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Citrus2011 [14]
3 years ago
10

if k is a constant, determine and state the value of kids such that the polynomial k^2x^3-6kx+9 divisible by x-1

Mathematics
1 answer:
ZanzabumX [31]3 years ago
5 0
By the polynomial remainder theorem, k^2x^3-6kx+9 will be divisible by x-1 if the value of the polynomial at x=1 is 0.

k^2(1)^3-6k(1)+9=k^2-6k+9=(k-3)^2=0

which occurs for k=3.
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3 years ago
Which equation does the graph represent?
Art [367]

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It is the second answer

Step-by-step explanation:

The standard form of an ellipse is

x^2/a^2 + y^2/b^2  or x^2/b^2 + y^2/a^2 = 1

If the x is the main axis we use the first form.  If the y is the main axis we use the second form.  We will use the second form.

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3 0
2 years ago
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)

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3 years ago
Please help! What is 1 1/2+1/6?<br><br> -giving out 20 points
Mice21 [21]

1  \frac{2}{3}

Hope it helps

7 0
3 years ago
Read 2 more answers
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