Can someone help me with reviewing for math fsa 7th grade...... Any topic

Mathematics

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Svetach [21]4 years ago

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Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

NemiM [27]

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$

3 0

4 years ago

Solve for k 5/8(3k+1)-1/4(7k+2)=1

lesya [120]

Answer:

Step-by-step explanation:

$\dfrac{5}{8}(3k+1)-\dfrac{1}{4}(7k+2)=1\qquad\text{multiply both sides by 8}\\\\8\!\!\!\!\diagup^1\cdot\dfrac{5}{8\!\!\!\!\diagup_1}(3k+1)-8\!\!\!\!\diagup^2\cdot\dfrac{1}{4\!\!\!\!\diagup_1}(7k+2)=8\cdot1\\\\5(3k+1)-2(7k+2)=8\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\(5)(3k)+(5)(1)+(-2)(7k)+(-2)(2)=8\\\\15k+5-14k-4=8\qquad\text{combine like terms}\\\\(15k-14k)+(5-4)=8\\\\k+1=8\qquad\text{subtract 1 from both sides}\\\\k=7$