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vesna_86 [32]
2 years ago
7

Thirty eight years ago George was 14. how old is he now

Mathematics
2 answers:
ahrayia [7]2 years ago
7 0
Lol I think it’s 38 I mean yeah right? It says thirty eight years ago which means he talking about the present
Anestetic [448]2 years ago
3 0

Answer:

I believe you need to add 38 and 14 together to get answer of 52 years old.

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Solve for x. ILL GIVE BRAINLIEST
Elodia [21]

Answer:

64

Step-by-step explanation:

the red height is 48 by pythagorean, and by 3:4:5 similarity x=64.

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http://www.ck12.org/geometry/Surface-Area-of-Triangular-Prisms/lesson/Surface-Area-of-Triangular-Pri...
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Find the slope of the graphed line.
spin [16.1K]

Answer:

-6/3

Step-by-step explanation:

Move down 6 from the top point.

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2 years ago
WILL GIVE BRAINIEST Which of the following is NOT a perfect square? Question 3 options: 36 25 49 81 35 144 100
jok3333 [9.3K]

Answer:

35

Step-by-step explanation:

The way I know is because I memorized every square up to 33.

BUT that's probably not helpful to you.

Remember that every perfect square has an odd number of factors. I'm not going to list them all out, but the factors of 35 are 1, 5, 7, and 35, giving it an even number of factors. All the rest have an odd number of factors because of the property of a perfect square: a number times itself gives a perfect square, but that number only counts as 1 factor.

4 0
3 years ago
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
2 years ago
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