} \over {\tan \theta }}}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}$
$= {{{{\tan }^2}\theta } \over {\tan \theta - 1}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}$
$= {{{{\tan }^2}\theta } \over {\tan \theta - 1}} - {1 \over {\tan \theta \left( {\tan \theta - 1} \right)}}$
= ${{{{\tan }^3}\theta - 1} \over {\tan \theta \left( {\tan \theta - 1} \right)}}$
= ${{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + \tan \theta + 1} \right)} \over {{{\tan }}\theta \left( {\tan \theta - 1} \right)}}$
$[Since,{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)]$
= ${{{{\tan }^2}\theta + \tan \theta + 1} \over {{{\tan }^2}\theta }}$
= ${{{{\tan }^2}\theta } \over {\tan \theta }} + {{\tan \theta } \over {\tan \theta }} + {1 \over {\tan \theta }}$
= $\tan \theta + 1 + \cot \theta = 1 + \tan \theta + \cot \theta$
= $1 + {{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}$
= $1 + {{{{\sin }^2}\theta + {{\cos }^2}\theta } \over {\cos \theta }}$
= $1 + {1 \over {\sin \theta \cos \theta }} = 1 + \cos ec\theta \sec \theta$
= R.H.S.
(iv) R.H.S. = ${{{{\sin }^2}A} \over {1 - \cos A}} = {{1 - {{\cos }^2}A} \over {1 - \cos A}}$
$[Since,{\sin ^2}A = 1 - {\cos ^2}A]$
= ${{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \over {1 - \cos A}} = 1 + \cos A$
$[Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$
= $1 + {1 \over {\sec A}} = {{1 + \sec A} \over {\sec A}} = L.H.S.$
(v) L.H.S. = ${{\cos A - \sin A + 1} \over {\cos A + \sin A - 1}} = {{{{\cos A - \sin A + 1} \over {\sin A}}} \over {{{\cos A + \sin A - 1} \over {\sin A}}}}$
= ${{\cos A - 1 + \cos ecA} \over {\cos A + 1 - \cos ecA}}$
$[Since,1 + {\cot ^2}A = \cos e{c^2}A]$
$= {{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)} \over {\cot A - \cos ecA + 1}}$
= ${{\cot A + \cos ecA - \left( {\cos ecA + \cot A} \right)\left( {\cos ecA - \cot A} \right)} \over {\cot A - \cos ecA + 1}}$
$[Since{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$
Taking common(cosecA + cotA)
= ${{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot } \right)} \over {\left( {\cot A - \cos ecA + 1} \right)}}$
= cosec A + cot A
= R.H.S.
(vi) We have,
L.H.S. = $\sqrt {{{1 + \sin A} \over {1 - \sin A}}} = \sqrt {{{1 + \sin A} \over {1 - \sin A}} \times {{1 + \sin A} \over {1 + \sin A}}}$
[Multiplying and dividing by ] $\sqrt {1 + \sin A}$
= $\sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {1 - {{\sin }^2}A}}} = \sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {{{\cos }^2}A}}}$ $[Since,\,{\sin ^2}A + {\cos ^2}A = 1]$
= ${\sqrt {\left( {{{1 + \sin A} \over {\cos A}}} \right)} ^2} = {{1 + \sin A} \over {\cos A}}$
= ${1 \over {\cos A}} + {{\sin A} \over {\cos A}} = \sec A + \tan A$
= R.H.S. $\left[ {Since,\,\tan A = {{\sin A} \over {\cos A}}} \right]$
(vii) We, have,
L.H.S. = ${{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = {{\sin \theta (1 - 2{{\sin }^2}\theta )} \over {\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2\left( {1 - {{\sin }^2}\theta } \right) - 1}}} \right]$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2 - 2{{\sin }^2}\theta - 1}}} \right]$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {1 - 2{{\sin }^2}\theta }}} \right] = \tan \theta \times 1$
= $\tan \theta$ = R.H.S.
(viii) We have,
L.H.S. = ${\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2}$
= $({\sin ^2}A + \cos e{c^2}A + 2\sin A\cos ecA) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A\sec A} \right)$
= $\left( {{{\sin }^2}A + \cos e{c^2}A + 2\sin A.{1 \over {\sin A}}} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A.{1 \over {\cos A}}} \right)$
= $\left( {{{\sin }^2}A + \cos e{c^2}A + 2} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2} \right)$
= ${\sin ^2}A + {\cos ^2}A + \cos e{c^2}A + {\sec ^2}A + 4$ $[Since,\,\,{\sin ^2}A + {\cos ^2}A = 1]$
= $1 + \left( {1 + {{\cot }^2}A} \right) + \left( {1 + {{\tan }^2}A} \right) + 4$
= $7 + {\tan ^2}A + {\cot ^2}A$
= $[Since,\,\cos e{c^2}A = 1 + {\cot ^2}A\,\,and\,{\sec ^2}A = 1\, + \,{\tan ^2}A]$
= R.H.S.
(ix) We have,
L.H.S. = (cosec A – sinA) (secA – cosA)
= $\left( {{1 \over {\sin A}} - \sin A} \right)\left( {{1 \over {\cos A}} - \cos A} \right)$
= $\left( {{{1 - {{\sin }^2}A} \over {\sin A}}} \right)\left( {{{1 - {{\cos }^2}A} \over {\cos A}}} \right)$
= ${{{{\cos }^2}A} \over {\sin A}} \times {{{{\sin }^2}A} \over {\cos A}}$
= sinA cosA
= ${{\sin A\cos A} \over {{{\sin }^2}A + {{\cos }^2}A}}$ $[Since,{\sin ^2}A + {\cos ^2}A = 1]$
Dividing Numerator and Denominator by sinA cosA
${{{{\sin A\cos A} \over {\sin A\cos A}}} \over {{{{{\sin }^2}A} \over {\sin A\cos A}} + {{{{\cos }^2}A} \over {\sin A\cos A}}}}$
= ${1 \over {{{\sin A} \over {\cos A}} + {{\cos A} \over {\sin A}}}}$
= ${1 \over {\tan A + \cot A}}$ = R.H.S.
(x) We have,
L.H.S. = $\left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {{{{\sec }^2}A} \over {\cos e{c^2}A}}$
= ${1 \over {{{\cos }^2}A}} \times {{{{\sin }^2}A} \over 1} = {\tan ^2}A$
R.H.S. = ${\left( {{{1 - \tan A} \over {1 - \cot A}}} \right)^2} = {\left( {{{1 - \tan A} \over {1 - {1 \over {\tan A}}}}} \right)^2}$
= ${\left( {{{1 - \tan A} \over {{{\tan A - 1} \over {\tan A}}}}} \right)^2} = {\left( { - \tan A} \right)^2} = {\tan ^2}A$
Therefore, L.H.S. = R.H.S.
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# Difference between revisions of "2009 AIME II Problems/Problem 1"
## Problem
Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.
## Solution
### Simple Solution
Let the stripes be $b, r, w,$ and $p$, respectively. Let the red part of the pink be $\frac{r_p}{p}$ and the white part be $\frac{w_p}{p}$ for $\frac{r_p+w_p}{p}=p$.
Since the stripes are of equal size, we have $b=r=w=p$. Since the amounts of paint end equal, we have $130-b=164-r-\frac{r_p}{p}=188-w-\frac{w_p}{p}$. Thus, we know that $$130-p=164-p-\frac{r_p}{p}=188-p-\frac{w_p}{p}$$ $$130=164-\frac{r_p}{p}=188-\frac{w_p}{p}$$ $$r_p=34p, w_p=58p$$ $$\frac{r_p+w_p}{p}=92=p=b.$$ Each paint must end with $130-92=38$ oz left, for a total of $3 \cdot 38 = \boxed{114}$ oz.
### Solution 1
After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$. The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$. The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \boxed{114}$.
### Solution 2
We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are $188 - r$ and $164 - r$ respectively. The pink stripe is also r ounces of paint, but let there be $k$ ounces of red paint in the mixture and $r - k$ ounces of white paint. We now have two equations: $164 - r - k = 188 - r - (r-k)$ and $164 - r - k = 130 - r$. Solving yields k = 34 and r = 92. We now see that there will be $130 - 92 = 38$ ounces of paint left in each can. $38 * 3 = \boxed{114}$
### Solution 3
Let the amount of paint each stripe painted used be $x$. Also, let the amount of paint of each color left be $y$. 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, $164 + 188 = 352$ and obtain the following equations : $352 - 3x = 2y$ and $130 - x = y$. Solve to obtain $x = 92$. Therefore $y$ is $130 - 92 = 38$, with three cans of equal amount of paint, the answer is $38 * 3 = \boxed{114}$.
### Solution 4
Let $x$ be the number of ounces of paint needed for a single stripe. We know that in the end, the total amount of red paint and white paint combined will be double the amount of blue paint (since it is given that Bill has equal amounts of each color left).
The total amount of red and white paint remaining is equal to $164+188-2x$, and then minus another $x$ for the pink stripe. The amount of blue paint remaining is equal to $130-x$. So, we get the equation $2\cdot(130-x)=164+188-3x$. Simplifying, we get $x=38$ and $3x=\boxed{114}$.
~LegionOfAvatars
## Solution 5
Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$. The answer is $130+164+188-4(92)=\boxed{114}$
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# Knots have been studied extensively by mathematicians for the last hundred years. One of the most peculiar things which emerges as you study knots is.
## Presentation on theme: "Knots have been studied extensively by mathematicians for the last hundred years. One of the most peculiar things which emerges as you study knots is."— Presentation transcript:
Knots have been studied extensively by mathematicians for the last hundred years. One of the most peculiar things which emerges as you study knots is how a category of objects as simple as a knot could be so rich in profound mathematical connections Knot Theory is the mathematical study of knots. A mathematical knot has no loose or dangling ends; the ends are joined to form a single twisted loop. Knots
The Reidemeister Moves Reidemeister moves change the projection of the knot. This in turn, changes the relation between crossings, but does not change the knot. 3. Slide a strand from one side of a crossing to the other: 2. Add or remove two crossings (lay one strand over another): 1.Take out (or put in) a simple twist in the knot:
Famous Knots TrefoilFigure EightUnknot In order to talk mathematically about knots, I have to show them in some kind of way, to have a method of describing them. I did this for the simplest knots by using a piece of string or rope, which nicely shows the 3-dimensional nature of the object. Here are some of the most famous knots, all known to be inequivalent. In other words, none of these three can be rearranged to look like the others. However, proving this fact is difficult. This is where the mathematics comes in.
Crossings – What are they? Each of the places in a knot where 2 strands touch and one passes over (or under) the other is called a crossing. The number of crossings in a knot is called the crossing number. A zero knot has 0 crossings A trefoil knot has 3 crossings. On the left shows a picture with 9 crossings
Prime Knots The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,.... Any number can be written as a product of a set of prime numbers. Here is an example: 60 = 2 x 2 x 3 x 5 = 2 x 5 x 3 x 2. The number 60 determines the list 2, 2, 3, 5 of primes, but not the order in which they are used. The same is true for knots. A prime knot is one that is not the sum of simpler knots. To work out the number of knots with a number of crossings, a table is given below where it compares the number of prime knots against crossing number n3456789101112 Number of prime knots with n crossings 1123721491655532176
Torus Knots – What are they? Torus is the mathematical name for an inner tube or doughnut. It is a special kind of knot which lies on the surface of an unknotted torus. Each torus knot is specified by a pair of coprime integers p and q. The (p,q)-torus knot winds q times around a circle inside the torus, which goes all the way around the torus, and p times around a line through the hole in the torus, which passes once through the hole, On the left/right shows a picture called (15,4) torus knot because it is wrapped 15 times one way and 4 times the other
(p,q) torus knots The (p,q)-torus knot can be given by the parameterization This lies on the surface of the torus given by (r − 2)2 + z2 = 1
Arithmetic of knots From this we can make a general rule about the addition of knots: K + L = L + K. This is called commutativity. Below shows how you add 2 knots together
Here is a collection of torus knots arranged according to crossing number
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# AP Statistics: Section 9.1 Sampling Distributions
## Presentation on theme: "AP Statistics: Section 9.1 Sampling Distributions"— Presentation transcript:
AP Statistics: Section 9.1 Sampling Distributions
What is the usual way to gain information about some characteristic of a population?
We must note, however, that the sample information we gather may differ from the true population characteristic we are trying to measure. Furthermore, the sample information may differ from sample to sample.
This sample-to-sample variability, called ____________________________ poses a problem when we try to generalize our findings to the population. We need to gain an understanding of this variability.
A parameter is A statistic is
In statistical practice, the value of a parameter is unknown since we cannot examine the entire population. In practice, we often use a statistic to estimate an unknown parameter.
The population mean is represented by the symbol ___ (Greek: Mu), the population standard deviation by ___(Greek: Sigma) and the population proportion by ___.
The sample mean is represented by the symbol ____ (x bar), the sample standard deviation by ____ and the sample population by ____ (p hat).
Example: Identify the number that appears in boldface type as a parameter or a statistic, and then write an equation using the proper symbol from above and the number from the statement
A department store reports that 84% of all customers who use the store’s credit plan pay their bills on time.
A consumer group, after testing 100 batteries of a certain brand, reported an average of 63 hr of use.
We can view a sample statistic as a random variable, because we have no way of predicting exactly what statistic value we will get from a sample, BUT, given a population parameter, we know how these sample statistics will behave in repeated sampling.
Before we continue, we need to discuss two important definitions:
The population distribution of a variable is the distribution of values of the variable among all individuals in the population.
The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the population.
Careful: The population distribution describes the individuals that make up the population. A sampling distribution describes how a statistic varies in many samples of size n from the population.
Consider flipping a coin 10 times
Consider flipping a coin 10 times. We would expect to get 5 heads out of the 10, but we realize that we could also get 4 or 6 or 7 or even 10. Let’s simulate this using our graphing calculators.
How many different samples of size 10 are possible in this situation?
Let’s increase our sample size to 25.
Hopefully, most of us found that as we increased the sample size from 10 to 25, the mean and the median of our sample proportions became closer together and both became closer to .5. Also we should find that the standard deviation grows smaller and our distribution of the sample proportions became closer to being a normal distribution.
Since a sampling distribution is a distribution, we can use the tools of data analysis to describe the distribution: ________, ________, __________ and __________.
Example: According to 2005 Nielsen ratings, Survivor: Guatemala was one of the most-watched TV shows in the US during every week that it aired. Suppose that the true proportion of US adults who watched Survivor: Guatemala was p = 0.37.
Describe the distribution of sample proportions at the right for samples of size n = 100 of people who watched Survivor: Guatemala.
Describe the distribution of sample proportions for samples of size n = 1000 of people who watched Survivor: Guatemala.
A statistic used to estimate a parameter is unbiased if the mean of its sampling distribution equals the true value of the population parameter. The statistic is called an unbiased estimator of the parameter.
An unbiased statistic will sometimes fall above the true value of the parameter and sometimes below. There is no tendency to overestimate or underestimate the parameter, hence the “unbiased.”
We will see in sections 9. 2 & 9
We will see in sections 9.2 & 9.3 that are both unbiased estimators of population parameters.
The variability of a statistic is described by the spread of its sampling distribution.
This spread is determined by the sampling design and the sample size
This spread is determined by the sampling design and the sample size. Larger samples give a ________ spread.
As long as the population is larger than the sample by at least a factor of 10, the spread of the sampling distribution is approximately the same for any population size.
This means that a statistic from an SRS of size 2500 from the more than 300 million residents of the US is just as precise as an SRS of size 2500 from the 750,000 inhabitants of San Francisco.
For a better understanding of bias and variability, think of the center of a target as the true population parameter and an arrow shot at the target as a sample statistic.
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# How much is 60 percent of nine tenths
What is % of
### Percentage Calculator 2
is what percent of ?
is % of what?
### Solution for 'What is 60% of nine tenths?'
In all the following examples consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
#### Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 60 percent of 0.9. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 0.9.
Step 2: determine the percentage, which is 60.
Step 3: Convert the percentage 60% to its decimal form by dividing 60 into 100 to get the decimal number 0.6:
60100 = 0.6
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
60.0 → 6.00 → 0.60
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.6 x 0.9 = 0.54 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 0.9 × 60100 = 0.9 × 0.6 = 0.54 (answer)
Therefore, the answer is 0.54 is 60 percent of 0.9.
### Solution for '60 is what percent of nine tenths?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
#### Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 0.9.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 600.9
X% = 60000.9
X% = 6,666.67 (answer)
So, the answer is 60 is 6,666.67 percent of 0.9
### Solution for '0.9 is 60 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
### Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 0.9.
Step 2: identify the percent, which is 60.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 0.960
W = 100 × 0.015
W = 1.5 (answer)
The answer, in plain words, is: 0.9 is 60% of 1.5.
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#### Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to $\frac{(a+c)(b+c-2 a)}{2(b-a)}$
Here
$a\textsubscript{1} = a, a\textsubscript{2} = b\\$
$\\ d = a\textsubscript{2} - a\textsubscript{1}\\ = b - a \\ a\textsubscript{n} = c \: \: \: \: \: \: \: \: \: \: \: \: \because a\textsubscript{n} =a + (n - 1)d \\ a\textsubscript{1} + (n - 1).d = c\: \: \: \: \: Put a\textsubscript{1} = a , d= b - a \\ (n - 1) (b - a) = c - a$
$\\ \mathrm{n}-1=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}+1 \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}+\mathrm{b}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right] \\ =\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{2(\mathrm{~b}-\mathrm{a})}[\mathrm{a}+\mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ =\frac{(\mathrm{c}+\mathrm{b}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right] \quad\left[\because \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{a}_{\mathrm{n}}\right] \\ \frac{(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})} \quad \quad\left[\because \mathrm{a}_{\mathrm{n}}=\mathrm{c}\right]$
Hence proved.
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# GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Gujarat Board GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Question 1.
A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean of plants per house.
Which method did you use for finding the mean and why?
Solution:
$$\bar{x}$$ = $$\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$$ = $$\frac{162}{20}$$ = 8.1 plants
Direct method for finding the mean as numerical values of x. and f are small.
Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Using the step-deviation method
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h= 150 + $$\left(\frac{-12}{50}\right)$$ x 20 = 150 – 4.8 = 145.20
Hence, the mean daily wages of the workers of the factory is 145.20.
Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.
Solution:
Using the direct method,
$$\bar{x}$$ = $$\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$$
⇒ 18 = $$\frac{20 f+752}{f+44}$$
⇒ 20f + 752 = 18(f + 44)
⇒ 20f + 752 = 18f + 792
⇒ 20f – 18f = 792 – 752
2f = 40
f = $$\frac {4}{10}$$ = 20
Hence, the missing frequency is 20.
Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heartbeat per minute for these women, choosing a suitable method.
Solution:
Using the step-deviation method
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h = 75.5 + ($$\frac {4}{30}$$) x 3 = 75.5 + 0.4 = 75.9
Hence, the mean heart beat per minute is 75.9.
Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.
Find the mean number of mangoes kept in a packing box.
Which method of finding the mean did you choose?
Solution:
Using the step-deviation method,
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h = 57 + $$\left(\frac{25}{400}\right)$$ x 3 = 57 + 0.19 = 57.19
Question 6.
The table below shows the daily expenditure on the food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Solution:
Using the step-deviation method
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h = 225 + ($$\frac {-7}{25}$$) x 50 = 225 – 14 = ₹ 211
Hence, the daily expeiiditure on food is 211.
Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain n city and is presented below. Find the mean concentration of SO2 in the air.
Solution:
Using the step-deviation method,
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h = 0.14 + ($$\frac {-31}{30}$$) x 0.04 = 0.14 – 0.041 = 0.099 ppm
Therefore, the mean concentration of SO2 in the air is 0.099 ppm.
Question 8.
A class teacher has the following absent record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Solution:
Using direct method
$$\bar{x}$$ = $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ = $$\frac {499}{40}$$ = 12.47
Hence, the mean number of days a student was absent is 12.47.
Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Solution:
Using the step-deviation method
$$\bar{x}$$ = a + $$\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$$ x h = 70 + $$\frac {-2}{35}$$ x 10 = 70 – $$\frac {4}{7}$$ = 70 – 0.57 = 69.43%
Hence, the literacy rate is 69.43%.
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# In an $A.P.$ the first term is $2$ and the sum of first $5$ terms is one-fourth of the next five terms. Find its $20^{th}$ term.
$\begin{array}{1 1}112 \\ -112 \\ 116 \\ -116 \end{array}$
Toolbox:
• Sum of first $n$ terms of an $A.P.$=$S_n=\large\frac{n}{2}$$[2a+(n-1)d] • t_n=a+(n-1)d Given: In the A.P. First term=a=2 and Sum of first five terms =one fourth of sum of the next five terms. Sum of next 5 terms=t_6+t_7+.....t_{10}=S_{10}-S_5 \Rightarrow\:S_5=\large\frac{1}{4}$$(S_{10}-S_5)$...........(i)
We know that $S_n=\large\frac{n}{2}$$[2a+(n-1)d] \Rightarrow\:S_5=\large\frac{5}{2}$$[2\times 2+(5-1)d]$
$\therefore\:S_5=10+10d$ and
$S_{10}=\large\frac{10}{2}$$[2\times 2+(10-1)d] S_{10}=5(4+9d) \therefore\: By substituting the values of S_5\:and\:S_{10} in (i) we get 10+10d=\large\frac{1}{4}$$(20+45d-10-10d)$
$\Rightarrow\:40+40d=10+35d$
$\Rightarrow\:d=-6$
We know that $t_n=a+(n-1)d$
$\therefore\:t_{20}=2+(20-1)(-6)=-112$
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# Question Video: Changing the subject of a Formula Mathematics
The variables 𝑡 and 𝑚 are related by the formula 𝑡 = 𝑚/(𝑚 + 1). Make 𝑚 the subject.
02:57
### Video Transcript
The variables 𝑡 and 𝑚 are related by the formula 𝑡 is equal to 𝑚 divided by 𝑚 plus one. Make 𝑚 the subject.
In order to rearrange or change the subject of a formula, we need to use our knowledge of inverse operations and the balancing method. In this question, we want to rearrange the formula so that 𝑚 is equal to an expression not containing 𝑚. As multiplication is the opposite of division, we begin by multiplying both sides of the equation by 𝑚 plus one. On the left-hand side, we have 𝑚 plus one multiplied by 𝑡, which can also be written as 𝑡 multiplied by 𝑚 plus one.
On the right-hand side, the 𝑚 plus ones cancel. So we are left with 𝑡 multiplied by 𝑚 plus one is equal to 𝑚. Distributing the parentheses, otherwise known as expanding the brackets, means we need to multiply 𝑡 by 𝑚 and 𝑡 by one. This gives us 𝑡𝑚 plus 𝑡. Therefore, 𝑡𝑚 plus 𝑡 is equal to 𝑚.
Our next step is to get all our terms with 𝑚 in on one side of the equals sign. We can do this by subtracting 𝑡𝑚 from both sides. We are now left with 𝑡 is equal to 𝑚 minus 𝑡𝑚. We notice that 𝑚 is common in both terms on the right-hand side. Therefore, we can factor or factorize this out. 𝑡 is therefore equal to 𝑚 multiplied by one minus 𝑡.
Our final step is to divide both sides of this equation by one minus 𝑡. We leave the left-hand side as 𝑡 divided by one minus 𝑡. Canceling the one minus 𝑡 on the numerator and denominator of the right-hand side gives us 𝑚. 𝑚 is therefore equal to 𝑡 divided by one minus 𝑡. This could also be written as negative 𝑡 divided by 𝑡 minus one. Fractions are equivalent if we multiply the numerator and denominator by the same value.
In this case, we have multiplied the numerator and denominator by negative one to switch between our two answers. 𝑚 is equal to 𝑡 divided by one minus 𝑡 or negative 𝑡 divided by 𝑡 minus one.
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Must read:Is there a real future in data analysis for self-learners without a math degree?
# Foundations for inference
If you see an error in the article, please comment or drop me an email.
## Getting confused with the basics of inference?
If you are a newbie to statistics, or just have not dealt with such kind of topic for a while, you might get confused by the many new terms, some of which sound the same while others actually mean the same.
Even the basics of inference are quite dense. If you read this article, you might have already started a class on statistical inference. If you need a clear wrap-up, this short read might help you cobble the different parts together.
As it seems, this is a crucial chapter in statistics, which we will have to master before moving on to other, more complex topics.
## About how many distributions are we actually talking?
When reading through class material, articles and forum discussions, always make sure to know with which distribution is being dealt. It helps understand which statistic pertains to which kind of distribution.
1 The (individual) sample distribution(s)
2 The distribution of sample means (=sampling distribution)
3 The population distribution (which we can only estimate)
## From the Sample to the Confidence interval
### Sample
First we take a sample of n observations from a population.
### Point estimate (sample means)
We calculate the mean of each sample. These mean values are called point estimates and form the sampling distribution.
### Sample Standard Deviation
Now, back to one of our samples. As we already calculated its mean we can now determine the Sample Standard Deviation, which is based on the distance of each observation from the sample mean.
### Estimate of the Population Standard Deviation
We do not have the Population Standard Deviation (sigma), but the Standard Deviation of the sampling distribution (which is NOT the sample standard deviation) provides a useful estimate of it.
According to the Central Limit Theorem, you can calculate the Standard Deviation of the Sampling distribution by using the standard deviation of the sample distribution and the sample size:
sdsd = \[sqrt(var)\] = \[sqrt(sigma^2/n)\] = \[ sigma/sqrt(n)\]
### Standard Error
The Standard Error depends on both the standard deviation of the sampling distribution (expressing how much the sample means vary from each other) and the size of the samples (number n of observations in each sample). Logically, the standard error decreases as the sample size increases.
### Confidence Interval
A confidence Interval measures how certain we can be that the population parameter is in a certain interval. We calculate the Confidence interval by multiplying the z-score by the Standard Error. We obtain the z score by using qnorm (provided the distribution is normal).
## Quick Reminders:
• The sampling distribution is in fact the distribution of means of various samples. The whole process goes a little like this : many samples are taken from a population -> Their respective means are calculated -> We call the sample means point estimates -> These point estimates form the sampling distribution -> Sampling distributions are normally-distributed.
• The standard error is short for the standard error of means (thus of the sampling distribution). A high standard error, for instance, means that when taking a sample of a population and calculating its mean, this statistic (the mean) is quite “far away” from the true mean (the correct mean of the population). Therefore, the whole sample might not be an accurate estimator of the population. The standard error of the sampling distributions is NOT its standard deviation, but rather another standard of deviation.* The standard error is likely to increase when taking small samples.
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# Sense-Making in Mathematics: Understanding Fractions
Fractions. A single word that deflates the confidence of our most competent students and adults alike. In fact, I am sure that many readers are reliving their own fraction experiences as they read this post– good or bad! I can’t say that I myself don’t have certain feelings about my own experiences with learning fractions, but over the years, I have developed a deeper understanding of fraction concepts through my experience as a teacher, a mathematics education graduate student and also as a math coach. Today, I want to highlight different fraction tools that can be used to help students make sense of fractions and address specific Common Core State Standards for Math.
Pattern Blocks
Common Core State Standard 3.NF.A.1- Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a quantity a/b as the quantity formed by a parts as size 1/b.
The CCSSM standard above requires students to understand how to relate an individual piece to the whole using the formal fraction notation. The yellow hexagon, red trapezoid, blue rhombus, and green triangle shapes in a pattern block set fit together perfectly to model fractional relationships. Pattern blocks can be used to help students make the connection between the size of the part, named fraction, and the number of parts that make up the whole. The “What’s the Relationship?” chart below illustrates how this can be accomplished.
Fraction Tiles
Common Core State Standards:
• 4.NB.B.3.- Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
• 4.NF.B.3.A.- Understand addition and subtraction of fractions as joining and separating parts referring to the same whole
• 4.NF.B.3.B- Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g. by using a visual fraction model.
Building a strong foundation for fractions in the early grades provides long-term support for the development of fraction operations in the upper grades. Students begin fraction operations by decomposing fractions into fractions with the same denominator. Understanding how to decompose fractions supports the following essential fraction understandings:
• understand the meaning and purpose of the numerator, denominator, and unit (or whole)
• addition of fractions is an extension of whole number addition and involves joining or combining disconnected parts
• addition sequences can represent different problem situations
• the interpretation of a problem situation can lead to different representations
• the denominators of the addends remain the same and are not added together
• addition and subtraction of fractions refer to the same unit, or whole
The activity I want to highlight involves multiple ways to decompose fractions using fraction tiles and an 8-part spinner labeled with fractional quantities. Students spin the spinner, model how to decompose the fraction with fraction tiles, and write a number sentence showing the decomposition of the fraction using fractions with the same denominator.
After spinning for the initial fraction and showing one decomposition (with an equation), students complete the following:
• show the decomposition another way
• write a story to match the equation
The poster below illustrates an efficient way for students to demonstrate their understanding.
Fraction Bars
Common Core State Standard 5.NF.A.2- Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers.
Students struggle to understand how to make sense of the value of a fraction. A few years back, in an effort to try to help my fourth graders really make connections between the value of a fraction and the formal fraction notation, I taught them how to compare the fraction to a benchmark. Each time I presented a fraction, I posed the question, “Is this fraction closer to zero, one-half, or one whole?” And, I often added, “How do you know?” Using benchmark fractions gave the students a reference to use in their comparisons.
It took some time, but I began to notice that my students’ understanding of fractions developed into the deeper understanding I had envisioned. (An understanding that prevents the dreaded one-third plus one-third equals two-sixths because using benchmark fractions will allow students to see that one-half (the benchmark for one-third) plus one-half equals one whole. Therefore, the solution for one-third plus one-third should be closer to one whole. Two-sixths is closer to zero, not one whole.)
It’s important to note that students don’t just develop this understanding without beginning with the conceptual models. Students need lots of opportunities to make the comparisons using fraction tools before being able to make a visual estimation from the formal notation. See the examples using fraction bars below.
After students have explored the conceptual model, they can record their learning on a graphic organizer. The example below shows a graphic organizer where students record the formal fraction notation in the correct column and then “sum up” their learning by answering a question related to what they notice about the relationship of the parts of the fraction (the numerator and the denominator) and the benchmark fraction.
Like the resource pages above?
You can find them in the Common Core Aligned Resources section of my file cabinet.
Sound Off! What are your favorite fraction tools? Respond in the comments section below.
## You may also like...
### 3 Responses
1. Vishna Patel says:
Dear Shametria,
I am a preservice math teacher and I believe many students may benefit if their teachers had a deeper understanding of fractions. I see in your post you have addressed the standards and multiple different activities students can participate in to gain a better understanding of fractions. How do you suggest teachers can help make sense of what it means when multiplying fractions? For example, how would you explain to students why you multiply across the top and multiply across the bottom when multiplying fractions?
Thank you,
Vishna Patel
1. Great question! A deep understanding of fractions begins with a strong conceptual approach. Working with students to understand what it means to multiply with fractions and how to create a model that illustrates the concept is most important. When students begin connecting the model to a written method, they will begin to understand the standard algorithm. For example, think about three groups of two-fifths (3/1 x 2/5). With a model, students will discover that three groups of two-fifth equals six-fifths. When you connect to the written method, students will see that 3 x 2 = 6 and 1 x 5 = 5, so six-fifths. After completing enough examples, especially with a fraction times a fraction, the algorithm develops itself!
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The quadratic formula is not as complicated as it may look. You can learn to tackle this formula and succeed in using it to help you solve your quadratic problems. Watch this video to see how.
I used to hate working with the quadratic formula until I realized that it actually helped me with my problems. And sometimes, it even made my problem shorter as it gave me what I needed to know early on in the solving process. Right now, the quadratic formula might look like a beast, but think of it as a prince in disguise. It’s here to help you. The quadratic formula allows you to solve quadratic equations in the form of ax^2 + bx + c = 0 and is defined as x = (-b +/- sqrt(b^2 – 4ac)) / 2a.
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The a, b, and c letters come from the quadratic equation that you are solving. Let’s see how this works.
## Two Solutions
The first problem we’re going to solve is x^2 + 5x + 6 = 0. How do we figure out our letters? Well, we compare this to the general form of a quadratic: ax^2 + bx + c. A good way to do this is to write the general form above your quadratic so you can see how everything lines up. What we are looking for is the location of our x^2, our x, and our number by itself. We can then figure out what our letters stand for.
In our general form, the number attached to the x^2 is the letter a. So, looking at my quadratic, I see that there is nothing in front of the x^2, which means there is a 1 there, so that means my a = 1. Good.
Next, I look for what is front of the x. I see a 5, so that means my b = 5. The number by itself is a 6, so my c = 6. Now I have all my letters. If, though, you have missing terms, then that means the letter that matches that missing term is equal to zero. At this point, the quadratic formula turns into a prince to help you. To solve, all you need to do is plug in all your letters where they belong and evaluate the formula.
Let’s do this.
We plug in our numbers to get x = (-5 +/- sqrt(5^2 – 4*1*6)) / 2*1. We begin evaluating by first calculating what is inside the square root.
x = (-5 +/- sqrt(25 – 24)) / 2*1
x = (-5 +/- sqrt (1)) / 2*1
Look at that. We have a 1 inside the square root. It is a positive number inside the square so we know we can calculate that down. If we know the square root off the top of our head, we can go ahead and write that number down. Otherwise, you can use a calculator to find the exact square root. In our case, we know that the square root of 1 is 1, so we’ll write that down.
x = (-5 +/- 1) / 2*1
Our formula includes a +/-, so we have to split our problem into two problems, one for the plus and one for the minus. So:
x = (-5 + 1) / 2*1 and x = (-5 – 1) / 2*1
Now, we can finish solving these two little problems.
x = -4 / 2 and x = -6 / 2
So:
x = -2 and x = -3
And yes, we have two solutions. Most times, you will end up with two solutions like this. If they aren’t whole numbers, then they’ll be decimal numbers. The way you can tell that you have two solutions like this is if you get a positive number inside your square root.
There are two more cases that will give you different numbers of solutions. Let’s see what happens when the square root ends up as a 0.
## One Repeated Solution
This time, our problem is x^2 + 4x + 4 = 0. Comparing this to our general form of ax^2 + bx + c, we see that our a = 1, our b = 4, and our c = 4. Plugging these values into our quadratic formula gives us x = (-4 +/- sqrt(4^2 – 4*1*4)) / 2*1. Evaluating the square root, we find that it equals zero.
x = (-4 +/- sqrt(16 – 16)) / 2*1
x = (-4 +/- sqrt(0)) / 2*1
So:
x = (-4 +/- 0) / 2*1
Now, we split up our problem into two and solve.
x = (-4 + 0) / 2*1 and x = (-4 – 0) / 2*1
x = -4 / 2 and x = -4 / 2
So:
x = -2 and x = -2
Hey, we have two solutions, but they’re the same answer! Pretty cool, huh? So, if your square root ends up as zero, then you know that you will get the same solution twice.
There is one more special case and that is when your square root is negative inside.
## No Real Solutions
When your square root is negative inside, it actually makes your problem a lot shorter. Let’s see how.
Our problem is x^2 + 2x + 3 = 0. We see that our a = 1, our b = 2, and our c = 3. Plugging these into our quadratic formula, we get x = (-2 +/- sqrt(2^2 – 4*1*3)) / 2*1.
x = (-2 +/- sqrt(4 – 12)) / 2*1
x = (-2 +/- sqrt(-8)) / 2*1
Evaluating inside our square root, we see that we have a negative inside. What do we know about square roots? We know that we can’t take the square root of a negative number. What does this mean for us? It means that we have to stop here, because there’s no such thing as the square root of a negative number in the real world. It means that our problem has no real solutions, and we are done. We put ‘no real solutions’ as our answer.
## Lesson Summary
In this lesson, we’ve learned quite a few interesting things about the quadratic formula. We know that the quadratic formula lets us solve problems in the general form of ax^2 + bx + c = 0 and is defined as x = (-b +/- sqrt(b^2 – 4ac)) / 2a.
We’ve learned that the square root part of the formula tells us how many solutions we will be getting. If it is positive, then we will have two different solutions. If it is zero, then we have two solutions, but they are the same, just repeated. And, if it is a negative inside the square root, then we have no real solutions. The quadratic formula is not such a horrible thing to work with, especially when it helps us to know if we have the right number of solutions or not.
## Learning Outcomes
You’ll have the ability to do the following after this lesson:
• Identify the quadratic formula and explain when you can use this formula
• Explain how to determine the number of solutions based off the square root term in the quadratic formula
• Solve problems using the quadratic formula
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Solving a quadratic equation (factorization method)
In polynomials, we studied that a polynomial of degree 1 is called a linear polynomials.
For example: x - 5, 7x, 3 - 2x are linear polynomials which may be monomials or binomials.
A polynomial of degree 2 (two) is called a quadratic polynomial.
For example: 3x², x² + 7 , x² – 3x + 4 are quadratic polynomials which may be monomials, binomials or trinomials.
What is known as quadratic equation?
When these quadratic polynomials are equated to zero, equation is formed and is known as a quadratic equation.
The standard form of quadratic equation is ax² + bx + c = 0. Here a, b, c are real numbers and a ≠ 0. The power of x in the equation must be a non-negative integer.
(i) 3x² - 6x + 1 = 0 is a quadratic equation.
(ii) x + (1/x) = 5 is a quadratic equation.
On solving, we get x × x + (1/x) × x = 5 × x
⇒ x² + 1 = 5x
⇒ x² - 5x + 1 = 0
(iii) √2x² - x - 7 = 0 is a quadratic equation.
(iv) 3x² - √x + 1 = 0 is not a quadratic equation, since the power of x must be a positive integer.
(v) x² - (1/x) + 7 = 0 is not a quadratic equation, since on solving it becomes an equation of degree 3.
(vi) x² - 4 = 0 is a quadratic equation.
(vii) x² = 0 is a quadratic equation.
Write the quadratic equation in the standard form, i.e.,
ax² + bx + c = 0.
Express it as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, S are real numbers and p, r are not equal to zero.
Then, ax² + bx + c = 0
(px + q) (rx + s) = 0
Put each of the linear factors equal to zero
i.e., px + q = 0 and rx + s = 0
⇒ px = - q ⇒ rx = - s
⇒ x = -q/p ⇒ x = -s/r
Thus, the two values of x are called the roots of the quadratic equation.
Therefore, the solution set = {-q/p, -s/r}
Worked-out problems on solving quadratic equation will help the students to understand the detailed explanation showing the step-by-step quadratic equation solution.
1. Solve: x² + 6x + 5 = 0
Solution:
x² + 6x + 5 = 0
⇒ x² + 5x + x + 5 = 0
⇒ x(x + 5) + 1(x + 5) = 0
⇒ (x + 1) (x + 5) = 0
⇒ x + 1 = 0 and x + 5 = 0
⇒ x = -1 and x = -5
Therefore, solution set = {-1, -5}
2. Solve: 8x² = 21 + 22x
Solution:
8x² = 21 + 22x
⇒ 8x² - 21 - 22x = 0
⇒ 8x² - 22x - 21 = 0
⇒ 8x² - 28x + 6x - 21 = 0
⇒ 4x (2x - 7) + 3(2x - 7) = 0
⇒ (4x + 3) (2x - 7) = 0
⇒ 4x + 3 = 0 and 2x - 7 = 0
⇒ 4x = -3 and 2x = 7
⇒ x = -3/₄ and x = ⁷/₂
Therefore, solution set = {-3/₄, ⁷/₂}
3. 1/(x + 4) - 1/(x - 7) = 11/30
Solution:
1/(x + 4) - 1/(x - 7) = 11/30
⇒ [(x - 7) - (x + 4)]/(x + 4) (x - 7) = ¹¹/₃₀
⇒ [x - 7 - x - 4]/(x² - 3x - 28) = ¹¹/₃₀
⇒ - 11/(x² - 3x - 28) = ¹¹/₃₀
⇒ -1/(x² - 3x - 28) = ¹/₃₀
⇒ -30 = x² - 3x - 28
⇒ x² - 3x + 2 = 0
⇒ x² - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 1) (x - 2) = 0
⇒ x - 1 = 0 and x - 2 = 0
⇒ x = 1 and x = 2
Therefore, Solution set = {1, 2}
4. Solve (2x - 3)/(x + 2) = (3x - 7)/(x + 3)
Solution:
(2x - 3)/(x + 2) = (3x - 7)/(x + 3)
⇒ (2x - 3) (x + 3) = (x + 2) (3x - 7)
⇒ 2x² + 6x - 3x - 9 = 3x² - 7x + 6x - 14
⇒ 2x² + 6x - 3x - 9 - 3x² + 7x - 6x + 14 = 0
⇒ 2x² - 3x² + 6x - 3x + 7x - 6x - 9 + 14 = 0
⇒ -x² - 4x + 5 = 0
⇒ x² + 4x - 5 = 0
⇒ x² + 5x - x - 5 = 0
⇒ x (x + 5) -1 (x + 5) = 0
⇒ (x - 1) (x + 5) = 0
⇒ x - 1 = 0 and x + 5 = 0
⇒ x = 1 and x = -5
Therefore, solution set = {1, -5}
5. Solve x² - 9/5 + x² = -5/₉
Solution:
x² - 9/5 + x² = -5/₉
⇒ 9(x² - 9) = -5 (5 + x²)
⇒ 9x² - 81 = -25 - 5x²
⇒ 9x² + 5 x² = -25 + 81
⇒ 14x² = 56
⇒ x² = 56/14
⇒ x² = 4
⇒ x² - 4 = 0
⇒ x² - 2² = 0
⇒ (x - 2) (x + 2) = 0
⇒ x - 2 = 0 and x + 2 = 0
⇒ x = 2 and x = -2
Therefore, solution set = {2, -2}
These are the above examples on quadratic equations which are explained to show the exact way to solve.
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