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2 years ago

15

Harold's kitten drinks 5/8 cup of water every day. One cup is approximately equal to 236.4 millilitres which measurement is clos

est to the number of millilitres of water Harold's kitten drinks every day

1 answer:

Lapatulllka [165]2 years ago

4 0

Answer:

147.75 milliliters

Step-by-step explanation:

Find 5/8 of 236.4

So 147.75

So Harold’s kitten drinks 147.75 milliliters Of water everyday.

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you have reading homework every 6 days and you have science homework every 15 days. If you have homework in both today when will

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David invested $1000.00. What would that money grow to in 18 months at a 5.5% interest rate?

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3 years ago

A cube has side length a. The side lengths are decreased to 20% of their original size. Write an expression in simplest form for

Travka [436]

Answer:

0.512a³ cubic units

Step-by-step explanation:

Given a cube of length, a, then when decreased by 20%, what's left is actually 80% of the original. So, for this case, we have a cube with a length of 80%(a) = 0.80a.

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2 years ago

Read 2 more answers

A quadrilateral whose consecutive sides measure 15, 18,15, and 18

anastassius [24]

Answer:

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Step-by-step explanation:

how to explain this

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2 years ago

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!

7 0

2 years ago