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Suppose the facility hole"s radius is $,r,$ , then
$$pi r^2=(0.0136)pi(53+r)^2implies 0.9824 r^2-1.4416r-38.2024=0$$
Now just solve the above straightforward albeit pretty stroked nerves quadratic in $,r,$ ...
Let the radius of the center hole it is in x. Climate from the conditions we have.
$$r = 53 + x, ext whereby r is the radius the the CD$$
For the area that the facility hole we have:
$$P_1 = x^2pi$$
and because that the totality disk we have:
$$P = r^2pi$$
From the problems we have
$$P_1 = frac1.36100 P$$$$x^2pi = frac13610000 r^2pi$$$$10000x^2 = 136(53+x)^2$$$$10000x^2 = 136(2809+106x+x^2)$$$$10000x^2 = 382024 + 14416x + 136x^2$$$$9864x^2 - 14416x - 382024 = 0$$
Solving this quadratic equation we obtain two solutions:
$$x = frac53 (17+25 sqrt34)1233$$$$x = frac53 (17-25 sqrt34)1233$$
Because the radius need to be confident number we have:
$$ x = frac53 (17+25 sqrt34)1233 approx 7$$
So the radius the the center hole is 7 mm
reply Aug 8 "13 in ~ 12:26
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Lets say that the radius that the disk is $r$.
So because of this $r=53+x$ where $x$ is radius that hole.
The area, $A$ of the disk is$$A_disk=pi r^2$$$$A_disk=pi (53+x)^2$$
We likewise know that
This is since we know that the area that the feet is what castle took out of the CD
So because $A_hole=pi x^2$, and also $A_hole=0.0136A_disk$, for this reason we deserve to make the statement
And due to the fact that we recognize the area that the disk,
Accounting because that the hopeful only,
X is about 6.9967 i beg your pardon is about 7 devices so as such the radius that the hole is 7 units.
edited Aug 9 "13 at 5:30
reply Aug 9 "13 in ~ 4:57
Keith AfasKeith Afas
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