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bulgar [2K]
3 years ago
12

What types of lines create alternate interior angles?

Mathematics
1 answer:
Softa [21]3 years ago
3 0
C should be the correct answer
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Which of the following correctly describes the interval shown?
adoni [48]
B is the correct one
5 0
3 years ago
7/4 x 16/5 as a inproper fraction if posibble
creativ13 [48]

Answer:

28/5

Step-by-step explanation:

6 0
2 years ago
Is EFGH below a rectangle
poizon [28]

Answer:

  D. No, because EFGH is a parallelogram but it’s diagonals are not congruent

Step-by-step explanation:

The differences between the end points of the diagonals are ...

  F -H = (1, 1) -(2, -5) = (1 -2, 1 -(-5)) = (-1, 6)

  G -E = (4, -2) -(-1, -2) = (4 -(-1), -2 -(-2)) = (5, 0)

The length of FH is more than 6, the length of GE is exactly 5. The diagonals are different length, so the figure cannot be a rectangle.

__

The midpoints of the diagonals will be in the same place if the sum of their end points is the same. (Dividing each sum by 2 gives the midpoint of that segment.)

  F+H = (1, 1) +(2, -5) = (3, -4)

  G+E = (4, -2) +(-1, -2) = (3, -4)

The diagonals bisect each other (have the same midpoint), so the figure is a parallelogram.

EFGH is a parallelogram, but not a rectangle: its diagonals are not congruent.

3 0
2 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
Which figures are polygons?
MA_775_DIABLO [31]
Im pretty sure the answer would be A, B & D
4 0
3 years ago
Read 2 more answers
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