$-8\dfrac{3}{4}+\left(-11\dfrac{1}{3}\right)=-\left(8\dfrac{3\cdot3}{4\cdot3}+11\dfrac{1\cdot4}{3\cdot4}\right)\\\\=-\left(8\dfrac{9}{12}+11\dfrac{4}{12}\right)=-\left[(8+11)+\left(\dfrac{9}{12}+\dfrac{4}{12}\right)\right]\\\\=-\left(19+\dfrac{13}{12}\right)=-\left(19+1\dfrac{1}{12}\right)=-20\dfrac{1}{12}$

7 0

3 years ago

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Line segment WX is the radius of circle X, and line segment ZY is the radius of circle Y. Points W, X, C, Y, and Z are all on li

Andrews [41]

Point D is the midpoint of the outer circle that we aim to find the area of

The circle has a diameter of WZ and radii of WC and CZ

We know that YZ=YD=10 cm

Let DC be $x$ and CY be $10-x$

The radius of the outer circle can be written as $8+8+x$ or $10+10-x$ which we can equate to find the value of $x$

$8+8+x=10+10-x$
$16+x=20-x$
$2x=4$
$x=2$

Therefore, the radius of the circle is $8+8+2=18$

And hence the area of the circle is $\pi (18^{2})$ =324 $\pi$

4 0

4 years ago

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