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k0ka [10]
2 years ago
15

3 gallons of paint cover 900 square feet. How many gallons will cover 300 square feet?

Mathematics
2 answers:
SVEN [57.7K]2 years ago
4 0

Answer:

1 gallon

Step-by-step explanation:

Since the relationship is direct, meaning as the number of square feet decreases, the number of gallons also decreases.

3:900 = x:300

solution:

3(300) = 900x

900=900x

x=1

Liono4ka [1.6K]2 years ago
4 0

Answer:

1

Step-by-step explanation:

3 gallons for 900 sqft

1 gallon for 300 sqft

900/300=3

3/3=1

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Finger [1]

Answer:

(9,72)

Step-by-step explanation:

Since both equations equal y, set the equations equal to each other and solve for x.

8x=2x+54 Subtract 2x from both sides

6x= 54 Divide both sides by 6

x=9

Substitute x=9 into either equation and solve for y

y=8(9)

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4 0
3 years ago
I am so confused help me plz !!
Kryger [21]

Answer:

1. -5.5   2. 5   3.  5, 5.5

Step-by-step explanation:

For #1 just look at what point it would be towards the y- axis.

For #2 you want to kinda estimate what it would be between.

And lastly #3 they want to see the exact coordinates for Q so, look at the x axis first they the y axis. Remember Run then Jump

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3 years ago
A) y=5x+5<br> B)y=x+5<br> C)y=5x+25<br> D)y=x+125
kirill115 [55]
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6 0
3 years ago
Write a polynomial that represents the area of the shaded region
marusya05 [52]
<h3>Answer: x^2-3x+36</h3>

=========================================

Explanation:

The larger rectangle has area of (x+1)(x+1) = x^2+2x+1 through the use of the FOIL rule or distribution

If you use distribution, then it might help to let y = x+1 so we'd have y(x+1) lead to xy+1y which becomes x(x+1)+1(x+1). From there it might be easier to see how to get x^2+2x+1 after everything distributes again and simplifies.

The smaller rectangle has area 5x-35 which is found by distributing 5(x-7)

To get the shaded area, we subtract the two rectangle areas found above

shaded area = (larger area) - (smaller area)

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6 0
3 years ago
PLEASE SHOW WORK
CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

V = \dfrac{\pi}{3}r^2h

Taking the derivative of V with respect to time, we get

\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)

\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)

Since r is always equal to h, we can set

\dfrac{dr}{dt} = \dfrac{dh}{dt}

so that our expression for dV/dt becomes

\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)

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Solving for dh/dt, we get

\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}

\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})

\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}

7 0
3 years ago
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