Answer:
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Answer:
15) K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]
19) P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]
20) Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)
Step-by-step explanation:
We are to find the derivative of the questions pointed out.
15) K(t) = 5(5^(t)) - 2(3^(t))
Using implicit differentiation, we have;
K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]
19) P(w) = 2e^(w) - (2^(w))/5
P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]
20) Q(W) = 3w^(-2) + w^(-2/5) - w^(¼)
Q'(w) = -6w^(-2 - 1) + (-2/5)w^(-2/5 - 1) - ¼w^(¼ - 1)
Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)
Answer:
-2, -1, 0, 1, 2
Step-by-step explanation: divide each of them by 2 because y = 1/2 of x
not sure this is correct this is just what I think it is.
Eq1) 2r+2s=50
eq2) 2r-s=17
solve for s in equation2 (eq2)
-s=17-2r
s=-17+2r
Substitute s into equation1 (eq1)
2r+2(-17+2r)=50
2r-34+4r=50
6r-34=50
6r=50+34
6r=84
r=14
Substitute into either equation and solve for s
2(14)-s=17
28-s=17
-s=17-28
-s=-11
s=11
