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2 years ago

6

Can u help me solve this showing the work please

1 answer:

Leviafan [203]2 years ago

8 0

4 divides by 2,626 equals 656.5

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A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

3 years ago

In the equation x^2+10x+24=(x+a)(x+b), b is an integer. Find algebraically all possible values of b

Vinil7 [7]

(x+4)(x+6)
(x+6)(x+4)

8 0

3 years ago

24 is 16% of what number

Eva8 [605]

Is/of = %/100 plug the numbers in 24/x = 16/100 then cross multiply then you get 2400 = 16x so you get x by itself so you would have to divide by 16. 2400/16 = 16x/16 the 16 on the x side cancel out so you get x = 150 (I used proportions)

8 0

2 years ago

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Indicate whether the following congruence statement is true or false.<br> 50 points

Nostrana [21]

Step-by-step explanation:

Since line NL is not necessarily congruent to MQ (labelled with different congruent marks), triangle NLM is not congruent to triangle MQP.

Instead, triangles NLM and QPM are congruent.

8 0

3 years ago

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Solve:<br> (a+b+c) (a+b-c)

Katen [24]

Answer:

<h2><u><em>a²+2ab+b²-c²</em></u></h2>

Step-by-step explanation:

Solve:

(a+b+c) (a+b-c)=

(a²+ab-ac+ab+b²-bc+ac+bc-c²)=

a²+ab-ac+ab+b²-bc+ac+bc-c²=

a²+2ab+0ac+b²+0bc-c²=

a²+2ab+b²-c²

7 0

2 years ago

Read 2 more answers