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2 years ago

F(x) = 6x+2, find f (x+3)

Mathematics

2 answers:

nikklg [1K]2 years ago

3 0

Answer:

6x+20

Step-by-step explanation:

F(x) = 6x+2

f(x+3) means replace x with x+3

F(x+3) = 6(x+3)+2

= 6x+18+2

= 6x+20

Alika [10]2 years ago

3 0

Answer:

A; F(x+3) = 6x^2 + 20x + 6

Step-by-step explanation:

To find it you just would multiply F(x) onto the other side. So it would look like this:

F(x+3) = 6x + 2 (x+3)

F(x+3) = 6x^2 + 2x + 18x + 6 *combine like terms getting you

F(x+3) = 6x^2 + 20x + 6

Hope this helped!!

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If the side of a square is doubled in length, what is the percentage increase in area

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Doubling the side length multiplies the area by 4

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3 years ago

What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4

Alecsey [184]

Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: $f(x)=\dfrac{3x^2}{x^2-4}$

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. $x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2$

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at $y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}$

i.e. $y=\dfrac{3}{1}=3$

Hence, f(x) will have horizontal asymptote at y=3.

3 0

3 years ago

Find the function value f (6) if f(x) = 4x-10

Olenka [21]

So first you plug 6 into the equation which is 4(6)-10.
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8 0

3 years ago

Can anyone please help

shusha [124]

$\text{Hi there! :)}$

$\large\boxed{f'(2) = -\frac{16}{27} }$

$f'(x) = \frac{(5 + x^{2} )(-2x) - (7 - x^{2} )(2x)}{(5+x^{2})^{2} } \\\\f'(x) = \frac{-10x-2x^{3}- 14x + 2x^{3} }{(5+x^{2})^{2} } \\\\f'(x) = \frac{-24x}{(5+x^{2})^{2} }\\\\ \text{Solve for the derivative at f'(2) using substitution:}\\\\f'(2) = \frac{-24(2)}{(5+2^{2})^{2} } \\\\f'(2) = \frac{-48}{81} \\\\\text{Simplify:}\\\\f'(2) = -\frac{16}{27}$

8 0

3 years ago

Read 2 more answers

What are the coordinates of the image of point B, after the segment has been dilated by a scale factor of 3 with a center of dil

shepuryov [24]

Answer: (–9, 9)

Step-by-step explanation:

if the original point (x,y) gets dilated by a scale factor 'k' with a center of dilation at the origin, then

The coordinates of the image point are (kx, ky).

Given: The coordinates of line segment A B are A(-6,8) and B(-3,3).

then , the coordinates of B after dilation by scale factor of 3 with a center of dilation at the origin,

$(-3,3)\to(3(-3),3(3))\\\\\Rightarrow\ (-3,3)\to(-9,9)$

Hence, the coordinates of the image of point B, after the segment has been dilated by a scale factor of 3 with a center of dilation at the origin = (–9, 9).

8 0

3 years ago