Unlike the previous problem, this one requires application of the Law of Cosines. You want to find angle Q when you know the lengths of all 3 sides of the triangle.
Law of Cosines: a^2 = b^2 + c^2 - 2bc cos A
Applying that here:
40^2 = 32^2 + 64^2 - 2(32)(64)cos Q
Do the math. Solve for cos Q, and then find Q in degrees and Q in radians.
Answer:
bbc
bbc
ssc
Step-by-step explanation:
hahah
Answer:
1-3x+8y
Step-by-step explanation:
Multiply y and 2
Multiply y and 1
The y just gets copied along.
2*y evaluates to 2y
Multiply x and 4
Multiply x and 1
The x just gets copied along.
The answer is x
4*x evaluates to 4x
2*y-4*x evaluates to 2y-4x
The answer is 2y-4x+8
2*y-4*x+8 evaluates to 2y-4x+8
Multiply y and 6
Multiply y and 1
The y just gets copied along.
The answer is y
6*y evaluates to 6y
2y + 6y = 8y
The answer is 8y-4x+8
2*y-4*x+8+6*y evaluates to 8y-4x+8
-4x + x = -3x
The answer is -3x+8y+8
2*y-4*x+8+6*y+x evaluates to -3x+8y+8
8 - 7 = 1
The answer is 1-3x+8y
2*y-4*x+8+6*y+x-7 evaluates to 1-3x+8y
Answer:
I will do that buddy. I'll have fun
of the shaded portion.