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3 years ago

10

When finding the least common denominator you have to multiply the number by other factors such as 1,2,3 etc. true or false help

please

1 answer:

Mashutka [201]3 years ago

4 0

Answer:

True

Step-by-step explanation:

One way to find the least common multiple of two or more numbers is to first multiply each by 1, 2, 3, etc.

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Jillian exercises 5 times a week she runs 3 miles each morning and bikes in the evening if she exercises a total of 30 miles for

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iVinArrow [24]

Answer:

Since Tracy is charged $29 for 250 free minutes so:

For x ≤ 250, f(x) = $29

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3 years ago

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Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

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3 years ago

If the ratio of circles to triangles you will only need to write ratios in one form of your choice

harina [27]

Answer:

Ratio of area of circles: = πr2 :πc2 = πr2 :π r2 4 = 4 :1. 1. The small equilateral triangle is rotated through 60° about O, the center of the circle. The arm of the rotation is the radius of the circle

Step-by-step explanation:

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