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never [62]
3 years ago
8

Please help I got the answer wrong-

Mathematics
1 answer:
puteri [66]3 years ago
8 0

Answer:

x {}^{2}  = 150

Step-by-step explanation:

What I did was

5 \times 5 = 25

then I did

5 \times 5 = 25 \times 5 = 150

and I'm hoping 150 is the answer. HOPE THIS HELPED

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Slips of paper are numbered 1 through 10. If one slip is drawn and replaced 40 times, how many times should the slip with number
Masja [62]

Hey there!

We have 10 pieces of paper. The probability of drawing a 10 is 1/10.  Therefore, if we were to draw ten times, the ten should probably be drawn once. If we multiplied our number of drawings by four (10*4=40), our outcome of seeing our slip with the ten should also quadruple and become four times. (1*4=4)

I hope that this helps! Have an awesome day!

6 0
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What’s the unit rate for 400 in 8 hours? Please help.
laila [671]
I think the answer is: 1/50
7 0
3 years ago
What is the answer to this problem?
emmasim [6.3K]
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A yellow ribbon is 56 centimeters long. It is twice as long as a green ribbon. A brown ribbon is 4 times as long as the green ri
Shkiper50 [21]

56 dived by 2 equals 28. 28=green ribbon. Brown ribbon is 4 times as long. 28 times 4 equals 112. The brown ribbon is 112 cm long.

6 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
3 years ago
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