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3 years ago

I need help on number nine. Please answer ASAP this is homework needing to be turned in tommorow.

Mathematics

1 answer:

Julli [10]3 years ago

6 0

45% ? I'm not completely sure I just guessed so...

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I don’t know how to do this. Can someone help?

Aleks04 [339]

Answer:

x1=8

x2=1/5

Step-by-step explanation:

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3 years ago

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Whoever answers first gets brainliest

777dan777 [17]

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Step-by-step explanation:

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3 years ago

Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π

Natasha2012 [34]

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$ . The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$

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3 years ago

30 miles per hour is how many feet per second?

Kisachek [45]

Answer: The answer is 44

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3 years ago

The smallest object visible with your eyes is similar to the width of a piece of hair, which is 1×10−4 meters wide. Using an opt

Sloan [31]

Answer:

B. $5\times10^{2}$

Step-by-step explanation:

We are told that the smallest object visible with our eyes is similar to the width of a piece of hair, which is $1\times 10^{-4}$ meters wide.

Using an optical microscope, we can see items up to $2\times 10^{-7}$ meters wide.

To find the objects we can see with our eyes are how much larger than the objects we can see with an optical microscope, we can set an equation as:

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1*10^{-4}}{2*10^{-7}}$

Using the exponent rule of quotient $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1}{2}*10^{-4-(-7)}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{-4+7}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{3}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10\times 10^{3-1}$

$\text{The object we can see with our eyes}=5\times10^{2}*\text{The objects we can see with microscope}$

Therefore, the objects we can see with our eyes are $5\times10^{2}$ times larger than the objects we can see with an optical microscope and option B is the correct choice.

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3 years ago