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qaws [65]
3 years ago
15

What is the derivative of 5cos^2(pi)(t)?

Mathematics
1 answer:
Eva8 [605]3 years ago
3 0
<span>i assume you mean 5cos^2((pi*t)) in which case; i would use the chain rule, subs u=cos(pi*t), so you have (i presume it's y), y=5u^2 dy/du = 10u and du/dt=-pi*sin(pi*t) Multiply them together, dy/dx = 10u * -pi*sin(pi*t) replace u=cos(pi*t) dy/dx=-10*pi*cos(pi*t)*sin(pi*t) And you could then use the relation that sin(2x)=2sin(x)cos(x) so dy/dx=-5*pi*sin(2*pi*t) You could alternatively use the relation that 2cos(2x)=cos^2(x)+1 and rearrange for cos^2(x) [x being pi*t] .. which is a better way if you know how to do it .. hope this helps :)</span>
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Answer:

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Step-by-step explanation:

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Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3​
Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

f(x)=x^4-2ax^3+ax^2- x+k

and

f(-1)=3

Answer:

k = -2 and a=1

Step-by-step explanation:

Given

f(x)=x^4-2ax^3+ax^2- x+k

Factor:\ x - 4

f(-1)=3

Required

Find a and k

For f(-1)=3

Substitute -1 for x

f(x)=x^4-2ax^3+ax^2- x+k

f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k

f(-1) = 1 - 2a *-1 + a*1 +1 + k

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Substitute 3 for f(-1)

3 = 2 +3a + k

Collect Like Terms

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1 = 3a + k

Also:

If x - 2 is a factor, then

f(2) = 0

Substitute 2 for x and 0 for f(x)

f(x)=x^4-2ax^3+ax^2- x+k

0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k

0 = 16 - 2a * 8 + a * 4 - 2 + k

0 = 16 - 16a + 4a - 2 + k

0 = 16 - 12a - 2 + k

Collect Like Terms

2 - 16 = - 12a + k

-14 = k - 12a

k = 12a - 14

Substitute 12a - 14 for k in 1 = 3a + k

1 = 3a + 12a - 14

1 = 15a - 14

Collect Like Terms

15a = 1 + 14

15a = 15

Solve for a

a = \frac{15}{15}

a=1

Substitute 1 for a in k = 12a - 14

k = 12 * 1 - 14

k = 12 - 14

k = -2

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