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AveGali [126]
3 years ago
6

Rewrite 6/25 in percent form

Mathematics
2 answers:
liberstina [14]3 years ago
5 0

Answer: 24%

Step-by-step explanation:

Divide and multiply by 100

\frac{6}{25} *100=24

Yuri [45]3 years ago
3 0
The answer would be 24%.
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The parking garage has 9 rows with 10 parking space in each row . There are 8 empty space . How many space are filled.
tatiyna

Answer:

82 .

Step-by-step explanation:

9x 10 = 90 then 90-8 = 82 that how many spaces are filled .

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3 years ago
Evan is thinking of a number that contains the digit 2. The digit 2 in Evan's number has 10 times the value of the digit 2 in th
Len [333]

Answer:

200,000

Step-by-step explanation:

The number Evan could be thinking about is 200,000. I got this number because the value of the digit 2 in the number 328,907 is 20,000. Also, the number Evan is thinking about is 10 times the value of 20,000. So, I did 20,000 times 10 and got 200,000.

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piper needs to buy 320 new ping-pong balls for the team ping pong balls come in 16 how many sets of ping-pong balls should piper
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20 sets

Step-by-step explanation:

320/16=20

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3 0
3 years ago
-x+ 3 = — 3х + 23 What is it?​
nekit [7.7K]

Answer:

x=10

Step-by-step explanation:

-x+ 3 = - 3x + 23\\2x+3=23 \\2x=20\\x=10

1. Subtract -3x from both sides of equation

2. Subtract 3 from both sides of equation

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4 0
3 years ago
Read 2 more answers
Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.
jeka94

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0
3 years ago
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