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3 years ago

ALGEBRA 1, SEM. 2; please show work!

Mathematics

1 answer:

hoa [83]3 years ago

8 0

The enlarged flower bed is a rectangle and its area is x^2+9x+14 m²

So before its enlargement by 14 m², the area of the rectangle = x^2+9x

As you know, the area of rectangle is equal length times width

so area = x^2+9x

factors:

x^2+9x = x (x + 9)

So the dimensions are x and (x + 9)

Hope it helps.

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$\large\begin{array}{l} \left\{\!\begin{array}{l} \mathsf{f(x)=x^2-6x+2}\\ \mathsf{g(x)=\sqrt{x}}\\ \end{array}\right. \end{array}$

$\large\begin{array}{l} \textsf{a) }\mathsf{(f\circ g)(x)}\\\\ =\mathsf{f\big[g(x)\big]}\\\\ =\mathsf{\big[g(x)\big]^2-6\cdot g(x)+2}\\\\ =\mathsf{\big[\sqrt{x}\big]^2-6\sqrt{x}+2}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(f\circ g)(x)=x-6\sqrt{x}+2} \end{array}}\qquad\checkmark \end{array}$

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$\large\begin{array}{l} \textsf{b) }\mathsf{(g\circ f)(-2)}\\\\ =\mathsf{g\big[f(-2)\big]}\\\\ =\mathsf{\sqrt{f(-2)}}\\\\ =\mathsf{\sqrt{(-2)^2-6\cdot (-2)+2}}\\\\ =\mathsf{\sqrt{4+12+2}}\\\\ =\mathsf{\sqrt{18}}\\\\ =\mathsf{\sqrt{3^2\cdot 2}}\\\\ =\mathsf{3\sqrt{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(g\circ f)(-2)=3\sqrt{2}} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope this helps. :-)}$

Tags: composite function composition evaluate algebra

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3 years ago