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3 years ago

13

Manny used 8 tenth- size parts to deliver 8/10. Ana used fewer parts to model an equivalent fraction. How does the size of a ten

th - size part? What size part did Ana use?

1 answer:

Dmitriy789 [7]3 years ago

8 0

Part of this question seems to be missing, but you could use four fifth-size parts to reach an equivalent fraction.

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The diagonal of a square is 27 cm long. The diagonal of a smaller square is 12 cm long. What is the difference between the lengt

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length of big square

$\sqrt{27} = 5.2$

length of small square

$\sqrt{12 } = 3.5$

difference between length of two square

=5.2-3.5

=1.7

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Explain how to use the inverse operations to solve for x in the equation x^2=4

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How many tens are there in 10 ones? <br>

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Jane Rose’s semimonthly salary is $750. Her yearly salary is A.$9000 B. $12,500 C. $18,000 D. $19,500 E. $21,000

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50 points! I understand A. and B. but I would really appreciate help with C.

FromTheMoon [43]

Answer:

$51.72\text{ cells per hour}$

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

C)

So, we are given that the quadratic curve of the trend is the function:

$P(t)=6.10t^2-9.28t+16.43$

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

$\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]$

Expand:

$P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]$

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

$P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]$

Differentiate. Use the power rule:

$P'(t)=6.10(2t)-9.28(1)$

Simplify:

$P'(t)=12.20t-9.28$

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

$P'(5)=12.20(5)-9.28$

Multiply:

$P'(5)=61-9.28$

Subtract:

$P'(5)=51.72$

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

And we're done!

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3 years ago