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nataly862011 [7]
3 years ago
13

Find the area of each figure to the nearest tenth.

Mathematics
1 answer:
sergey [27]3 years ago
6 0

Answer:

Area of triangle ABC=7 square units

Area of triangle A'B'C'=7 square units

Step-by-step explanation:

Triangle ABC is the preimage of triangle A'B'C' under a reflection in the line x=0.

The two triangles are congruent so they have the same area.

We can calculate the area by counting the unit squares the triangle covered.

Each full box is 1 units square.

Each triangle covered

1 full box and 12 partially covered boxes.

Each partial box covered is 0.5 square units.

Therefore the area is

Area=1+0.5\times12

Area=1+6

Area=7 square units

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Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
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\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
6 0
3 years ago
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Dmitriy789 [7]

Suppose

S=50+51+52+\cdots+100+101

At the same time, we can write

S^*=101+100+99+\cdots+51+50

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S+S^*=(50+101)+(51+100)+(52+99)+\cdots+(100+51)+(101+50)

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

S+S^*=52\cdot151

But S=S^*, as we pointed out, so

2S=52\cdot151\implies S=\dfrac{52\cdot151}2=3926

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wolverine [178]
The maximum weight on a roller coaster is 400 pounds (Max = 400).
If your friend weighs 120 pounds (y = 120), what is the most you can weigh so you can ride together (x + y < 400).

x + 120 <_ 400.
y = 120

Subtract 120 from both sides.

x <_ 280.

You can weigh at most 280 pounds (x <_ 280).

I  hope this helps!
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What are the multiples of 4
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