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2 years ago

How to draw a non-congruent triangle that has the same area

Mathematics

1 answer:

Vera_Pavlovna [14]2 years ago

4 0

I have included a picture with the answer.

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I feel like something is wrong, is there?

Phoenix [80]

Answer:

# 4 (-5,2) should be placed one more place to the right

#5 (5,1) should be moved once up

Step-by-step explanation:

8 0

2 years ago

Find the volume of the cone. Round your answer to the nearest tenth.

Sergeu [11.5K]

9514 1404 393

Answer:

115.5 cubic yards

Step-by-step explanation:

The volume is given by the formula ...\

V = 1/3πr²h

Here the diameter is 7 yards, so the radius is 3.5 yards. The volume is then ...

V = (1/3)π(3.5 yd)²(9 yd) = 36.75π yd³ ≈ 115.454 yd³

The volume of the cone is about 115.5 cubic yards.

3 0

2 years ago

Help ASAP

kramer

Answer:

x=19/5

Step-by-step explanation:

x*2+8x=38

10x =38

5 0

2 years ago

8 Taylor has a bag of 10 fruit loops. Five loops are red, two are yellow, and three are green. If she has a larger bag that cont

Svetllana [295]

Answer:

Step-by-step explanation:

There are 2 yellow out of 10.

2/10 = 1/5

1/5 of the total number is yellow.

1/5 of 60 =

= 1/5 * 60

= 12

Answer: 12

7 0

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

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

4 0

2 years ago