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3 years ago

What is the value of x? Enter your answer in the box.

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

15/5= 3

8*3= 24

Since the congruent angles are on opposite sides, the side with 8 in. would be congruent to x.

x= 24.

I hope this helps!
~kaikers

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3 years ago

(08.07 HC)

andreev551 [17]

Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

$\implies ac=2 \cdot -10=-20$

$\implies b=-1$

Therefore, the two numbers are -5 and 4.

Rewrite $b$ as the sum of these two numbers:

$\implies f(x)=2x^2-5x+4x-10$

Factor the first two terms and the last two terms separately:

$\implies f(x)=x(2x-5)+2(2x-5)$

Factor out the common term (2x - 5):

$\implies f(x)=(x+2)(2x-5)$

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

Therefore:

$\implies (x+2)=0 \implies x=-2$

$\implies (2x-5)=0 \implies x=\dfrac{5}{2}$

So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

The x-value of the vertex is:

$\implies x=\dfrac{-b}{2a}$

Therefore, the x-value of the vertex of the given function is:

$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

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Plot the vertex found in Part B.

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The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).

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Answer:

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I am joyous to assist you anytime.

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