If f(x) = 3x + 2 and g(x) = x^2 + 1, which expression is equivalent to (g(f(x))?

1 answer:

7 0

<span>Whenever we are given one function and must calculate a funciton of the funciton, such as g(f(x)) in this case, we simply substitute the second function, f(x) in this case, in the first function, g(x) in this case, wherever the first function has a variable. Therefore, g(f(x)) = (3x + 2)^2 + 1 g(f(x)) = 9x^2 + 12x + 4 + 1 g(f(x)) = 9x^2 + 12x + 5</span>

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Graph the soloution x+2y=8

ELEN [110]

Answer:

Im 100% sure that it’s (A)

5 0

3 years ago

Please help with this question

ioda

Answer:

$25$

Step-by-step explanation:

$\frac{5^8 \times 5^{-2}}{5^4}$

$\frac{5^{8+-2}}{5^4}$

$\frac{5^{6}}{5^4}$

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$=25$

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%20x%5E%7B2%7D-10x%2B25%7D%2B25%2B12%5Csqrt%7Bx%7D%20%3D15%5Csqrt%7Bx%7D" id="TexFor

vodomira [7]

Step-by-step explanation:

$\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x} \\ \\ \therefore \: \sqrt{ x^{2}-10x+ {5}^{2} }+25 =15\sqrt{x} - 12\sqrt{x}\\ \\ \therefore \: \sqrt{( x - 5)^{2}}+25 =3\sqrt{x} \\ \\ \therefore \: x - 5+ 25 = 3 \sqrt{x} \\ \\ \therefore \: x + 20 = 3 \sqrt{x} \\ \\ squaring \: both \: sides \\ (x + 20)^{2} = ( {3 \sqrt{x} })^{2} \\ \therefore \: {x}^{2} + 40x + 400 = 9x \\ \therefore \: {x}^{2} + 40x + 400 - 9x = 0 \\ \therefore \: {x}^{2} + 31x + 400 = 0 \\$

8 0

3 years ago

Write the standard form of the line that passes through (-1,-3) and (2,1).

kobusy [5.1K]

Answer:

Step-by-step explanation:

$\mathrm{Line\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Find\:the\:line\:}\\\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Compute\:the\:slope\:}\left(-1,\:-3\right),\:\left(2,\:1\right):\quad \\m=\frac{4}{3}\\\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(-1,\:-3\right),\:\left(x_2,\:y_2\right)\\\left\\$

$\\m=\frac{1-\left(-3\right)}{2-\left(-1\right)}\\\\Refine\\\\m=\frac{4}{3}\\\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-\frac{5}{3}\\\\y=\frac{4}{3}x-\frac{5}{3}\\$

Convert equation to standard form

$y - \frac{4}{3} x + \frac{5}{3} = 0\\\\Multiply \:through\: by\: 3;\\\\3\times (y - \frac{4}{3} x + \frac{5}{3} )=0\\\\3y -4x+5 = 0\\Arrange\:n\:form\:of \:;\\Ax +By = C\\\\-4x +3y =-5$