Explain how System 1 becomes equivalent to System 2.

Mathematics

2 answers:

Nostrana [21]3 years ago

5 0

Answer:

$\left\{\begin{array}{ccc}(A+L)x+(B+M)y=C+N\\Ax+By=C\end{array}\right$

<h3>A.The first equation in System 2 is the sum of the equations in System 1. The second equation in System 2 is the first equation in System 1.</h3>

Step-by-step explanation:

$\underline{+\left\{\begin{array}{ccc}Ax+By=C\\Lx+My=N\end{array}\right}\qquad\text{add both sides of the equations}\\(Ax+Lx)+(By+My)=C+N\qquad\text{distributive}\\(A+L)x+(B+M)y=C+N$

DedPeter [7]3 years ago

3 0

The correct answer is A) The first equation in System 2 is the sum of the equations in System 1. The second equation in System 2 is the first equation in System 1.

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Find (f*g)(x) when f(x)=x^2-7x+12 and g(x)=3/x^2-16

IgorC [24]

The answer is: $(f*g)(x)= \frac{3x-9}{x+4}$

<u><em>Explanation</em></u>

Given functions are.......

$f(x)=x^2-7x+12$

$g(x)= \frac{3}{x^2-16}$

For finding $(f*g)(x)$ , we just <u>need to multiply the two functions</u> $f(x)$ and $g(x)$ . So......

$(f*g)(x)= f(x)*g(x)\\ \\ (f*g)(x)= (x^2-7x+12)(\frac{3}{x^2-16})\\ \\ (f*g)(x)= [(x-3)(x-4)][\frac{3}{(x+4)(x-4)}]\\ \\ (f*g)(x)= \frac{3(x-3)}{x+4} = \frac{3x-9}{x+4}$