1 = 51
2= 61
3= 29
Step-by-step explanation:
1) 61+68= 129
180-129= 51
2) 61 is across it so they have the same number
3) 61 +90= 151
180-151= 29
Answer:
Minimum 8 at x=0, Maximum value: 24 at x=4
Step-by-step explanation:
Retrieving data from the original question:
![f(x)=x^{2}+8\:over\:[-1,4]](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B2%7D%2B8%5C%3Aover%5C%3A%5B-1%2C4%5D)
1) Calculating the first derivative

2) Now, let's work to find the critical points
Set this
0, belongs to the interval. Plug it in the original function

3) Making a table x, f(x) then compare
x| f(x)
-1 | f(-1)=9
0 | f(0)=8 Minimum
4 | f(4)=24 Maximum
4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
Answer:
it would be 3.28125
Step-by-step explanation:
do base times height to get the area