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UNO [17]
3 years ago
15

HELP ME SOMEONE PLEASE!!!!!!!!!!!

Mathematics
1 answer:
Lelu [443]3 years ago
4 0

Answer:

The probability that they will both break down would be 8%

Step-by-step explanation:

I believe it is 8% (sorry if its wrong) because tractor 1 is 3% and tractor 2 is 5% add them together and that is where you get 8%

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the acceleration of an object due to gravity is 32 feet per second squared.what is acceleration due to gravity in inches per sec
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4 years ago
Plz help me it's very important I pass
Elis [28]
The answer is 30%) because all the angles of a triangle needs to add up to 180 so 180-75-75= 30
6 0
4 years ago
What is the length of the figure?
nadezda [96]
There are two ways to do this. One is by adding up the squares, which takes a while. The other way is if you notice that the length along the bottom is the same as that long the top, and the same is true for the sides. While it does not appear this way at first, imagine that that was the floor plan of a house. If you looked at it from the side, you wouldn't see the dent in the corner, only one side. Since the length of the top is 13 units, from -7 to 6, and the side is also 13 units, from -6 to 7, the answer is 13*2 + 13* 2 = 26 + 26 = 52

52 units long.
4 0
3 years ago
Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
4 years ago
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