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marissa [1.9K]
3 years ago
9

a pool is being filled with a large water hose. the height of the water in a pool is determined by 7g^2+3g-4. Previously the poo

l had been filled with a different hose. then the height was determined by 8g^2+2g-2. Enter an expression that determines the height of the water in the pool if both hoses are on the same time simplify the expression.
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

If both the hoses are on the same time then the height of the water in the pool is given by the expression  15g^2 + 5g - 6

Step-by-step explanation:

(7g^2 + 3g - 4) + (8g^2 + 2g - 2)

= (7g^2 + 8g^2) + ( 3g + 2g) + (-4 -2)

with the help of distribution law,

= (7 +8)g^2 + (3 + 2)g - (4 + 2)

= 15g^2 + 5g - 6

Therefore if both the hoses are on the same time then the height of the water in the pool is given by the expression  15g^2 + 5g - 6

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How many solutions does the system have?
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Answer:

A; infinitely many solutions

Step-by-step explanation:

first, let's simplify the first equation to make it simpler.

divide both sides by 5

y=3x-8

new system

y=3x-8

y=3x-8

as you can see, they are the same.

not mandatory but useful!⬇⬇⬇⬇

just to be more clear, let's substitute them. since they're both already in y=, we can make them both set to each other.

3x-8=3x-8

add 8 to both sides; subtract 3x from one side

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3 years ago
Figure EFGHK is to be transformed to Figure E'F'G'H'K' using the rule (x, y)→(x + 8, y + 5):
IceJOKER [234]
The rule that has been given means that x coordinate of some point we increase by 8 and y coordinate of some point we increase by 5.

Coordinates of K point on original Figure are:
(-2,-3)

once we implement rule on this we get K':

K' ( -2+8,-3+5) 
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Answer is third option.

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3 years ago
Read 2 more answers
The radius of a sphere is 3 inches. Which represents the volume of the sphere?
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Answer:

b

Step-by-step explanation:

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Here r = 3, thus

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3 years ago
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Answer:

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(b)g^{-1}(n)=\log_{2}20g(n)

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Step-by-step explanation:

<u>Part A</u>

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n

<u>Part B</u>

<u />g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)<u />

<u />

<u>Part C</u>

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43

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Angle QST=52°+47°

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