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xenn [34]
1 year ago
15

Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal

to f(x) · g(t)?OA.41 - 81 + 61OB.SIKIAOC.21 + 6I + 2D.6

Mathematics
1 answer:
lisabon 2012 [21]1 year ago
4 0

Given the following functions below,

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}

Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

Multiplying both functions,

undefined

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3 years ago
Please help. Thank you​
love history [14]

Answer:

The length of side x is x = \sqrt{6}

Step-by-step explanation:

Pythagorean Theorem:

In a right triangle, with sides a and b, and hypothenuse h, we have that:

a^2 + b^2 = h^2

In this question:

The vertical line on the sides means that both are equal, that is, measuring x, so a = b = x.

Hypothenuse is the square root of 12. So h = \sqrt{12}

x^2 + x^2 = (\sqrt{12})^2

2x^2 = 12

x^2 = \frac{12}{2}

x^2 = 6

x = \sqrt{6}

The length of side x is x = \sqrt{6}

8 0
2 years ago
What is th he sum of the first seven terms of the series -3+6-12+24-...?
ohaa [14]

Answer:

-129

Step-by-step explanation:

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