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3 years ago

Diagram this statement. The answer the questions that follow.

Mathematics

1 answer:

konstantin123 [22]3 years ago

5 0

Answer:

helo guys sorry i dont know answer like please

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Ms. Robinson gave her class 12 minutes to read. Carrie read 5 1/2 pages in that time. At what rate, in pages per hour, did Carri

cricket20 [7]

The answer is: " 27.5 pages " ; or, write as: " 27 $\frac{1}{2}$ pages " .
____________________________________________________________

5.5 pages / 12 minute = ? pages / 60 minutes;

12 * 5 = 60 ;

5.5 * 5 = ?

5.5 * 5 = 27.5
____________________________________________________________
The answer is: " 27.5 pages " ; or, write as: " 27 $\frac{1}{2}$ pages " .
____________________________________________________________

4 0

3 years ago

After the party, Jane washed half of the glasses. Alex washed half of the remaining glasses, and then Sasha Alex washed half of

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

Let's take a look.

So we have 3*2*2*2*2=48 glasses

if half used 2 and other half used 1... we have a sum of 2+1=3

so 48/3 = 16

4 0

3 years ago

What is the approximated answer for:t{39}

WITCHER [35]

It’s 40 because it’s closer to 40 than 30

3 0

3 years ago

Read 2 more answers

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

4 years ago

Can someone please help? Thanks!

Paul [167]

Hey, I have already answered these all questions except the last question in your previous question. Make sure to see them.

For the last question, the answer would be the second choice as when times keep flowing, the distance still remains the same (not to be confused with speed.)

Basically when times keep going, the distance will stay the same without any changes.

7 0

3 years ago