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3 years ago

Please help with this question.

Mathematics

1 answer:

Gnesinka [82]3 years ago

4 0

Answer:

teqawg veqwargth4ve fqhvreh

Step-by-step explanation:

aferd thgaReghaergheargh

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2m = 1 + m what do I need to do to solve

devlian [24]

So all you need to do is subtract m from both sides.

Then you have your answer

m = 1

I hope this helps

-ayden

4 0

2 years ago

Read 2 more answers

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.

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

What is negative 1/4 to the 3rd power

Y_Kistochka [10]

Answer:

-1/64

Step-by-step explanation:

Just do 4 to the 3rd and make that the denominator

4^3 = 64

-1 / 64

7 0

2 years ago

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Simplify 7.4z - 5 (-1.6z + 2.4) please

Sergeu [11.5K]

The answer is 15.4z-12

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2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

Mazyrski [523]

We can use the fact that, for $|x|$ ,

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

Notice that

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}$

so that

$f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right]$

$f(x)=\displaystyle5\sum_{n=0}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=1}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=0}^\infty(n+1)x^n$

By the ratio test, this series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|$

so the series has radius of convergence $R=1$ .

5 0

2 years ago