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Valentin [98]
3 years ago
12

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=3, u'(0)= -4, v(0)= 2, v'(0)= -7. Find the value

s of the following derivatives at x=0. A.) d/dx(uv)
Mathematics
2 answers:
3241004551 [841]3 years ago
7 0
\bf \cfrac{d}{dx}[u(x)\cdot v(x)]\implies \cfrac{du}{dx}\cdot v(x)+u(x)\cdot \cfrac{dv}{dx}\quad 
\begin{cases}
u(0)=3\qquad u'(0)=-4
\\\\
v(0)=2\qquad v'(0)=-7
\end{cases}
\\\\
\left[ \cfrac{du}{dx} \right]_{x=0}\cdot v(0)+u(0)\cdot \left[ \cfrac{dv}{dx} \right]_{x=0}

so.. hmm pretty sure you know what that is
grin007 [14]3 years ago
6 0

Answer:

none of these are useful

Step-by-step explanation:

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(A) We let y = the cost and are told x = the number of people. Since you pay $20 per person, the cost is 20x. That is, y=20x

(B) Again, let the cost =y and the number of people is given as x. You pay $10 per person or 10x plus an additional $50 for the room. That is, y=10x+50

(c) Link to graphs: https://www.desmos.com/calculator but if that doesn't work see the attachment for a screen shot. You just have to put the equations (type them) at left and the graph comes automatically.

(D) The admission price is the same when the two equations are equal. You can find this by setting them equal to each other as such: 20x = 10x+50 and solving for x. However, since you just graphed them the point of intersection (where the lines share/have the same point) gives the information. Remembers that (x,y) = (people, cost). The graphs intersect at (5, 100) so for 5 people the cost is the same and the cost is $100.

(E) For the regular rate we let x = 6 and solve for y (the cost). We get y = 20x which is y = (20)(6)=120. It costs $120 using the regular rate to take 6 people. Now let's use the equation for the group rate again with x = 6. Here we get y = 10x +50 or y = 10(6)+50 = $110. The group rate costs $110.

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