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1 year ago

11

The table shows the rental cost for a moving truck. Graph the data on the coordinate plane and explain whether the relationship

between the number of miles and the total cost is proportional or nonproportional.

1 answer:

shtirl [24]1 year ago

4 0

For the shape of the line we conclude that there is a proportional relationship. Both numbers increase that's why there is a proportion.

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Match the example on the left with the corresponding property on the right.

Orlov [11]

Answer:

1 = C

2 = A

3= B

4 = B

Step-by-step explanation:

4 0

3 years ago

Draw a rectangle that has a perimeter of 20 units and an area of 25 square units

marissa [1.9K]

<h2>Explanation:</h2><h2></h2>

For any rectangle the dimensions are given by:

$w:width \\ \\ l:Length$

So the perimeter (P) is given by the following equation:

$P=2(w+l)$

And the area is:

$A=w\times l$

We know that:

$P=20units \\ \\ A=25units^2$

By substituting:

$20=2(w+l) \\ \\ \therefore w+l=10 \ \ldots eq1 \\ \\ \\ 25=w\times l \ldots eq2$

Solving for $w$ from eq1:

$w=10-l$

Substituting into eq2:

$25=(10-l)\times l$

Then:

$\text{Solving for l:} \\ \\ \\ 10l-l^2=25 \\ \\ 10l-l^2-25=25-25 \\ \\ -l^2+10l-25=0 \\ \\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ a=-1,\:b=10,\:c=-25 \\ \\ \\ l_{1,\:2}=\frac{-10\pm \sqrt{10^2-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)} \\ \\ l_{1,\:2}=\frac{-10\pm \sqrt{0}}{2\left(-1\right)} \\ \\ l=5$

Solving for w:

$w=10-5=5$

So this rectangle is a square and is shown below.

8 0

3 years ago

Determine the equation for the given line in slope intercept form.

Vera_Pavlovna [14]

Answer:

y = - $\frac{3}{5}$ x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (5, - 4) and (x₂, y₂ ) = (0, - 1) ← 2 points on the line

m = $\frac{-1+4}{0-5}$ = $\frac{3}{-5}$ = - $\frac{3}{5}$

Note the line crosses the y- axis at (0, - 1) ⇒ c = - 1

y = - $\frac{3}{5}$ x - 1 ← equation in slope- intercept form

7 0

3 years ago

How would the average rate of change over years 1 to 5 and years 6 to 10 be affected if the population increased at a rate of 8%

Sever21 [200]

Answer:

(A): As year increases the number of pikas reduces.

(B): As year increases the number of pikas increases as opposed to when the rate reduces.

Step-by-step explanation:

<em>See comment for complete question</em>

<u>Given</u>

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

<u>(A) WHEN THE RATE DECREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1-8\%)^x$

Convert % to decimal

$f(x) = 144(1-0.08)^x$

$f(x) = 144(0.92)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(0.92)^x$

$f(1) = 144*0.92^1 =144*0.92=132.48$

$f(5) = 144*0.92^5 =144*0.66=95.04$

$Rate = \frac{95.04 - 132.48 }{4}$

$Rate = \frac{-37.44}{4}$

$Rate = -9.36$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(0.92)^x$

$f(6) = 144*0.92^6 =144*0.61=87.84$

$f(10) = 144*0.92^{10} =144*0.43=61.92$

$Rate = \frac{61.92-87.84}{4}$

$Rate = \frac{-25.92}{4}$

$Rate = -6.48$

So, we have:

$Rate = -9.36$ for year 1 to 5

This means that the number of pikas reduces by 9.36 yearly

$Rate = -6.48$ for year 6 to 10

This means that the number of pikas reduces by 6.48 yearly

So, we can say that, as year increases the number of pikas reduces.

<u>(B) WHEN THE RATE INCREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1+8\%)^x$

Convert % to decimal

$f(x) = 144(1+0.08)^x$

$f(x) = 144(1.08)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(1.08)^x$

$f(1) = 144(1.08)^1 = 144*1.08= 155.52$

$f(5) = 144(1.08)^5 = 144*1.47= 211.68$

$Rate = \frac{211.68 - 155.52}{4}$

$Rate = \frac{56.16}{4}$

$Rate = 14.04$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(1.08)^x$

$f(6) = 144(1.08)^6 = 228.52$

$f(10) = 144(1.08)^{10} = 310.89$

$Rate = \frac{310.89-228.52}{4}$

$Rate = \frac{82.37}{4}$

$Rate = 20.59$

So, we have:

$Rate = 14.04$ for year 1 to 5

This means that the number of pikas increases by 14.04 yearly

$Rate = 20.59$ for year 6 to 10

This means that the number of pikas increases by 20.59 yearly

So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.

5 0

3 years ago

Help help help help help

storchak [24]

Answer is

x > 5

Tell me if it helps

8 0

3 years ago