Mathematics, 12.07.2019 10:00 tainy98

The water height of a pool is determined by 8g2 + 3g - 4, the rate that the pool is filled, and 9g2 - 2g - 5, the rate that water leaves the pool, where g represents the number of gallons entering or leaving the pool per minute. answer the folowing questions: a. write an expression that determines the height of the water in the pool. b. what will be the height of the water if g = 1, 2, 3, and 4? c. to the nearest tenth, at which value for g will the water reach its greatest height? explain. i need i need to turn this in for tomorrow!

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Answer from: xxtonixwilsonxx

Part A

From the question, we know that

Rate at which pool is filled = 8g² + 3g - 4

Rate at which water leaves the pool = 9g² - 2g - 5

Hence, height of the pool would be = Rate at which pool is filled - Rate at which water leaves the pool

⇒ Height of the pool = 8g² + 3g - 4 - (9g² - 2g - 5)

⇒ Height of the pool = -g² + 5g + 1

Part B

Let water height be H

H (g) = -g² + 5g + 1

When g=1, H(1) = -(1)² + 5 (1) + 1 = 5 units

When g=2, H(2) = -(2)² + 5 (2) + 1 = -4+11 = 7 units

When g=3, H(3) = -(3)² + 5 (3) + 1 = -9+16 = 7 units

When g=4, H(4) = -(4)² + 5 (4) + 1 = -16+21 = 5 units

Part C

Now, we need to determine the value for g for which height would be maximum.

Looking at the expression that determines height, -g² + 5g + 1, we see that the coefficient of g² is negative. Hence the equation would represent a downward facing parabola, which means that the function H(g) will have a maxima point.

To find out the maxima point, differentiate H(g) with respect to g, and equate the resulting expression to zero.

$\frac{d(H(g)}{dg}$ = -2g + 5 = 0

⇒ 2g = 5

⇒ g = 2.5

So at g = 2.5, the height of the water in the pool is maximum. [Note: nearest tenth means rounding till the first decimal point, hence the answer is g=2.5]

Answer from: makeithappen60

$(a)H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g\\(b)H(1)=3.17\\H(2)=9.33\\H(3)=16.5\\H(4)=22.67\\(c)g=5.19 gallons$

Step-by-step explanation:

(A)

Rate at which pool is filled $= 8g\² + 3g - 4$

Rate at which water leaves the pool $= 9g\² - 2g - 5$

The rate at which the height of the pool is changing =Rate In-Rate Out

$\alpha = 8g\² + 3g - 4 -( 9g\² - 2g - 5)\\=8g^2+3g-4-9g^2+2g+5\\\frac{dH(g)}{dg} =-g^2+5g+1\\H(g)=\int(-g^2+5g+1)dg\\H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g$

(B)

Let water height be H

$H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g\\H(1)=-\frac{1^3}{3}+ \frac{5^2}{2}+1=3.17\\H(2)=-\frac{2^3}{3}+ \frac{5(2)^2}{2}+2=9.33\\H(3)=-\frac{3^3}{3}+ \frac{5(3)^2}{2}+3=16.5\\H(4)=-\frac{4^3}{3}+ \frac{5(4)^2}{2}+4=22.67$

(C)The water will reach its greatest height when the derivative of the Height Function is zero.

$\frac{dH(g)}{dg} =-g^2+5g+1=0\\-g^2+5g+1=0\\g=-0.19,5.19\\$

At g=5.19 gallons the height of the water will reach its greatest height.

Answer from: Quest

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

Can someone plz me 1 to 37 . i will earn 10 points

Answer from: Quest

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

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The water height of a pool is determined by 8g2 + 3g - 4, the rate that the pool is filled, and 9g2...