All categories

3 years ago

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

You might be interested in

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,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

50+51+52+...+101=...<br>Can you help me? I do not understand

Dmitriy789 [7]

Suppose

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

At the same time, we can write

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

Note that $S=S^*$ (just reverse the sum). Let's pair the first terms of $S$ and $S^*$ , and the second, and the third, and so on:

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

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

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

5 0

3 years ago

60 minutes is 20% of blank

leva [86]

60 = 0.20x
60/0.20 = x
300 = x <== so 60 minutes is 20% of 300 minutes

6 0

3 years ago

The maximum weight on a roller coaster is 400 pounds. If your friend weighs 120 pounds, what is the most that you can weigh so y

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!

7 0

3 years ago

What are the multiples of 4

N76 [4]

1 and 2 are the factors its a compatable number

8 0

3 years ago

Read 2 more answers