Which statement about the lines below is true?

In the diagram, which is not drawn to scale, DE | FG || BC.

AysviL [449]

Answer:

FG || BC.

Step-by-step explanation:

In the diagram, which is not drawn to scale, DE | FG || BC.

3 0

2 years ago

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

Find the midpoint of the segment with endpoints of −11 + i and −4 + 4i.

Y_Kistochka [10]

The midpoint is the average of the endpoints.
((-11+i) + (-4+4i))/2 = -15/2 +5/2i

4 0

2 years ago

Read 2 more answers

(01.02 LC)Simplify 9 − (−4).

levacccp [35]

Answer:

13

Step-by-step explanation:

When we subtract a negative, it is like adding

9 - -4

9 +4

13

3 0

2 years ago

Read 2 more answers

−12x−5y=40 12x−11y=88 x= y=

Nezavi [6.7K]

Add the two equations and use the elimination method. The x term will "cancel out" to be able to solve for the y term.

STEP 1
−12x−5y=40
12x−11y=88
x terms "cancels out"; add y terms

-16y= 128
divide both sides by -16

y= -8

STEP 2
substitute y value into either original equation

12x−11y=88

12x - 11(-8)= 88

12x + 88= 88
subtract 88 from both sides

12x= 0
divide both sides by 12; 0 divided by any number is 0

x= 0

ANSWER: x= 0; y= -8

Hope this helps! :)

6 0

2 years ago