Answer: The answer is 18
Step-by-step explanation:
The figures are the same shape, just one is longer. As you can see, the smallest side on the smaller figure is 3 inches. If you look at the larger figure, the smallest side is 6 inches. From this, you can see that you are supposed to multiply the lengths on the smaller figure by 2, to the lengths for the larger figure.
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.
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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))
Answer:
the answer is 3
Step-by-step explanation:
because if you do the methoes method you know that it rules out 1 and 2 after you do that its just between 3 and 4, for the caunususicon there has to be one, that is why i think its 3. hope this helps!
9514 1404 393
Answer:
A) ∠DHG = 96°
B) arc EF = 134°
Step-by-step explanation:
<u>Part A</u>:
As you have shown in your work, angle DHG is supplementary to given angle FHG:
∠DHG = 180° -∠FHG
∠DHG = 180° -84°
∠DHG = 96°
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<u>Part B</u>:
∠DHG is half the sum of arcs EF and DG.
96° = (1/2)(arc EF + 58°) . . . . . fill in known values
192° -58° = arc EF . . . . . . . . . . multiply by 2, subtract 58°
arc EF = 134°
