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serg [7]
3 years ago
7

Plz help

Mathematics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

p=-1

Step-by-step explanation:

3x^2+6x-6=0

Add 6 on both sides

3x^2+6x=6

Divide through by 3

x^2+2x=2

Complete the square on the right side

If a quadratic equation is given by

ax^2+bx+c=0

then c=b/2

x^2+2x=2

c=2/2=1

Adding 1 on both sides

x^2+2x+1=2+1

x^2+x+x+1=3

Combining

x(x+1)+1(x+1)=3

(x+1)(x+1)=3

(x+1)^2=3

p=-1

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An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
Factor the expression below. Pls help!
amm1812
I think this is right (x-2)(2x+y) 
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https://percentagecalculator.net/

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That is an interesting fact thank you for sharing your information

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