9514 1404 393
Answer:
- 2nd force: 99.91 lb
- resultant: 213.97 lb
Step-by-step explanation:
In the parallelogram shown, angle B is the supplement of angle DAB:
∠B = 180° -77°37' = 102°23'
Angle ACB is the difference of angles 77°37' and 27°8', so is 50°29'.
Now, we know the angles and one side of triangle ABC. We can use the law of sines to solve for the other two sides.
BC/sin(A) = AB/sin(C)
AD = BC = AB·sin(A)/sin(C) = (169 lb)sin(27°8')/sin(50°29') ≈ 99.91 lb
AC = AB·sin(B)/sin(C) = (169 lb)sin(102°23')/sin(50°29') ≈ 213.97 lb
None because it is impossible for any triangle to have more than one obtuse angle.
∵ ΔSRQ is a right triangle
and RT⊥SQ
∴ x² = ST * TQ
∴ x² = 9 *16 = 144
∴ x = √144 = 12
Answer: idk
Step-by-step explanation:
idk
You have to distribute!
6(x) = 6x and 6(-6)=-36
So now you have 6x-36=-3(-x+3)
Now do the same on the other side.
-3(-x)=3x -3(3)=-9
So now, 6x-36=3x-9
Combine the like terms:
Subtract 3x in both sides.
6x-3x=3x
Now: 3x-36=-9
Add 36 in both sides.
3x=27
Divide by 3 on both sides to get x alone.
3x/3= x 27/3=9
X=9