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2 years ago

7

△JKL has vertices at J(−2, 4), K(1, 6), and L(4, 4). Determine whether △JKL is a right triangle. Explain. please with full calcu

lations

1 answer:

KATRIN_1 [288]2 years ago

3 0

Answer:

below

Step-by-step explanation:

basically find the line that connects JK and KL, and since its 2/3 and -2/3 which are not opposite reciprocals, the points dont form a right triangle

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Answer: 3 cakes were at the party

Step-by-step explanation:

Multiply 1/8*24 to find the exact number of cakes that were at the party

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2 years ago

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How do I find x? I am not sure how to get it because of the lines

JulsSmile [24]

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3 years ago

The equation V equals pi r squared h, where r represents radius and h represents height, is used to find the volume of a _______

baherus [9]

Answer:

Volume of a cylinder.

Step-by-step explanation:

The given equation is :

$V=\pi r^2 h$

Where

r is the radius

h is the height

This equation is used to find the volume of a cylinder.

For example, a cylinder having radius of 7 units and height of 10 units, its volume will be :

$V=\dfrac{22}{7}\times 7^2\times 10\\\\V=1540\ \text{unit}^2$

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3 years ago

The total cost for 6 mangoes and four apples is $46. Which equation represent this information?

Eva8 [605]

Answer:

3x + 2y = 23

Step-by-step explanation:

mangoe = x

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8 0

2 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

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3 years ago