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4 years ago

If a point is equidistant from the endpoints of a segment, then it is:

Mathematics

2 answers:

Rom4ik [11]4 years ago

4 0

It's c . On the perpendicular bisector of the segment

Burka [1]4 years ago

4 0

Answer:

Step-by-step explanation:

A point equidistant from the endpoints of a segment not necessarily is on the segment, then, options A and D are discarded (the midpoint of the segment is on the segment). The angle bisector divides an angle in two equal angles, and is not applied to a segment; then, option B is discard. The perpendicular bisector of a segment divide a segment in two equal parts, then, all of its points are equidistant from the endpoints of the divided segment.

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Solve 5x+2y=4 x-y=5 (by elimination)I really need help

Ray Of Light [21]

I hope this helps you

x=5+y

5 (5+y)+2y= 4

25+5y+2y=4

7y= -21

y= -3

x=5+y

x=5-3 x= 2

6 0

3 years ago

Write an equation for the line parallel to the given line that contains <br> C(4,8);y=-3x+4

Brrunno [24]

Parallel line means they have the same slope
y = mx + b
where m = slope
y = -3 + 4
here the slope is negative 3
y = -3x + b
plug in (4, 8)
8 = -3(4) + b
b = 8 + 12
b = 20

The answer is y = -3x + 20

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3 years ago

9y = 3y Combine the like terms

-BARSIC- [3]

9y-3y=0
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Hope that helps

5 0

4 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago

Please please answer this correctly I have to finish this today as soon as possible

Simora [160]

Answer:

Step-by-step explanation:

Look at where -2 is on the x-axis (the left side of the horizontal line), and go another half a box in, that is -2.5. But then go down one box from there, and plot the point. that is (-2.5,-1).

Next go half a box to the right and two boxes up, and plot your point there. that is (.5,2).

8 0

3 years ago