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Umnica [9.8K]
3 years ago
7

Alan makes a fruit salad using strawberries and blueberries. He uses 5 cups of strawberries for every 3 cups of blueberries. Whi

ch measure represents the amount of strawberries Alan uses for every 1 cup of fruit salad? A. 3/8 cup B. 3/5 cup C. 5/8 cup D. 5/3 cup PLEASE HURRY I NEED THIS.
Mathematics
1 answer:
fredd [130]3 years ago
4 0

Answer:

C. 5/8 cup

Step-by-step explanation:

  1. 5 + 3 = 8
  2. Strawberries/total amount = 5/8

I hope this helps!

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FOIL

Step-by-step explanation:

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3 years ago
Find the measure of one interior angle in each regular polygon. round your answer to the nearest 10th if necessary.
dem82 [27]

Answer:

140 degree or Answer D

Step-by-step explanation:

140 degrees is the answer for a Nonagon

8 0
2 years ago
Solve for the indicated variable.<br> 2a + 2b = c for b
AysviL [449]

Answer:

2a+2b=c

2b=c-2a

b=(c-2a)÷2

5 0
3 years ago
Marneshia walked three over eight of a mile in three over five of an hour. What equation can be used to calculate her unit rate
Rama09 [41]
3/8 mile in 3/5 hour
\frac{ \frac{3}{8} }{ \frac{3}{5} }
we want to find miles per hour or m/1h
convert 3/5 to 1 by multiplying it by 5/3
keep the fraction the same by multiply ing the whole thing by
\frac{ \frac{5}{3} }{ \frac{5}{3} } to get

\frac{ \frac{15}{24} }{ \frac{15}{15} }=
\frac{ \frac{5}{8} }{ 1}=5/8 miles per hour


7 0
3 years ago
Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
=-2\pi

So, the flux over the paraboloid alone is

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
6 0
3 years ago
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