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ZanzabumX [31]
2 years ago
15

HELP!! Jamaal bounces on a trampoline. His height, as a function of time, is modeled by y= -16x2(power of 2) + 20x + 4.

Mathematics
2 answers:
Harrizon [31]2 years ago
7 0

Answer:

B the function is nonlinear

Step-by-step explanation:

zhenek [66]2 years ago
3 0

Answer: B, the function is nonlinear

Step-by-step explanation:

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4. What other information is needed to prove the two triangles similar? : Please step by step
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Step-by-step explanation:

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The weight of an object on the moon is of its weight on Earth. Write 1/6 as a decimal.​
xxTIMURxx [149]

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1/6 as an decimal would be, 0.1666 Continued

Step-by-step explanation:

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3 years ago
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Which of the following inequalities represents "the car has no more than 5 gallons of gas left in the gas tank"?
MrRissso [65]
The answer is B. Hope this helps :)
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An online furniture store sells chairs for $50 each and tables for $550 each. Every day, the store can ship a maximum of 32 piec
Hunter-Best [27]

Answer:

you have to sell 6 tables to meet all requirements

Step-by-step explanation:

chairs=$50x

tables=$550x

24 chairs×50= $1200

4100-1200= $2900

take $2900 and divide by $550 to find the exact number of tables which is 5 but selling 5 tables and 24 chairs doesnt reach the $4100 mark so I rounded up to 6 tables which doesnt surpass the maximum number of furniture(32) but beats the $4100 mark

3 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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