The perimeter of a semicircle is doubled when the radius is increased by 3. Find the radius of the semicircle.

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The perimeter of a semicircle is doubled when the radius is increased by 3. Find the radius of the semicircle.

Algebra
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perimeter of a semicircle is r*pi + 2r so the old perimeter is r*pi + 2r and the new perimeter is (r+3)*pi + 2(r+3) we know that the new perimeter is twice the old perimeter so: 2*(r*pi + 2r) = (r+3)*pi + 2(r+3) solve for r
did you understand
\[2(\pi r+2r)=(r+3)\pi +2(r+3)\] does it look like this

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yes, do you know where to go from there?
\[\pi r =3\pi - 2r+6\] i end up to this :(
i distributed first
hmm let's take it from the top 2*(r*pi + 2r) = (r+3)*pi + 2(r+3) 2*r*pi + 4r = pi*r + 3*pi + 2r + 6 pi*r + 2r = 3*pi + 6 we can factor r on the left side, and 3 on the right side giving us: r(pi+2) = 3(pi + 2) then we divide both sides by (pi + 2) r = 3 is that clear?
oh, i did not factor... thanks. its now clear to me
uhmmmm can i ask 5 more problems?
sure
Joy is an instructional assistant in one of the college labs. She is on a very tight budget. She loves Jamaican Blue Mountain coffee, but it costs $15 a pound. She decides to blend this with regular coffee beans that costs $1.67 a pound. If she spends $14.25 on 3 pounds of coffee, how many pounds of each did she purchase? Round your answers to two decimal places.
let's let b = number of pounds of blue mountain and r = regular r + b = 3 since there are 3 pounds of coffee 15b + 1.67r = 14.25 know where to go from there?
no :(
sorry
ok, that's fine, no need to apologize we have 2 equations that will help us find what r and b are r + b = 3 15b + 1.67r = 14.25 we can multiply the first equation by (-15), giving us -15r - 15b = -45 now we can put the two equations together again -15r - 15b = -45 15b + 1.67r = 14.25 now we can add the two equations together -15r + 1.67r - 15b + 15b = -45 + 14.25 solve for r
i got 2.31 for r. that is for the regular? what about the pounds of jamaican blue mountain coffee?
i'll multiply r+b=3 to -1.67?
thanks!!! i got it right... thanks a lot! :)))
A patient requires an IV of 0.9% saline solution, also known as normal saline solution. How much distilled water, to the nearest milliliter, must be added to 100 milliliters of a 6 % saline solution to produce normal saline?

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