![f(x)=(x+4)(x-2)(x+4)(x+4)\\f(x)=(x+4)^3(x-2)=0\\(x+4)^3=0~|~(x-2)=0\\(x+4)(x+4)(x+4)=0~|~x=2\\x=-4,2](https://tex.z-dn.net/?f=f%28x%29%3D%28x%2B4%29%28x-2%29%28x%2B4%29%28x%2B4%29%5C%5Cf%28x%29%3D%28x%2B4%29%5E3%28x-2%29%3D0%5C%5C%28x%2B4%29%5E3%3D0~%7C~%28x-2%29%3D0%5C%5C%28x%2B4%29%28x%2B4%29%28x%2B4%29%3D0~%7C~x%3D2%5C%5Cx%3D-4%2C2)
The multiple zero is x = -4.
The multiplicity is 2 because there are two values of x.
Answer:
OF COURSE SHE ISNT A MORNING PERSON. Anyway....
8:45-25 minutes is 8:20. -30 minutes from that, and you get 7:50. PRESUMING she wants to get to class the SECOND the bell rings, she should leave at 7:50!
:))
Answer:
3/2
Step-by-step explanation:
sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =
![\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B4%7D-%5Cbeta%29%3D%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B4%7D%29%5Ccos%5Cbeta-%5Ccos%5Cfrac%7B3%5Cpi%7D%7B4%7D%5Csin%5Cbeta%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Ccos%5Cbeta-%28-%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%29%5Csin%5Cbeta%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%28%5Ccos%5Cbeta%20%2B%5Csin%5Cbeta%29%5C%5C%5C%5C%5Csin%5E2%28%5Cfrac%7B3%5Cpi%7D%7B4%7D-%5Cbeta%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ccos%5Cbeta%20%2B%5Csin%5Cbeta%29%5E2%3D%5Cfrac%7B1%7D%7B2%7D%28%5Ccos%5E2%5Cbeta%20%2B%5Csin%5E2%5Cbeta%2B2%5Csin%5Cbeta%5Ccos%5Cbeta%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%281%2B%5Csin2%5Cbeta%29%3D%5Cfrac%7B1%7D%7B2%7D%281-%5Cfrac%7B1%7D%7B5%7D%29%20%3D%20%5Cfrac%7B2%7D%7B5%7D%5C%5C)
Use ![\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1](https://tex.z-dn.net/?f=%5Ccot%5E2x%3D%5Ccsc%5E2x-1%3D%5Cfrac%7B1%7D%7B%5Csin%5E2x%7D-1)
so
![\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}](https://tex.z-dn.net/?f=%5Ccot%5E2%28%5Cfrac%7B3%5Cpi%7D%7B4%7D-%5Cbeta%29%3D%5Cfrac%7B1%7D%7B%5Csin%5E2%28%5Cfrac%7B3%5Cpi%7D%7B4%7D-%5Cbeta%29%7D-1%20%3D%20%5Cfrac%7B1%7D%7B%5Cfrac%7B2%7D%7B5%7D%7D-1%3D%5Cfrac%7B5%7D%7B2%7D-1%3D%5Cfrac%7B3%7D%7B2%7D)
It is given in the question that
![h(x) = (fog)(x) = x^2 -81](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%28fog%29%28x%29%20%3D%20x%5E2%20-81)
And ![x^2 -81](https://tex.z-dn.net/?f=x%5E2%20-81)
can also be written as
![(x^2 -40) -41](https://tex.z-dn.net/?f=%28x%5E2%20-40%29%20-41)
Therefore,
![(fog)(x) = (x^2 -40)-41](https://tex.z-dn.net/?f=%28fog%29%28x%29%20%3D%20%28x%5E2%20-40%29-41)
![f(g(x)) = (x^2 -40)-41](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%28x%5E2%20-40%29-41)
That gives,
![g(x) = x^2 -40, f(x) = x-41](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%5E2%20-40%2C%20f%28x%29%20%3D%20x-41)
ST/VT = UT/WT6+14/14 = (x+4)+21/21X= 20*21/14 - 21x=420/14-25X=30-25x=5