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4 years ago

12

A cubic water storage facility stores water for a upcoming construction site. The required volume of the water storage facility

is dependent on the dimensions of the region that needs to be watered. For a certain construction project, suppose that the length of one side of the cubic water storage must be 5/6 y^3 What should the total volume of the water storage facility be?

2 answers:

Nikolay [14]4 years ago

8 0

I would say ((5/6)y^3)^3 = (5/6)^3 * y^9

marin [14]4 years ago

3 0

For anyone looking for the correct answer, it's D. 125/216^y9. I just took the test.

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in the endpoints of `bar (AB) have the cordinates A(9,8) and B(-1,-2), what is the midpoint of `bar (AB)`?

yanalaym [24]

Midpoint of AB:
$( \frac{9 + ( - 1)}{2} . \frac{8 + ( - 2)}{2} )$
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4 years ago

What is the equation of the parabola in vertex form?

Snowcat [4.5K]

The vertex form of a parabola:

$y=a(x-h)^2+k$

You have y². It's not a function. OK. But away, the graph is a parabola but it is horizontal parabola.

$y^2-x+2y-3=0\ \ \ \ |+x+3\\\\y^2+2y=x+3\\\\y^2+2\cdot y\cdot1=x+3\ \ \ \ |+1^2\\\\\underbrace{y^2+2\cdot y\cdot1+1^2}_{use:(a+b)^2=a^2+2ab+b^2}=x+4\\\\\boxed{(y+1)^2=(x+4)}$

Your answer is:

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4 years ago

Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).

storchak [24]

Answer:

$P=16.53\ units$

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(0)^{2}}$

$d=\sqrt{1}$

$dPQ=1\ unit$

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

$d=\sqrt{(-2-3)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-5)^{2}+(-4)^{2}}$

$dQR=\sqrt{41}\ units$

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

$d=\sqrt{(3+2)^{2}+(-2+2)^{2}}$

$d=\sqrt{(5)^{2}+(0)^{2}}$

$dRS=5\ units$

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(-4)^{2}}$

$dPS=\sqrt{17}\ units$

step 5

Find the perimeter

$P=PQ+QR+RS+PS$

substitute the values

$P=1+\sqrt{41}+5+\sqrt{17}$

$P=6+\sqrt{41}+\sqrt{17}$

$P=16.53\ units$

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4 years ago

Multiple choose math question

guapka [62]

B
( A P E X )................

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3 years ago

If g(x) = 725, which of the following is true?

kotykmax [81]

Answer:

B and D

Step-by-step explanation:

The g(x) = 725 makes no sense to me....

however g(x) = -2x/(x-3) can be solved by plugging x = value into function

a) g(x) = -2x/(x-3)

g(0) = -2(0)/(0-3) = -3

= 0/-3 = 0 ≠ -3 FALSE

b) g(x) = -2x/(x-3) = 1

g(1) = -2(1)/(1-3) = 1

= -2 / -2 = 1 equals 1 TRUE

c) g(x) = -2x/(x-3)

g(2) = -2(2)/(2-3) < g(4) = -2(4)/(4-3)

= -4 / -1 < -8 / 1

= 4 is greater than - 8 FALSE

d) g(x) = -2x/(x-3) both values were already calcuated

g(1) = -2(1)/(1-3) = 1 < g(2) = -2(2)/(2-3) = 4

1 < 4 TRUE

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3 years ago