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2 years ago

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Polygon ABCD is plotted on a coordinate plane. If the polygon translates 3 units to the left and 1 unit down, the length of diag

onal is units. If the polygon translates 3 units further to the left and four units down, the length of diagonal .

1 answer:

torisob [31]2 years ago

5 0

Answer:

2 and remains the same

Step-by-step explanation:

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Describe in words the surface whose equation is given. (assume that r is not negative.) θ = π/4

Delvig [45]

An equation in the form $\theta=\dfrac{\pi}{4}$ is the line
that goes through the origins and whose tangent equates $\dfrac{\pi}{4}$ . In general, any equation in the form $\theta=\theta_0$
is the equation of a line.

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2 years ago

Peter is at a lumber yard. He gets 2 free boxes of nails for every 10 boards he buys. Write an expression for the number of boxe

Sati [7]

A. The number of 10-boards Peter bought is equal to n divided by 10. Then, each of the 10-boxes will get two boxes of nails. The number of boxes of nails that Peter will have after buying n boards will be,

N = (2)(n/10)

Simplifying,

<em> N = n/5</em>

b. If the number of boards are 90 then,

N2 = (90/10)(2)(100 nails/box)
N2 = 1800

Answer: 1800

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2 years ago

Read 2 more answers

Consider the graph of f(x) given below.

Andrej [43]

Answer: d

Step-by-step explanation:

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2 years ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

Ugo [173]

By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

$f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2$

Evaluating for x = 3:

$f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18$

Then, the AVR is:

$AVR = \frac{18-(-2)}{3-(-2)}$

$AVR = \frac{18+2}{3+2}$

$AVR = \frac{20}{5}$

$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

$f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4$

Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

$AVR = \frac{3}{1}$

$AVR = 3$

For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

Evaluating for x = 4:

$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

$AVR = -1$

For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

4 0

2 years ago

A law school administrator was interested in whether a student's score on the entrance exam can be used to predict a student's g

frutty [35]

Answer:

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the $r^2$ who represent the determination coefficient and we got:

$r^2 = 0.934^2= 0.872$

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the $r^2$ who represent the determination coefficient and we got:

$r^2 = 0.934^2= 0.872$

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

8 0

2 years ago