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3 years ago

Quadrilateral is a rectangle with vertices D(–8, 2), E(2, 7), F(5, 1), and G(–5, –4). Find the area of the rectangle.

Mathematics

1 answer:

Katarina [22]3 years ago

8 0

The answer would be C

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Simple word problem. 40 POINTS!!!!Thank you.

Mila [183]

Answer:

$50

Step-by-step explanation:

Hello There!

We are given that for 1 hour of work 250 dollars is charged and for 3 hours of work 350 dollars is charged

This could also be represented in two points (1,250) and (3,350)

The question wants us to find the hourly charge rate (slope)

we can easily find the slope ( hourly charge rate ) by using the slope formula

$slope=\frac{y_2-y_1}{x_2-x_1}$

we have our two points so all we need to do is plug in the values (remember y values go on top and x values go on the bottom.)

$slope=\frac{350-250}{3-1} \\350-250=100\\3-1=2\\slope=\frac{100}{2} or50$

So we can conclude that the hourly charge rate is $50

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3 years ago

What is h(12) equal to?

dem82 [27]

Answer:

Human Adenovirus Type 12 (microbiology)

Step-by-step explanation:

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3 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]

Zina [86]

Answer: Absolute minimum: f(-1) = -2 $\sqrt{6}$

Absolute maximum: f( $\sqrt{12.5}$ ) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = $t\sqrt{25-t^{2}}$

f'(t) = $1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)$

f'(t) = $\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }$

f'(t) = $\frac{25-2t^{2}}{\sqrt{25-t^{2}} }$ = 0

For this function to be zero, only denominator must be zero:

$25-2t^{2} = 0$

t = ± $\sqrt{2.5}$

$\sqrt{25-t^{2}}$ ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = $-\sqrt{2.5}$ and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = $t\sqrt{25-t^{2}}$

f(-1) = $-1\sqrt{25-(-1)^{2}}$

f(-1) = $-\sqrt{24}$

f(-1) = $-2\sqrt{6}$

f( $\sqrt{12.5}$ ) = $\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}$

f( $\sqrt{12.5}$ ) = 12.5

f(5) = $5\sqrt{25-5^{2}}$

f(5) = 0

Therefore, absolute maximum is f( $\sqrt{12.5}$ ) = 12.5 and absolute minimum is

f(-1) = $-2\sqrt{6}$ .

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3 years ago

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Nitella [24]

Answer:

-272

Step-by-step explanation:

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3 years ago

I need help with semetry

Dafna11 [192]

Answer:

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