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2 years ago

A cat has 24 whiskers by its nose. A manatee has 25 times as many whiskers.

Mathematics

2 answers:

elena55 [62]2 years ago

8 0

Answer:

600

Step-by-step explanation:

24*25 is 600

Brums [2.3K]2 years ago

5 0

Answer:

24 times 25 = 600 whiskers

step-by-step explanation:

since the question says 25 times as many whiskers as a cat

you would just multiply

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4/5 divided 8/7 HELP ASAP !!!

Leno4ka [110]

Answer:

7/10

Step-by-step explanation:

You can either use a calculator, or make both fractions have the denominator of 35 and change the two fractions to be,

$\frac{28}{35}$ and $\frac{40}{35}$ then divide.

Hope that helps and have a great day!

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2 years ago

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soldi70 [24.7K]

Answer:

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Step-by-step explanation:

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What is the answer to 8x10^6 + 5x10^9=

Lesechka [4]

Answer:

13x10^15

Step-by-step explanation:

8+5 = 13

6+9 = 15

3 0

3 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

2 years ago

a driver won a recent race with a record average of about 200 mph. This record is about 2 3/11 times the average speed of the fi

Oksi-84 [34.3K]

Answer:
300
——
11

Step-by-step solution

200/2*3/11
100*3/11
300/11

300/11 or 27 3/11 or 27.27

5 0

2 years ago