Patrick is constructing the circumscribed circle for △RST. Which construction could be his first step?

Mathematics

2 answers:

labwork [276]2 years ago

8 0

Answer:

1. Construct the perpendicular bisector of any side of the given triangle. Let it be RS.

Step-by-step explanation:

In order to construct the circumscribed circle for △RST, follow the following steps:

1. Construct the perpendicular bisector of any side of the given triangle. Let it be RS.

2. Construct the perpendicular bisector of another side of the given triangle.Let it be ST.

3. The center of the Circumscribed circle is the point at which they both cross each other.

4. We can draw Circumscribed circle by placing compass on the center point, adjusting its length to reach any corner of the triangle.

Gnom [1K]2 years ago

7 0

Construct the perpendicular bisector of ST

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ICE Princess25 [194]

Answer:

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Step-by-step explanation:

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2 years ago

Let f(x) = 7x^2-5x+3 and g(x) = 2x^2+4x-6

Troyanec [42]

Answer:

$\large\boxed{A.\ f(x)+g(x)=9x^2-x-3}\\\boxed{B.\ f(x)-g(x)=5x^2-9x+9}\\\boxed{g(x)-f(x)=-5x^2+9x-9}$

Step-by-step explanation:

$f(x)=7x^2-5x+3,\ g(x)=2x^2+4x-6\\\\A:\\f(x)+g(x)=(7x^2-5x+3)+(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3+2x^2+4x-6\qquad\text{combine like terms}\\f(x)+g(x)=(7x^2+2x^2)+(-5x+4x)+(3-6)\\f(x)+g(x)=9x^2-x-3$

$B:\\f(x)-g(x)=(7x^2-5x+3)-(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3-2x^2-4x+6\qquad\text{combine like terms}\\f(x)-g(x)=(7x^2-2x^2)+(-5x-4x)+(3+6)\\f(x)+g(x)=5x^2-9x+9$

$C:\\g(x)-f(x)=(2x^2+4x-6)-(7x^2-5x+3)\\g(x)-f(x)=2x^2+4x-6-7x^2+5x-3\qquad\text{combine like terms}\\g(x)-f(x)=(2x^2-7x^2)+(4x+5x)+(-6-3)\\g(x)-f(x)=-5x^2+9x-9$