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3 years ago

Solid: What is the surface area of the pyramid? 7

Mathematics

2 answers:

telo118 [61]3 years ago

8 0

Answer:

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Step-by-step explanation:

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baherus [9]3 years ago

5 0

Answer:

i am so sorry if its wrong

Each of the triangles making up the sides and base have a base of 8 cm and a height of 6.9 cm. Then the area of the pyramid is that of 4 such triangles.

A = 4×(1/2)×b×h

A = 2×(8 cm)×(6.9 cm)

A = 110.4 cm²Step-by-step explanation:

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Please help. Please. I will mark brainliest

kolbaska11 [484]

Answer:

what type of math is this i never seen this before

7 0

3 years ago

One common factor of two numbers is 40. Another common factor is 10. If both numbers are less than 100, what are the two numbers

Viefleur [7K]

Only two numbers less than 100 have a factor of 40: 40 (1 x 40) and 80 (2 x 40). Since 40 is a factor of both 40 and 80, 10 must also be a factor - indeed, 40 = 4 x 10 and 80 = 8 x 10.

The two number must be 40 and 80.

5 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago

what would be the equation of a line that passes through (3,-4) with slope 2/3

Len [333]

Answer:

y = 2/3x - 6

Step-by-step explanation:

Use the slope intercept equation, y = mx + b

Plug in the slope and given point, then solve for b

y = mx + b

-4 = 2/3(3) + b

-4 = 2 + b

-6 = b

Plug in the slope and b into the equation

y = 2/3x - 6

So, the equation of the line is y = 2/3x - 6

7 0

3 years ago

How do I simplify this?

Anastasy [175]

3/4 + 1 = 1.75 which is equivalent to 7/4
1- 3/4 = 0.25 which is equivalent to 1/4
the problem now looks like 7/4 / 1/4
you cancel out both 4
this problem simplified looks like 7/1
which is equivalent to 7 if you divide

6 0

1 year ago