Can someone plz help me with this one problem plzzzzz I’m being timed!!!!

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

3.

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

4.

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

3 years ago

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Use the linear combination method to solve the system of equations. Explain each step of your solution. 2x-3y=13 4x-y=-9

Tems11 [23]

2x -3y = 13

4x -y = -9

Multiply the second equation by -3 to make the coefficient of Y opposite the first equation.

4x -y = -9 x -3 = -12x + 3y = 27

Now add this to the first equation:

2x -12x = -10x

-3y +3y = 0

13 +27 = 40

Now you have :

-10x = 40

Divide each side by -10:

x = 40 / -10

x = -4

Now you have a value for x, replace that into the first equation and solve for y:

2(-4) - 3y = 13

-8 - 3y = 13

Add 8 to both sides:

-3y = 21

Divide both sides by -3:

y = 21/-3

y = -7

Now you have X = -4 and y = -7

(-4,-7)

3 0

3 years ago

What is the answer. to.

nydimaria [60]

I believe it is (2) 60

7 0

3 years ago

A Square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches.The area of the result

LekaFEV [45]

Let s represent the length of any one side of the original square. The longer side of the resulting rectangle is s + 9 and the shorter side s - 2.

The area of this rectangle is (s+9)(s-2) = 60 in^2.

This is a quadratic equation and can be solved using various methods. Let's rewrite this equation in standard form: s^2 + 7s - 18 = 60, or:

s^2 + 7s - 78 = 0. This factors as follows: (s+13)(s-6)=0, so that s = -13 and s= 6. Discard s = -13, since the side length cannot be negative. Then s = 6, and the area of the original square was 36 in^2.

4 0

3 years ago

Toms Toy car went 2 meters without stopping Roberts went 145 centimeters whose car went farther

r-ruslan [8.4K]

Answer:

Tom's car went further

Step-by-step explanation:

Since there are 100 centimeters in 1 meter - Roberts car went nearly 1 and a half meters or Tom's car went 200 centimeters.

6 0

3 years ago

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