Variable Relationships Flashcards - Algebra II

Question

What is the next number in this sequence: 8, 27, 64, 125 ?

Answer

Find the pattern of the sequence:

This pattern is $2^{3}, 3^{3}, 4^{3}, 5^{3}$ so the next number in the sequence would be $6^{3}=216$

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Question

Find the next 2 numbers in this sequence: 33, 46, 72, 111

Answer

Find the pattern in this sequence of numbers:

In this case, the pattern is adding 13n to the previous number where n= how many numbers came before the current number.

$33+13, 46+(13 \times 2), 72+(13 \times 3)$

so the first number we are looking for would be:

$111+ (13 \times 4)=163$

the second number we are looking for would be:

$163 + (13\times 5)=228$

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Question

The amount of water inside of a leaky boat is measured periodically after the boat has been in the water in different periods of time and are found to have a linear relationship. The results are given in the following chart:

Time in water (mins) Amount of water in the boat (gal)
0 0
6 4.8
19 15.2
28 22.4

Using the method of linear extrapolation based on the data from the table, how much water would you expect to be in the boat after minutes?

Time in water (mins)	Amount of water in the boat (gal)
0	0
6	4.8
19	15.2
28	22.4

Answer

To extrapolate the results of the study out to 53 minutes, first we have to determine an equation representing the relationship between time passed and amount of water; we can write our equation in slope-intercept form:

Where our y-axis represents amount of water and the x-axis represents time. We can pick 2 points and label them Point 1 and Point 2; looking back at the table:

Time in water (mins) Amount of water in the boat (gal)
0 0
6 4.8
19 15.2
28 22.4

We label Point 1 as and Point 2 as ; we plug these points into the slope formula as follows:

$\frac{y_{2}-y_{1}}{x_2-x_1}=\frac{15.2-4.8}{19-6}=\frac{10.4}{13}=0.8$

So, the slope of our line that describes how much water is in the boat is ; to find our term, the y-intercept, we need to pick a point on the graph and plug in our slope to solve for y-intercept. Let's once again choose the point :

Simplify the expression and we find that b=0, so our slope-intercept equation is:

Plugging in a value of 53 for , we find that:

So the answer is 42.4 gallons of water in the boat after 53 minutes.

Time in water (mins)	Amount of water in the boat (gal)
0	0
6	4.8
19	15.2
28	22.4

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Question

varies directly with the square root of . If , then . What is the value of if ?

Answer

If varies directly with the square root of , then for some constant of variation ,

$\small y = K \sqrt{x}$

If , then ; therefore, the equation becomes

$\small \small 48 = K \sqrt{25}$ ,

or

.

Divide by 5 to get , making the equation

$\small \small y = 9.6 \sqrt{x}$ .

If , then $\small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4$ .

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Question

If varies directly with $x^{3}$ and when , due to the effect of a constant, what is the value of when ?

Answer

Since varies directly with $x^{3}$ , $y=Kx^{3}$ where is a constant.

1. Solve for when and .

$y=Kx^{3}$

$61.125=K(5^{3})$

$K=\frac{61.125}{125}=0.489$

2. Use your equation to solve for when .

$y=0.489x^{3}$

$y=0.489(11^{3})=650.859$

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Question

If varies indirectly with $\sqrt{x}$ and when , due to the effect of a constant, what is the value of when ?

Answer

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when and .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

2. Use the equation you found to solve for when .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

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Question

varies directly with $x^{2}$ . If , what is if ?

Answer

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

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Question

varies inversely with $x^{3}$ . If , then what is equal to when ?

Answer

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

$2=\frac{K}{8}$

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

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Question

varies directly with $\sqrt[3]{x}$ and when . What is when ?

Answer

1. Since varies directly with $\sqrt[3]{x}$ :

$y=K\sqrt[3]{x}$

2. Use the values given for x and y to solve for K:

$y=K\sqrt[3]{x}$

$3=K\sqrt[3]{8}$

$K=\frac{3}{2}$

3. Use your new equation with the K you solved for to solve for y when x=27:

$y=(\frac{3}{2})\sqrt[3]{27}$

$y=(\frac{3}{2})(3)=4.5$

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Question

varies inversely with $\sqrt{x}$ . When $x=4,\ y=0.5$ . What is the value of when ?

Answer

1. Since y varies indirectly with $\sqrt{x}$ :

$y=\frac{K}{\sqrt{x}}$

2. Solve for K using the x and y values given:

$0.5=\frac{K}{\sqrt{4}}$

$0.5=\frac{K}{2}$

3. Using the equation you created by solving for K, find y when x=100:

$y=\frac{1}{\sqrt{x}}$

$y=\frac{1}{\sqrt{100}}$

$y=\frac{1}{10}$

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Question

Given the two following points, use interpolation to determine the best estimate for the value

,

Answer

Using our two known points, we can use interpolation to determine the value at any point between them with the following formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}$

Where is our first given point, is our second given point, and is the point we want to find. We know our two given points, as well as the x value of our unknown point, so now all we must do is plug in all of our known values and solve for y, our only unknown:

$\frac{93-81}{25-24}=\frac{y-81}{24.75-24}$

$12=\frac{y-81}{0.75}$

$y-81=9\rightarrow y=f(24.75)=90$

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Question

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if workers are present.

Answer

We want to do a linear interpolation since the relationship between workers and units can be assumed to be linear. This means there is a constant slope between the points, so the slope between two known points will be equal to the slope between the point we are trying to find and some known point. This is expressed in the relation:

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and and are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

and .

Plugging these into our interpolation formula and knowing , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

.

So there are units produced when the number of workers is .

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Question

Given the points $\left ( 30,51\right )$ and $\left ( 20,36\right )$ , use linear interpolation to find the value of when .

Answer

Use the formula for interpolation to determine the value of y:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 51) as our x2 and y2 and (20, 36) as our x1 and y1 and we will solve for y using 26.5 for x.

$\frac{51-36}{30-20}=\frac{y-36}{26.5-20}$

$\frac{15}{10}=\frac{y-36}{6.5}$

$\frac{3}{2}(6.5)=y-36$

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Question

Given $\left ( 15,10\right )$ and $\left ( 30,15\right )$ use linear interpolation to find when .

Answer

Use the formula for interpolation:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 15) as x2 and y2, (15, 10) as x1 and y1, and solve for y when x=17.9:

$\frac{15-10}{30-15}=\frac{y-10}{17.9-15}$

$\frac{5}{15}=\frac{y-10}{2.9}$

$(\frac{5}{15})(2.9)=y-10$

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Question

Mary measures her height every year on her birthday, starting at 11 until she turns 16. She wants to make a table with all the information gathered, but discovers she lost the piece of paper on which she wrote her height down on her 14th birthday. Her incomplete table looks like this:

Age (years) Height (inches)
11 47.5
12 50.25
13 53
14 ?
15 58.5
16 61.25

Using the method of linear interpolation, which of the following is the closest estimate of Mary's height on her 14th birthday?

Age (years)	Height (inches)
11	47.5
12	50.25
13	53
14	?
15	58.5
16	61.25

Answer

Using linear interpolation means that we draw a line between the points on our data set and use that line to estimate a value that lies between two data points; in this case, we have the data from Mary's 13th and 15th birthdays, so we can describe a line between those two points and estimate her height at 14. Our line will be written in slope-intercept form:

Where the variable represents Mary's Age in years and the variable represents her height in inches. First, we need to find the slope. Using 2 points on our table and as point 1 and point 2, respectively, we plug these values into our slope formula:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{(58.5-53)}{(15-13)}=\frac{5.5}{2}=2.75$

Next, we find the y-intercept by plugging in our slope (which we just found) and a point from our table (we'll stick with ) and solving for :

Simplify:

Subtract 35.75 from both sides to solve for :

The equation of our interpolation line is:

So, to get an estimate of Mary's height on her 14th birthday, we plug in and solve for :

Our estimate of Mary's height at is

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Question

Find the value of when given the points and .

Answer

Write the interpolation formula.

$y=y_1 + \frac{(x-x_1)(y_2-y_1)}{(x_2-x_1)}$

Identify and substitute the values.

$y=1 + \frac{(-2-8)(9-1)}{(4-8)}$

Simplify the fraction.

$y=1 + \frac{-10(8)}{-4} = 1+\frac{-80}{-4} =1+20 = 21$

The answer is:

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Variable Relationships - Algebra II

What is the next number in this sequence: 8, 27, 64, 125 ?

Find the pattern of the sequence: This pattern is so the next number in the sequence would be

Find the next 2 numbers in this sequence: 33, 46, 72, 111

Find the pattern in this sequence of numbers: In this case, the pattern is adding 13n to the previous number where n= how many numbers came before the current number. so the first number we are looking for would be: the second number we are looking for would be:

varies directly with the square root of . If , then . What is the value of if ?

If varies directly with the square root of , then for some constant of variation , If , then ; therefore, the equation becomes , or . Divide by 5 to get , making the equation . If , then .

If varies directly with and when , due to the effect of a constant, what is the value of when ?

Since varies directly with , where is a constant. 1. Solve for when and . 2. Use your equation to solve for when .

If varies indirectly with and when , due to the effect of a constant, what is the value of when ?

Since varies indirectly with , 1. Solve for when and . 2. Use the equation you found to solve for when .

varies directly with . If , what is if ?

1. Since varies directly with : with K being some constant. 2. Solve for K using the x and y values given: 3. Use the equation you solved for to find the value of y:

varies inversely with . If , then what is equal to when ?

1. Since varies indirectly with : 2. Use the given x and y values to determine the value of K: 3. Using the equation along with the value of K, find the value of y when x=5:

varies directly with and when . What is when ?

1. Since varies directly with : 2. Use the values given for x and y to solve for K: 3. Use your new equation with the K you solved for to solve for y when x=27:

varies inversely with . When . What is the value of when ?

1. Since y varies indirectly with : 2. Solve for K using the x and y values given: 3. Using the equation you created by solving for K, find y when x=100:

Given the two following points, use interpolation to determine the best estimate for the value ,

Given the points and , use linear interpolation to find the value of when .

Use the formula for interpolation to determine the value of y: We will use (30, 51) as our x2 and y2 and (20, 36) as our x1 and y1 and we will solve for y using 26.5 for x.

Given and use linear interpolation to find when .

Use the formula for interpolation: We will use (30, 15) as x2 and y2, (15, 10) as x1 and y1, and solve for y when x=17.9:

Find the value of when given the points and .

Write the interpolation formula. Identify and substitute the values. Simplify the fraction. The answer is:

Find the pattern of the sequence:

This pattern is $2^{3}, 3^{3}, 4^{3}, 5^{3}$ so the next number in the sequence would be $6^{3}=216$

If varies directly with $x^{3}$ and when , due to the effect of a constant, what is the value of when ?

Since varies directly with $x^{3}$ , $y=Kx^{3}$ where is a constant.

1. Solve for when and .

$y=Kx^{3}$

$61.125=K(5^{3})$

$K=\frac{61.125}{125}=0.489$

2. Use your equation to solve for when .

$y=0.489x^{3}$

$y=0.489(11^{3})=650.859$

If varies indirectly with $\sqrt{x}$ and when , due to the effect of a constant, what is the value of when ?

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when and .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

2. Use the equation you found to solve for when .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

varies directly with $x^{2}$ . If , what is if ?

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

varies inversely with $x^{3}$ . If , then what is equal to when ?

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

$2=\frac{K}{8}$

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

varies directly with $\sqrt[3]{x}$ and when . What is when ?

varies inversely with $\sqrt{x}$ . When $x=4,\ y=0.5$ . What is the value of when ?

Given the two following points, use interpolation to determine the best estimate for the value

,

Given the points $\left ( 30,51\right )$ and $\left ( 20,36\right )$ , use linear interpolation to find the value of when .

Given $\left ( 15,10\right )$ and $\left ( 30,15\right )$ use linear interpolation to find when .

Use the formula for interpolation:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 15) as x2 and y2, (15, 10) as x1 and y1, and solve for y when x=17.9:

$\frac{15-10}{30-15}=\frac{y-10}{17.9-15}$

$\frac{5}{15}=\frac{y-10}{2.9}$

$(\frac{5}{15})(2.9)=y-10$

Write the interpolation formula.

$y=y_1 + \frac{(x-x_1)(y_2-y_1)}{(x_2-x_1)}$

Identify and substitute the values.

$y=1 + \frac{(-2-8)(9-1)}{(4-8)}$

Simplify the fraction.

$y=1 + \frac{-10(8)}{-4} = 1+\frac{-80}{-4} =1+20 = 21$

The answer is: