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Serggg [28]
2 years ago
7

In the diagram below, OPis circumscribed about quadrilateral ABCD. What is the value of x?

Mathematics
1 answer:
trasher [3.6K]2 years ago
4 0

Answer:

x+10+88=180

x. =180-98

x. =82

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Answer:

\large\boxed{y=\dfrac{1}{4}x^2-x-4}

Step-by-step explanation:

The equation of a parabola in vertex form:

y=a(x-h)^2+k

<em>(h, k)</em><em> - vertex</em>

The focus is

\left(h,\ k+\dfrac{1}{4a}\right)

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using k+\dfrac{1}{4a}

<em>k = -5</em>

-5+\dfrac{1}{4a}=-4        <em>add 5 to both sides</em>

\dfrac{1}{4a}=1           <em>multiply both sides by 4</em>

4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4

\dfrac{1}{a}=4\to a=\dfrac{1}{4}

Substitute

a=\dfrac{1}{4},\ h=2,\ k=-5

to the vertex form of an equation of a parabola:

y=\dfrac{1}{4}(x-2)^2-5

The standard form:

y=ax^2+bx+c

Convert using

(a-b)^2=a^2-2ab+b^2

y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5

<em>use the distributive property: a(b+c)=ab+ac</em>

y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4

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