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4 years ago

8

Evaluate composite functions

1 answer:

Ber [7]4 years ago

5 0

As shown in the pic, you place the given value into g first, which results in 3 (as shown in the pic). Next, you place the result you got from placing the given value into g into f now, so that would be (3)^2-6(3)+3. Evaluating this, so far we'd get 9-18+3. When adding, we'd get 12-18, which is -6, so f(g(11)) is -6

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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6

kenny6666 [7]

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

Step-by-step explanation:

Given the simultaneous equations

$y=9-x$

$y\:=\:2x^2\:+\:4x\:+\:6$

Subtract the equations

$y=9-x$

$-$

$\underline{y=2x^2+4x+6}$

$y-y=9-x-\left(2x^2+4x+6\right)$

$\mathrm{Refine}$

$x\left(2x+5\right)=3$

$\mathrm{Solve\:}\:x\left(2x+5\right)=3$

$2x^2+5x=3$ ∵ $\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$2x^2+5x-3=3-3$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$=\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{4}$

$=\frac{-5+7}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Similarly,

$x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{2},\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

3 0

4 years ago

The elevator paused for 10 seconds after you stepped on before beginning to rise at a constant rate of 8 feet per seconds. Propo

notsponge [240]

Nonproportional;The graph of this situation will extend from (0, 0) to (10, 0) on the x-axis before rising.

3 0

4 years ago

Read 2 more answers

at a book fair, Joe bought 24 books at 3 for $5 and had $2 left. How much money did he have at first?

wel

Answer: $42 maybe?

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

A line goes through the points (-6, -8) and (12, 7). a) What is the slope of the line? Show your work. b) Write the equation of

kompoz [17]

m = $\frac{5}{6}$ , y - 7 = $\frac{5}{6}$ ( x - 12), y = $\frac{5}{6}$ x - 3

(a) calculate the slope using the gradient formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 6, - 8) and (x₂, y₂) = (12, 7)

m = $\frac{7+8}{12+6}$ = $\frac{15}{18}$ = $\frac{5}{6}$

(b) the equation of a line in point-slope form is

y - b = m(x - a )

where m is the slope and (a, b) a point on the line

using m = $\frac{5}{6}$ and (a , b) = (12, 7 )

y - 7 = $\frac{5}{6}$ (x - 12)

(c) the equation of a line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange (b ) into this form

y - 7 = $\frac{5}{6}$ - 10

y = $\frac{5}{6}$ x - 3

3 0

4 years ago

Which equations are correct? Select each correct answer.

Juliette [100K]

Answer:

Equations 2 and 3 are correct

3 0

3 years ago

Read 2 more answers