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3 years ago

Please answer helpppp !!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!

Mathematics

1 answer:

disa [49]3 years ago

7 0

Answer:

Step-by-step explanation:

Remember that TR is 1/3 the length of MR

MR = 54

TR = 1/3 *MR

TR = 1/3 * 54

TR = 18

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How much is 36 nickels 24 dimes 19 quarters and 15 fifty cent?

FromTheMoon [43]

Fifty Cent = 50 cents

Quarter = 25 cents

Dime = 10 cents

Nickel = 5 cents

__________________________________________________________________

15 Fifty Cent coins = 7.50 dollars

19 Quarters = 4.75 dollars

24 Dimes = 2.40 dollars

36 Nickels = 1.80 dollars

__________________________________________________________________

Total: $16.45

Hope that helps!

3 0

4 years ago

The table shows the distance that two trains are from center station at various times of the day

VLD [36.1K]

Answer: Train 2 is traveling 4 miles per hour faster than train 1.

Hope this helps..... Stay safe and have a Merry Christmas!!! :D

3 0

3 years ago

Helppppppppppp plzzz

RideAnS [48]

60 degrees, hope this helps

7 0

3 years ago

Pls help, will mark brainliest

SashulF [63]

Step-by-step explanation:

$\frac{\pi}{2} = 90$

its 1/4 of the circle

circle circumference:

$8 \times \pi$

we need 1/4 of it, so its

$2 \times \pi = 6.28$

5 0

3 years ago

F(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 3)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 18 that l

Cerrena [4.2K]

$\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k$
$\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1$

By the divergence theorem, the flux of $\mathbf F$ across the *closed* surface $\mathcal S$ combined with the plane $z=2$ is given by a volume integral over the closed region:

$\displaystyle\iint_{\mathcal S}\mathbf F\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}\nabla\cdot\mathbf F\,\mathrm dV$

So in fact, to find the flux over $\mathcal S$ alone, we'll need to subtract the flux of $\mathbf F$ over the planar portion, oriented outward. First, compute the volume integral by converting to cylindrical coordinates:

$x^2+y^2+z=18$
$z=2\implies x^2+y^2=16\implies r^2=16\implies r=4$

$\displaystyle\iiint_{\mathcal R}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=2}^{z=18-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=128\pi$

If the surface does actually contain $z=2$ , then you can stop here; otherwise, continue.

Now, parameterize the part of the *closed* surface in $z=2$ by

$\mathbf s(r,\theta)=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j+2\,\mathbf k$

where $0\le r\le4$ and $0\le\theta\le2\pi$ . We get a surface element

$\mathrm d\mathbf S=(\mathbf s_r\times\mathbf s_\theta)\,\mathrm dr\,\mathrm d\theta=(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$

We don't need to worry about the first two components of

and so the surface integral over this region is

$\displaystyle\iint_{z=2\,\land\,x^2+y^2\le16}\mathbf F\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}2r\,\mathrm dr\,\mathrm d\theta=32\pi$

Then the total flux over $\mathcal S$ alone is $(128-32)\pi=96\pi$ .

4 0

3 years ago