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:
multiply the first equation by 2
Step-by-step explanation:
24x + 36y = 72
-16x + 8y = 80
48x + 72y = 144
-48x + 24y = 240
96y = 384
y = 4
24x + 36(4) = 72
24x + 144 = 72
24x = -72
x = -3
(-3, 4)
Answer:
Step-by-step explanation:
If MQ also bisects angle ∠LMP, then;
<LMQ + <QMP = <LMP
Also <LMQ = <QMP
The equation becomes <QMP + <QMP = <LMP
2<QMP = <LMP
Let <LMP be 8x+40° (This is an assumed value)
2(5x+10) = 8x+40
Open the parenthesis
10x+20 = 8x+40
10x-8x = 40-20
2x = 20
x = 20/2
x = 10
Find <QMP
Since <QMP = 5x+10
<QMP = 5(10)+10
<QMP = 50+10
<QMP = 60 degrees
<em>Note that the angle <LMP was assumed. Any other function can be used depending on the question</em>
