For the given expressions we will have:
y = exp(x - 4) →we have a shift of 4 units to the right.
y = exp (x +9) → we have a shift of 9 units to the left.
y = exp(x) + 7 → we have a shift of 7 units up.
y = exp(x) - 6 → we have a shift of 6 units down.
<h3>
How to work with vertical and horizontal shifts?</h3>
Remember that the shifts work as follows.
For a function f(x), we define a vertical shift of N units as:
g(x) = f(x) + N
- If N > 0, the shift is upwards.
- If N < 0, the shift is downwards.
For a function f(x), we define a horizontal shift of N units as:
g(x) = f(x + N)
- If N > 0, the shift is to the left.
- If N < 0, the shift is to the right.
Then, if we have:
exp(x - 4) we have a shift of 4 units to the right.
exp (x +9) we have a shift of 9 units to the left.
exp(x) + 7 we have a shift of 7 units up.
exp(x) - 6 we have a shift of 6 units down.
Learn more about translations
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d) neither parallel or perpendicular
The answer is 91 ............
46/14. Hoped this helped!
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C
![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)
LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
