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3 years ago

How to work the problems and explain step by step by showing the work

Mathematics

1 answer:

Norma-Jean [14]3 years ago

5 0

Well, you're asked to simply grab the second version of that amount, or balance, and provide a common factored version of it, is all

so $\bf A_2=(P+Pr)+(P+Pr)r
\\\\
\textit{now, let's take common factor}
\\\\

A_2=\underline{(P+Pr)}+\underline{(P+Pr)} r\impliedby \textit{see the common factor?}
\\\\

A_2=\underline{(P+Pr)}(1+r)$

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Which of the following statements are true?

NNADVOKAT [17]

Answer:

Both graphs have been shifted and flipped

Both graphs have exactly one asymptote

Step-by-step explanation:

Choice A is incorrect since both graphs are not logarithmic functions. The graph of f(x) appears to be an exponential function.

Choice B is also incorrect since both graphs are not exponential functions. The graph of g(x) does not belong to the exponential family since y can take on negative values.

When the graph of f(x) is shifted downwards and flipped, we can obtain the graph of g(x).

Both graphs have exactly one horizontal asymptote

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3 years ago

9. Write the equation of a line going<br>through the points (9, -2) and (-3,2).

hammer [34]

I dont know if it needs to be in slope intercept or general form so i'll put it in both for you

Answer:

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4 years ago

a cylinder has a volume of 83 in3 and is 10 in. tall what is the radius of the cylinder in inches use 3.14 and round to the near

lara [203]

Check the picture below.

5 0

3 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$