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frosja888 [35]
3 years ago
13

Need help with these. 11th grade algebra 2 class. thanks ​

Mathematics
1 answer:
OlgaM077 [116]3 years ago
3 0

15) x=31/10
16)x=2
17)x=16/3
18)x=-2
19)2.5 hours
20) 32 hours
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h(t)=(t+3) 2 +5 h, left parenthesis, t, right parenthesis, equals, left parenthesis, t, plus, 3, right parenthesis, squared, plu
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Answer:

1

Step-by-step explanation:

If I understand the question right, G(t) = -((t-1)^2) + 5 and we want to solve for the average rate of change over the interval −4 ≤ t ≤ 5.

A function for the rate of change of G(t) is given by G'(t).

G'(t) = d/dt(-((t-1)^2) + 5). We solve this by using the chain rule.

d/dt(-((t-1)^2) + 5) = d/dt(-((t-1)^2)) + d/dt(5) = -2(t-1)*d/dt(t-`1) + 0 = (-2t + 2)*1 = -2t + 2

G'(t) = -2t + 2

This is a linear equation, and the average value of a linear equation f(x) over a range can be found by (f(min) + f(max))/2.

So the average value of G'(t) over −4 ≤ t ≤ 5 is given by ((-2(-4) + 2) + (-2(5) + 2))/2 = ((8 + 2) + (-10 + 2))/2 = (10 - 8)/2 = 2/2 = 1

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3 0
4 years ago
Solve for v.<br>2 = 11 - 3v<br>v=​
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What number must you add to complete the square?<br> X2+8x=-3<br> A. 8<br> B. 16<br> C. 4<br> D. 64
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3 years ago
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In this graph, it is unclear as to if the function goes on to infinity, but if it does, the answer is quite easily:

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Area of park = 324 Square yards

Dimensions  \: of  \: the  \: park  \\  =  \sqrt{Area  \: of \:  park }  \\  =  \sqrt{324}  \\  = 18 \: yards

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