Answer:
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b. Let h (t) = ( sin (t) + cos (t)) t 4. Find h ′(t).
c. Determine the slope of the tangent line to the curve y = f (x) at the point where a = 1 if f is given by the rule f(x) = e^x sin (x).
d. Find the tangent line approximation L(x) to the function y = g (x) at the point where a = − 1 if g is given by the rule g (x) = ( x^2 + x ) 2^x .
Answer:
A) M'(w) = w^16 * 4^w [ 51 + 3w In4 ]
B) h'(t) = [ cos (t) - sin (t) ] t^4 + [ sin(t) + cos (t) ] 4t^3
C) f'(1) = e' [sin(1) + cos(1) ]
D) g'(a) = 0 - 1/2
L(x) = - 1/2 ( x + 1 )
Step-by-step explanation:
Attached below is the detailed solution of the problem
A) m(w) = 3w^17 * 4^w
M'(w) = w^16 * 4^w [ 51 + 3w In4 ]
B) h(t) = [sin(t) + cos(t) ] t^4
h'(t) = [ cos (t) - sin (t) ] t^4 + [ sin(t) + cos (t) ] 4t^3
C) f(x) = e^x sin (x). at a = -1
f'(1) = e' [sin(1) + cos(1) ]
D) g (x) = ( x^2 + x ) 2^x .
g'(a) = 0 - 1/2
L(x) = - 1/2 ( x + 1 )
Answer:
Step-by-step explanation:
Proportion of Heat Loss Between sundown and midnight
Proportion of Heat Loss between midnight and 4 AM
Proportion of Total Heat Already Lost
Proportion of Remaining Heat
Between 4 AM and 5 AM, five-eighths of the remaining heat is lost.
Proportion of Heat Loss between 4 AM and 5 AM=
Therefore, Proportion of Remaining Heat Left
We therefore say that:
Answer:53.8%
Step-by-step explanation:
270° counterclockwise about the origin
([?], [])
Enter the number that belongs in
the green box.
9514 1404 393
Answer:
P'(-1, -2)
Step-by-step explanation:
The transformation for 270° CCW rotation is ...
(x, y) ⇒ (y, -x)
Then the image of the given point is ...
P(2, -1) ⇒ P'(-1, -2)