Use the equation 28/.70 and you should get your answer
Answer:
164
Step-by-step explanation:
You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.
The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").
By definition of inverse functions, we have
f(finv(x)) = finv(f(x)) = x
Differentiating both sides of the second equality with respect to x using the chain rule gives
finv'(f(x)) * f'(x) = 1
When x = a, we get
finv'(b) * f'(a) = 1
or
finv'(b) = 1/f'(a)
Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).
So we take a = π/3 and b = √3/2. Then
arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2
Answer:
a. Yes
b. VT
c. Segment RQ
Step-by-step explanation:
a. Find the slope of RS and UV
Slope = rise/run
Slope of RS = rise/run = RQ/QS
Slope of RS = 6/6
Slope of RS = 1
Slope of UV = rise/run = UT/TV
Slope of UV = 3/3
Slope of UV = 1
Thus, TS and UV have equivalent slopes
b. Slope of VT:
VT is an horizontal line.
It has no rise. But only run.
Therefore, it's rise = 0, while run = VT = 3
Slope of VT = rise/run = 0/3
Slope of VT = 0
c. Vertical lines have undefined slope.
Segment RQ is vertical line and therefore has an undefined slope.
RQ has rise but no run.
Thus:
Rise = 6
Run = 0
Slope of Segment RQ = 6/0 (this can't divide)
Therefore, slope of Segment RQ is undefined.
When solving, you will combine all your terms then divide the coefficient over.
2z+z+4z-4z+4z= 7z
7z = 0
/7 /7
z = 0