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Akimi4 [234]
2 years ago
15

Which choices are solutions to the following equatior Check all that apply. X^2 -5x=-9/4

Mathematics
1 answer:
svlad2 [7]2 years ago
7 0

The solution of the equation 4x² - 20x + 9 = 0 will be 0.5 and 4.5.

<h3>What is a quadratic equation?</h3>

The quadratic equation is given as ax² + bx + c = 0. Then the degree of the equation will be 2. Then we have

The equation is given below.

x² - 5x = -9/4

Then the equation can be written as

4x² - 20x + 9 = 0

Then the factor of the equation will be

4x² - 20x + 9 = 0

4x² - 18x - 2x + 9 = 0

2x(2x - 9) -1 (2x - 9) = 0

(2x - 1)(2x - 9) = 0

Then the solution of the equation 4x² - 20x + 9 = 0 will be

x = 0.5, 4.5

More about the quadratic equation link is given below.

brainly.com/question/2263981

#SPJ1

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What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of
notsponge [240]

Answer:

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

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What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, (x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)

<u>So, The coordinates of the triangle A'B'C' are:</u>

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

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