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3 years ago

What is the volume of a cube with length 7 cm?

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

3 0

343 is the answear :)

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From first principles, find the indicated derivatives

LenaWriter [7]

By definition of the derivative,

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)$

$\displaystyle\frac{dr}{ds} = \frac{3s^2}2$

6 0

3 years ago

Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v

Juliette [100K]

Compute the surface element:

$\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}$

$\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}$

So the integral is

$\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)$

$=\dfrac23(37^{3/2}-1)$

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3 years ago

16y2 – 64<br> What is the answer to this problem

Elodia [21]

Answer:

448 is the answer to 16y2 – 64.

8 0

3 years ago

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Please help don’t understand

const2013 [10]

The first one is congruent by ASA. The second does not have to be congruent, we don’t know enough.

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3 years ago

How many 1/3 inch cubes does it take to fill a box with width 2 2/3 inches, lengths 3 1/3 inches, and height 2 1/3 inches?

Nataly [62]

Answer:

560 cubes

Step-by-step explanation:

width: 8 cubes

length: 10 cubes

height: 7 cubes

volume: 8*10*7=560

divided l, w, and h by 1/3 to get them in cubes

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3 years ago

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