The solution for proving the identity is as follows:
sin(2A) = sin(A + A)
As sin(a + b) = sinacosb + sinbcosa,
<span>sin(A + A) = sinAcosA + sinAcosA
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<span>Therefore, sin(2A) = 2sinAcosA
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can also be written as 2i
where i = √-1
Rationalizing denominator,
I=prt
20=p X .05 X 5
20 = p X .25
20/.25 = p
80 = p
She started with $80
You can reword the two equations as:
-5x-y=15 (Divide original value by 3)
-2x+6y=6
Then use elimination to find x:
-30x-6y=90 (Multiply by 6 to get y values to be same to cancel out)
-2x+6y=6
You're left with:
-32x=96. Which can then be solved to find x which is -3.
Then plug back in
-2x+6y=6
Now to: -2(-3)+6y=6.
Which reduces to 6+6y=6. So y=0.
To graph them, just reword the equations (yes once again) so that y is in front.
y=-5x-15 and y=(1/3)x+1
