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: 
Step-by-step explanation:
See attached image, it's direct formula substitution.
Answer:
9/40
Step-by-step explanation:
3/5 * 3/8
Multiply the numerators
3*3 = 9
Multiply the denominators
5*8 = 40
9/40
The fraction does not simplify
Answer:
B
Step-by-step explanation:
B
x - 4 squared provides 2 real zeros that are the same. These two are not distinct.
The other two come from x^2 - 7x + 10 which has a discriminate of
sqrt(7^2 - 4*1*10) = sqrt(9) = +/- 3 leading to something real and different.
Answer:
144
Step-by-step explanation:
On this shape we are going to splt up the 4 triangles for the square to make it simple.
So, firstly, plug is the h an w of one of the triangles into the formula for the area of a triangle which is hxw/2
H=9
W=6
9x6/2=27
Then multiply the area of the triangle, 27, to get the total area of all 4 triangles (since they all have the same measurements)
27x4=108
108 is the area of all 4 triangles
Next, just put the l and w into the equation for the area of a square which is lxw
L=6
W=6
6x6=36
36 is the area of the square
Lastly, just add the areas of the triangles and the square together
108+36=144
144 is the total area/surface area, of the pyramid
