Algebra II - Graphing Data

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

Explanation

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

Explanation

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

Explanation

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 ?

Explanation

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:

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

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

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.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

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 .

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.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

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

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 .

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

Explanation

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

Explanation

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:

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

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

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.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

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

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 .

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