-72yexponet73 + 2xexponet2 :)
The tension in both cables are calculated as; T₁ = 127.63 N and T₂ = 185.48 N
<h3>How to find the Tension in Suspended Cables?</h3>
Resolving forces in the y direction gives us;
T₁ sin 29° + T₂ sin 53° = 210 ------(eq 1)
Resolving forces in the horizontal direction gives;
T₁ cos 29° = T₂ cos 53° ----(eq 2)
T₁ = T₂ cos 53°/cos 29°
Thus;
(T₂ cos 53°/cos 29°)sin 29° + T₂ sin 53° = 210
0.3336T₂ + 0.7986T₂ = 210
T₂ = 210/1.1322
T₂ = 185.48 N
Thus;
T₁ = 185.48 cos 53°/cos 29°
T₁ = 127.63 N
Read more about Tension in Cables at; brainly.com/question/11653585
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We just plug in, simplify, and solve.
x/2 + 3x - y
8/2 + 3(8) - 2
4 + 24 - 2
28 - 2
26
Option C should be your answer.
The correct answer is for this is 1820 i completed this a day ago
Please mark as brainiest :)
14) -5(x+1)>20;
-5x-5>20;
-5x>20+5;
-5x>25 ;
-x>5
& x< - 5
15-1) 2.5+y < 7.5
y< 7.5 - 2.5
y<5 (answer c)
15-2) y/18<1/6 (cross multiplication):
6y<18
y < 3 (answer d)
15-3) y-1.5> - 0.5
y> - 0.5 +1.5
y> 1 (answer b)
15-4) -1.5y<4.5
- y < 4.5/1.5
- y< 3
y> -3 (answer a)
