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iogann1982 [59]
2 years ago
11

Given C(x, 16), D(2,-4), E(-6, 14), and F(-2, 4), find the value of x so that CD || EF.

Mathematics
1 answer:
Andrew [12]2 years ago
8 0

Answer:

x=-6

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are the same

In this problem

slope CD=slope EF

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

step 1

Find the slope EF

we have

E(-6,14) and F(-2,4)

substitute the values in the formula

m_E_F=\frac{4-14}{-2+6}\\m_E_F=\frac{-10}{4}\\m_E_F=-2.5

step 2

Find the slope CD

we have

C(x, 16) and D(2, -4)

substitute the values in the formula

m_C_D=\frac{-4-16}{2-x}\\m_C_D=\frac{-20}{2-x}

Remember that

m_C_D=m_E_F

so

\frac{-20}{2-x}=-2.5

-20=-5+2.5x\\2.5x=-20+5\\2.5x=-15\\x=-6

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3 0
2 years ago
Read 2 more answers
F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 ​ x 2
34kurt

Answer:

f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

g(f(x)) = 3x^4 + 18x^2 + 18

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

f(x) = 3x^2 + 9

g(x) = \dfrac{1}{3}x^2 - 9

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

f(x) = 3x^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9

Expand Brackets

f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9

f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9

f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9

f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

Calculating g(f(x))

g(x) = \dfrac{1}{3}x^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9

g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9

Expand Brackets

g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9

g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9

g(f(x)) = 3x^4 + 18x^2 + 18

Checking for inverse functions

f(x) = 3x^2 + 9

Represent f(x) with y

y = 3x^2 + 9

Swap positions of x and y

x = 3y^2 + 9

Subtract 9 from both sides

x - 9 = 3y^2 + 9 - 9

x - 9 = 3y^2

3y^2 = x - 9

Divide through by 3

\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}

y^2 = \frac{x}{3} - 3

Take square root of both sides

\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}

y = \sqrt{\frac{x}{3} - 3}

Represent y with g(x)

g(x) = \sqrt{\frac{x}{3} - 3}

Note that the resulting value of g(x) is not the same as g(x) = \dfrac{1}{3}x^2 - 9

<em>Hence, f(x) and g(x) and not inverse functions</em>

4 0
2 years ago
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