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Goryan [66]
3 years ago
13

Circle Q is centered at the origin with radius r. Point P(x, y) lies on circle Q. Make a conjecture. How can you find an equatio

n relating the radius to the coordinates of point P? Check all that apply. Notice that ΔPQS forms a right triangle.
Because ΔPQS is a right triangle,

apply the Pythagorean theorem.

x² + y² = r²

Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0

Answer:

x2=y2=r2

Step-by-step explanation:

edge

salantis [7]3 years ago
5 0

Answer:

The relation x^{2} +y^{2} =r^{2} is explained below.

Step-by-step explanation:

P(x, y) is a point on the circle. Q is the origin. Join PQ.

PQ is the radius.

Therefore, PQ = r

Draw PS perpendicular to the x-axis.

Now, ΔPQS is a right triangle.

By Pythagoros theorem,

PQ^{2}=PS^{2} +QS^{2}

r^{2}=x^{2} +y^{2}

x^{2} +y^{2} =r^{2}

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49x^2=-21x-2 quadratic functions
lara [203]

Answer: 49x^2=-21x-2 quadratic functions -1/7and -2/7    



Step-by-step explanation:

Quadratic function:

In elementary algebra, the quadratic formula is a formula that provides the solution to a quadratic equation. There are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring, completing the square, graphing and others.

Move terms to the left side

49x^{2}  =-21x-2

49x^{2}  -(-21x-2) =0

 Distribute

49x^{2}  -(-21x-2) =0

49x^{2}+21x+2=0

Use the quadratic formula



 x=(-b±√b^{2}  -4ac  ) / 2a

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

 49x^{2}+21x+2=0

let, a=49

b=21

c=2

 Replace with values in this equation

X=(-b±√b^{2}  -4ac  ) / 2a

Simplify

Evaluate the exponent

Multiply the numbers

Subtract the numbers

Evaluate the square root

Multiply the numbers

x=(-21±7) /98

Separate the equations

To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus.Separate

x=(-21+7) /98

x=(-21-7) /98

Solve

Rearrange and isolate the variable to find each solution

x=-1/7

x=-2/7



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__

2. 6^2 +20.1^2 = 440.01 ≠ 22.9^2 = 524.41 . . . . . this is an obtuse triangle

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