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kaheart [24]
2 years ago
7

Can u please help me with this

Mathematics
1 answer:
zhannawk [14.2K]2 years ago
5 0

Answer:

1/4

1/6

7/10

3 7/8

2 5/7

3 1/9

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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
4 years ago
Please help me with this <br><br><br> also good morning!
Doss [256]

Answer:

Divide both sides by 6 :)

8 0
3 years ago
Read 2 more answers
Help please with this question.
alexandr1967 [171]

Answer:

the classmate is incorrect

Step-by-step explanation:

when you do the math for both inequalities they are equal to each other, despite the fact that one has multiplication and the other has addition.

7 0
3 years ago
A flowchart can be a(n)<br> O Hamilton circuit<br> Euler les<br> tree graph
Fiesta28 [93]

Answer:

I need points sry plz dont dellete

Step-by-step explanation:

6 0
3 years ago
A paved walkway around a garden has an outer diameter of 80 ft and an inner diameter of 40 ft. A diagram is shown below.
k0ka [10]

Given:

  • Outer diameter = D = 80 ft
  • Inner diameter = d = 40 ft

So,

Outer radius = R = D/2

=> R = (80 ft)/2

=> R = 40 ft

Inner radius = r = d/2

=> r = (40 ft)/2

=> r = 20 ft

Area of paved pathway = πR² - πr²

= π(R² - r²)

= π(R + r)(R - r)

= π(40 + 20)(40 - 20) ft²

= π(60)(20) ft²

= 3.14 * 1200 ft²

= 3768 ft²

6 0
3 years ago
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