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3 years ago

Triangles ABC and FED are congruent right triangles. Find the length of FE. 8 m 6 m

Mathematics

1 answer:

const2013 [10]3 years ago

4 0

Sawyer and Emily is so much scary

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Round 2.9 to the nearest whole number.

-Dominant- [34]

Answer:

it would be 3

Step-by-step explanation:

anything above 5, goes to the nearest, (in this case, whole number)

so it would be 3. hope this helped

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3 years ago

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Find the area of the semicircular window. Round your answer to the nearest hundredth.

Fittoniya [83]

Answer:

108

Step-by-step explanation:

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4 years ago

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Write an equation passing through the point and parallel to the given line. (4,7);y=3x+4

vlabodo [156]

Answer:

$y=3x-5$

Step-by-step explanation:

step 1

Find the slope

we know that

If two lines are parallel, then their slopes are equal

The slope of the given line is $m=3$

The slope of the line parallel to the given line is also $m=3$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=3$

$point\ (4,7)$

substitute

$y-7=3(x-4)$

step 3

Convert to slope intercept form

isolate the variable y

$y-7=3x-12\\y=3x-12+7\\y=3x-5$

4 0

4 years ago

Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t

Nimfa-mama [501]

Integrate the force field along the given path (call it <em>C</em>):

$W=\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=\int_0^{2\pi}\mathbf F(x(t),y(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\bigg((t-\sin t)\,\mathbf i+(6-\cos t)\,\mathbf j\bigg)\cdot\bigg((1-\cos t)\,\mathbf i+\sin t\,\mathbf j\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(t-t\cos t+5\sin t)\,\mathrm dt=\boxed{2\pi^2}$

5 0

3 years ago

Please help the question is in the picture.

xxMikexx [17]

Answer:

$\text{A. }\frac{5}{6}\pi x^2$

Step-by-step explanation:

The area of a sector with measure $\theta$ is equal to $r^2\pi\cdot \frac{\theta}{360}$

Since there are 360 degrees in an angle, the measure of sector K must be 360-60=300 degrees.

What we know:

$\theta$ of $300^{\circ}$
$r$ of $x$

Thus, the area of the sector is $x^2\pi\cdot \frac{300}{360}=\boxed{\text{A. }\frac{5}{6}\pi x^2}$

3 0

3 years ago