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Likurg_2 [28]
4 years ago
8

WILL GIVE BRAINLIEST PLEASE HELP!!

Mathematics
1 answer:
Anastasy [175]4 years ago
5 0
Hi.

Your answer is going to be

C.) 33

Hope this helps! :)
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I need help on #37!!
Nataliya [291]

Answer:

A) 12.1 , 12.01

B) infinitely large

Step-by-step explanation:

Give that the function f(x)=x2 describes the area of the square, f(x), in square inches, whose sides each measure x inches. If x is changing. 

Using the formula

(f(a + h) - f(a))/h

as x changes from 6 inches to 6.1 inches

a = 6, h = 0.1

= [f(6.1) - f(6)]/0.1

= (6.1^2 - 6^2)/0.1

= (37.21 - 36)/0.1

= 1.21/0.1

= 12.1

and when x changes from 6 inches to 6.01 inches. 

a = 6, And h = 0.01

Using the same formula

(f(a + h) - f(a))/h

= [f(6.01) - f(6)]/0.01

= (6.01^2 - 6^2)/0.01

= (36.120 - 36)/0.01

=0.1201/0.01

= 12.01

B) the instantaneous rate of change of the area with respect to x at the moment when x = 6 inches is infinitely large since h = 0

5 0
3 years ago
Compare the functions below:
rjkz [21]
f(x)\to y_{max}=4\\\\g(x)=4\cos(2x-\pi)-2\to y_{max}=2\\\\h(x)=-(x-5)^2+3\to y_{max}=3

Answer: a. f(x)
6 0
3 years ago
Read 2 more answers
Britney bought 2 apples, 5 bananas, and 8 oranges at the grocery store.
Sladkaya [172]

Answer:

2 apples: 5 bananas: 8 oranges

Final answer: 1 apple: 2,5 bananas ( divide by 2 )

5 0
3 years ago
Read 2 more answers
4) Find the slope of the following<br> graph.<br> у<br> -2<br> o<br> B
Strike441 [17]

Answer:

-6/13

Step-by-step explanation:

7 0
3 years ago
webassign If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate at which the diameter dec
drek231 [11]

Answer:

The answer is \frac{3}{10\pi }  cm/min

Step-by-step explanation:

Assuming the snowball is a perfect sphere, then if A denotes the surface area and D the diameter then:

A=4\pi r^{2} = 4\pi (\frac{D}{2} )^{2} =\pi  D^{2}

Differentiating wrt r we have:

\frac{dA}{dD} =2\pi D

We are told that \frac{dA}{dt}= -6 and we want to find \frac{dD}{dt}

By the chain rule we have:

\frac{dA}{dD}=\frac{dA}{dt}.\frac{dt}{dD}=\frac{\frac{dA}{dt} }{\frac{dD}{dt} }

∴2\pi D=-\frac{6}{\frac{dD}{dt} }

∴\frac{dD}{dt}=-\frac{6}{2\pi D}

When D=10 then

\frac{dD}{dt} =-\frac{6}{10*2\pi } =-\frac{3}{10\pi }

The sign (-) shows that the D is decreasing.

3 0
3 years ago
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