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2 years ago

If UT is a midsegment of QRS, find SQ.

Mathematics

1 answer:

serg [7]2 years ago

8 0

If UT is a midsegment,
SR = 2·UT
7x-1 = 2(3x+1)
We can add 1-6x to solve this
x = 3

Then the length SQ is twice the length of SU.
SQ = 2(2x+2) = 2(2·3+2)
SQ = 16

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1 year ago

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Center: (-5,10)

Radius: 2

Step-by-step explanation:

The equation of the circle in center-radius form is:

$(x-h)^2+(y-k)^2=r^2$

Where the point (h,k) is the center of the circle and "r" is the radius.

Subtract 121 from both sides of the equation:

$x^2+y^2+121-20y-121=-10x-121\\x^2+y^2-20y=-10x-121$

Add 10x to both sides:

$x^2+y^2-20y+10x=-10x-121+10x\\x^2+y^2-20y+10x=-121$

Make two groups for variable "x" and variable "y":

$(x^2+10x)+(y^2-20y)=-121$

Complete the square:

Add $(\frac{10}{2})^2=5^2$ inside the parentheses of "x".

Add $(\frac{20}{2})^2=10^2$ inside the parentheses of "y".

Add $5^2$ and $10^2$ to the right side of the equation.

Then:

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$(h,k)=(-5,10)$

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