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lana [24]
3 years ago
10

Carly can paint 6/7 of a picture in 3/14 of an hour how many pictures can she paint in a full hour

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
8 0
6/7 divide by 3/14
Divide = multiply reciprocal
6/7 * 14/3 = 84/21 = 4
She can paint 4 pictures in a full hour
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Step-by-step explanation:

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Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an
Leya [2.2K]
Parameterize the lateral face T_1 of the cylinder by

\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v

where 0\le u\le2\pi and 0\le v\le3, and parameterize the disks T_2,T_3 as

\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)
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The integral along the surface of the cylinder (with outward/positive orientation) is then

\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS
=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr
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=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr
=136\pi
7 0
3 years ago
at five pm a four foot mailbox casts a shadow six feet long at the same time a tree cast a shadow 36 feet Long bhow tall is the
Andrej [43]

So the 4ft mailbox’s shadow was six feet. That 2ft longer. According to my logic if the tree had a showdown of 36ft we subtract 2ft. The trees actual size is 34ft.

I might be wrong don’t hate me :(

5 0
3 years ago
A population of insects, in thousands, can be molded using the function (t)= 1.75(0.97)x
Dominik [7]

Question:

A population of insects, in thousands, can be modeled using the function

p(t) = 1.75(0.97)^t, where t is time in months. Which statement best

describes the population of insects?

A. The population is decaying at a rate of 3% each month.

B. The population is decaying at a rate of 25% each month.

C. The population is growing at a rate of 75% each month.

D. The population is growing at a rate of 97% each month.

Answer:

A. The population is decaying at a rate of 3% each month.

Step-by-step explanation:

Given

p(t) = 1.75(0.97)^t

Required

True statement about the function

From the options, we can see that we are to answer the question on the basis of decay and growing rates.

An exponential form is:

y=ab^x

Compare to p(t) = 1.75(0.97)^t

b= 0.97\\

If b > 1, then b = 1 + r r represents growth rate

else, b= 1-r r represents decay rate

Since b < 0.97:

0.97= 1-r

r = 1 - 0.97

r = 0.03

r = 3\%

r = 0.03

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vesna_86 [32]

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