1 because if you were to get h rn e measure of the problem right here it would be a cosine of 1
Answer:
there's a minimum at (4, -1)
Step-by-step explanation:
Please use " ^ " to denote exponentiation: y= x^2 - 8x+ 15.
To "complete the square," take half of the coefficient of x (which is -8). Square this result, obtaining 16.
In y= x^2 - 8x+ 15, add 16, and then subtract 16, between -8x and +15:
y = x^2 - 8x + 16 - 16 + 15
This becomes:
y = (x - 4)^2 -1
Reading off the coordinates of the vertex, we get (4, -1). Because the coefficient of the (x - 4)^2 term is positive, we know .there's a minimum at (4, -1)
X=number; then:
9x+5=38+6x
9x-6x=38-5
3x=33
x=33/3
x=11
Answer:
Step-by-step explanation:
![sin (\frac{\pi}{2} - x) = cos x \\\\ sin (a - b) = sin a.cos b - sin b.cos a \\\\ sin (\frac{\pi}{2} - x) = sin \frac{\pi}{2}.cos x - sin x.cos \frac{\pi}{2} \\\\ sin \frac{\pi}{2} = 1; cos \frac{\pi}{2} = 0 \\\\ sin (\frac{\pi}{2} - x) = 1.cos x - sin x.0 \\\\ sin (\frac{\pi}{2} - x) = cos x](https://tex.z-dn.net/?f=sin%20%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20x%29%20%3D%20cos%20x%20%5C%5C%5C%5C%20sin%20%28a%20-%20b%29%20%3D%20sin%20a.cos%20b%20-%20sin%20b.cos%20a%20%5C%5C%5C%5C%20sin%20%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20x%29%20%3D%20sin%20%5Cfrac%7B%5Cpi%7D%7B2%7D.cos%20x%20-%20sin%20x.cos%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5C%5C%5C%5C%20sin%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%201%3B%20cos%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%200%20%5C%5C%5C%5C%20sin%20%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20x%29%20%3D%201.cos%20x%20-%20sin%20x.0%20%5C%5C%5C%5C%20sin%20%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20x%29%20%3D%20cos%20x)
I hope I helped you.