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2 years ago

A parabola is graphed such that the axis of symmetry is x = 3. One of the x-intercepts is (5, 0). What is the other x-intercept?

Mathematics

1 answer:

Stells [14]2 years ago

6 0

The axis of symmetry is the fold line that splits the parabola down the middle.

Now, since a parabola is symmetrical, every point, except the vertex, will have a mirror image of another point if we folded the parabola over the axis of symmetry.

So if we know the axis of symmetry is x = 3 and one of our points

has the coordinates (5, 0), the other point will have the coordinates (3, 0).

The axis of symmetry is always halfway between the x-intercepts.

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Find the measure of each angle.

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Given:

Given that the isosceles trapezoid JKLM.

The measure of ∠K is 118°

We need to determine the measure of each angle.

Measure of ∠L:

By the property of isosceles trapezoid, we have;

$\angle K+\angle L=180^{\circ}$

$118^{\circ}+\angle L=180^{\circ}$

$\angle L=62^{\circ}$

Thus, the measure of ∠L is 62°

Measure of ∠M:

By the property of isosceles trapezoid, we have;

$\angle L \cong \angle M$

Substituting the value, we get;

$62^{\circ}=\angle M$

Thus, the measure of ∠M is 62°

Measure of ∠J:

By the property of isosceles trapezoid, we have;

$\angle J \cong \angle K$

Substituting the value, we get;

$\angle J =118^{\circ}$

Thus, the measure of ∠J is 118°

Hence, the measures of each angles of the isosceles trapezoid are ∠K = 118°, ∠L = 62°, ∠M = 62° and ∠J = 118°

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2 years ago

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vaieri [72.5K]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28

Nataliya [291]

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$ , and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$ , we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$

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