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Degger [83]
4 years ago
7

The endpoints of `bar(EF)` are E(xE , yE) and F(xF , yF). What are the coordinates of the midpoint of `bar(EF)`?

Mathematics
1 answer:
Hoochie [10]4 years ago
6 0

For a better understanding of the solution provided here please find the diagram attached.

Please note that in coordinate geometry, the coordinates of the midpoint of a line segment is always the average of the coordinates of the endpoints of that line segment.

Thus, if, for example, the end coordinates of a line segment are (x_{1}, y_1) and (x_2, y_2) then the coordinates of the midpoint of this line segment will be the average of the coordinates of the two endpoints and thus, it will be:

(\frac{(x_1+x_2)}{2}, \frac{(y_1+y_2)}{2})

Thus for our question the endpoints are (x_E, y_E) and (x_F, y_F) and hence the midpoint will be:

(x_M, y_M)=(\frac{(x_E+x_F)}{2}, \frac{(y_E+y_F)}{2})

Thus, Option C is the correct option.

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Answer: D

Step-by-step explanation:

It seems there is two ways to solve this, remember that this is just one of them.

The triangle to the left is an isosceles triangle by definition. It is given that two of its sides are equal.

The angles opposite the equal sides in an isosceles triangle are congruent. It is given that one of these angles is 70 degrees, so the other one must be 70 degrees as well.

This angle is opposite to Angle Y. Vertical angles are congruent, so Angle Y must be 70 degrees.

D is the answer.

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3 years ago
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Step-by-step explanation:

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3 years ago
If a certain cannon is fired from a height of 8.8 meters above the​ ground, at a certain​ angle, the height of the cannonball ab
Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

h(t)=-4.9t^2+30.5t+8.8

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

-4.9t^2+30.5t+8.8=0

Multiply both sides by 10.

-49t^2+305t+88=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a = -49, b = 305 and c=88

t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}

Ignore the negative value of t as time can't be a negative number.

Thus,

t=\frac{305+\sqrt{110273}}{98}\approx6.50

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

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3 years ago
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