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3 years ago

10

Can you please help me with this, I have no idea

1 answer:

sergejj [24]3 years ago

6 0

I have no clue sorry

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Given f(x)=1/x+5 and g(x)=x-2

vladimir1956 [14]

For this case we have the following functions:

$f (x) = \frac {1} {x + 5}\\g (x) = x-2$

We must find f (g (x)). So:

$f (g (x)) = \frac {1} {(x-2) +5} = \frac {1} {x-2 + 5} = \frac {1} {x + 3}$

Finally we have to:

$f (g (x)) = \frac {1} {x + 3}$

For the function to be defined the denominator must be different from 0. That is x different from -3.

Answer:

Option B

3 0

3 years ago

The measure of the supplement angle is fourteen times greater than its supplement

disa [49]

Let's assume you have supplement angle
$x$
Now when you talk about multiplication you can simply say...
$14x = y$
And that will give you the answer.

6 0

3 years ago

Question 1 of 5

PilotLPTM [1.2K]

Answer:

I think the answer is D.) 7/8 mile

Step-by-step explanation:

If you make the mixed numbers improper fractions, 2 and 1/8 is 17/8, 1 and 1/4 (2/8) is 10/8, then you subtract the numerators and keep the denomitators the same to get 7/8 mile (sorry if I spelt something wrong)

5 0

2 years ago

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

8 0

2 years ago

Plz answer quick... It's not that hard...also 30 points...

Aleksandr-060686 [28]

Your answer is 52.95 :)

5 0

2 years ago

Read 2 more answers