FOR THE FIRST PROBLEM:
Differentiating p^2 + 2q^2 = 1100 with respect to time => 2p(dp/dt) + 4q(dq/dt) = 0
<span>dp/dt = -2q/p(dq/dt) </span>
<span>R = pq </span>
<span>dR/dt = p(dq/dt) + q(dp/dt) = -p²/2q(dp/dt) + q(dp/dt) = (2q² - 1100)/2q * (dp/dt) + q(dp/dt) </span>
<span>A = (2q - 550/q)
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So, what’s the question, I probably can answer it if you include the question
Answer:
1/3
Step-by-step explanation:
if you move up 1 over three, the line intersects the graph
well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)
Answer:
15
Step-by-step explanation:
2/3 of 30 is 20, so 1/2 of 30 is 15