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2 years ago

10

Hello can you please help me with this question

2 answers:

Papessa [141]2 years ago

8 0

The answer for this is a

rusak2 [61]2 years ago

4 0

Answer:

A

Step-by-step explanation:

...

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Adam rides his bike from his home to the school's bike lot at point F each day. Assuming he takes the same route home, how far d

Wewaii [24]

I'm saying the answer is D.

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3 years ago

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(-72)/(-6)<br> Awnser <br> Plz&Thx

KIM [24]

(-72)/(-6) equals to 12

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3 years ago

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2/5(25y - 50) + 25y simplified

Vitek1552 [10]

Answer:

$35y - 20$

Step-by-step explanation:

here's the solution :-

=》

$\frac{2}{5} (25y - 50) + 25y$

=》

$10y - 20 + 25y$

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$35y - 20$

3 0

3 years ago

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Please helpppppp, i'm stuck

nata0808 [166]

Answer:

The operations you see with you eyes are multiplication and addition.

4 times X plus 2 = 10

The operations you don't see are Subtraction and Division.

4x+2=10

Subtract 2 from both sides.

4x+2−2=10−2

4x=8

Divide both sides by 4.

4x/4 = 8/4

x=2

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3 years ago

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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