Assuming you mean f(t) = g(t) × h(t), notice that
f(t) = g(t) × h(t) = cos(t) sin(t) = 1/2 sin(2t)
Then the difference quotient of f is
![\dfrac{\frac12 \sin(2(t+h)) - \frac12 \sin(2t)}h = \dfrac{\sin(2t+2h) - \sin(2t)}{2h}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cfrac12%20%5Csin%282%28t%2Bh%29%29%20-%20%5Cfrac12%20%5Csin%282t%29%7Dh%20%3D%20%5Cdfrac%7B%5Csin%282t%2B2h%29%20-%20%5Csin%282t%29%7D%7B2h%7D)
Recall the angle sum identity for sine:
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
Then we can write the difference quotient as
![\dfrac{\sin(2t)\cos(2h) + \cos(2t)\sin(2h) - \sin(2t)}{2h}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csin%282t%29%5Ccos%282h%29%20%2B%20%5Ccos%282t%29%5Csin%282h%29%20-%20%5Csin%282t%29%7D%7B2h%7D)
or
![\boxed{\sin(2t)\dfrac{\cos(2h)-1}{2h} + \cos(2t)\dfrac{\sin(2h)}{2h}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csin%282t%29%5Cdfrac%7B%5Ccos%282h%29-1%7D%7B2h%7D%20%2B%20%5Ccos%282t%29%5Cdfrac%7B%5Csin%282h%29%7D%7B2h%7D%7D)
(As a bonus, notice that as h approaches 0, we have (cos(2h) - 1)/(2h) → 0 and sin(2h)/(2h) → 1, so we recover the derivative of f(t) as cos(2t).)
In quadrant I, when sin angle is 1/2, cos angle is (sqrt3)/2
Answer:
![\huge\boxed{\text{Width: } 24 \ \ \ \text{Length: } 32}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Ctext%7BWidth%3A%20%7D%2024%20%5C%20%5C%20%5C%20%5Ctext%7BLength%3A%20%7D%2032%7D)
Step-by-step explanation:
When we scale an object by a scale factor, the side lengths become the scale factor times bigger than the previous sides. Therefore, with a scale factor of 6, the side lengths would be multiplied by 6.
![4 \cdot 6 = 24\\\\4 \cdot 8 = 32](https://tex.z-dn.net/?f=4%20%5Ccdot%206%20%3D%2024%5C%5C%5C%5C4%20%5Ccdot%208%20%3D%2032)
Hope this helped!
Answer:
x²+6x+9=0
x²+3x+3x+9=0
x(x+3)+3(x+3)=0
(x+3)(x+3)=0
either
x+3=0
x=-3
or
x+3=0
x=-3
Step-by-step explanation:
next method
x²+6x+9=0
x²+2×x×3+3²=0
it is in formula of (x+y)²
(x+3)²=0
x+3=√0
x+3=0
x=-3
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