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scoray [572]
3 years ago
9

What is the value of X in the figure? Enter your answer in the box. X=

Mathematics
2 answers:
bonufazy [111]3 years ago
7 0

Answer:

x = 29 degrees

Step-by-step explanation:

First we know that x+61 = 90

Then we can subtract 61 by both sides x = 29

Therefore, your answer would be x = 29

alukav5142 [94]3 years ago
7 0

Answer:

29 degrees

Step-by-step explanation:

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Aloiza [94]

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Step-by-step explanation:

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7 0
2 years ago
A+b+c=31 2a+3b+4c=105 who's a,b,c
Alik [6]

Answer:

begin{aligned}

   & x = \frac{D_x}{D} = \frac{-616}{-154} =  4 \\

   & y = \frac{D_y}{D} = \frac{ 616}{-154} = -4 \\

   & z = \frac{D_z}{D} = \frac{-770}{-154} =  5

         \end{aligned}begin{aligned}    & x = \frac{D_x}{D} = \frac{-616}{-154} =  4 \\    & y = \frac{D_y}{D} = \frac{ 616}{-154} = -4 \\    & z = \frac{D_z}{D} = \frac{-770}{-154} =  5          \end{aligned}

Step-by-step explanation:

4 0
3 years ago
What is the factor of 3x2-9x-12?<br> pleaseeee help me guys !!
8090 [49]

Answer:3x² - 9x - 12 =

3(x² - 3x - 4) =

3(x² + x - 4x - 4) =

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Step-by-step explanation:

8 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
3 years ago
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