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Sholpan [36]
3 years ago
9

Whats this question im confused 9/250

Mathematics
1 answer:
romanna [79]3 years ago
8 0
Its probably a division question. 250 divided by nine which is 27.7 which rounds to 28.
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*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts
Nataly [62]

Answer: Assuming the function is g(x)=\frac{2x+1}{x-3}:

The x-intercept is (\frac{-1}{2},0).

The y-intercept is (0,\frac{-1}{3}).

The horizontal asymptote is y=2.

The vertical asymptote is x=3.

Step-by-step explanation:

I'm going to assume the function is: g(x)=\frac{2x+1}{x-3} and not g(x)=2x+\frac{1}{x}-3.

So we are looking at g(x)=\frac{2x+1}{x-3}.

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

0=\frac{2x+1}{x-3}

A fraction is only 0 when it's numerator is 0.  You are really just solving:

0=2x+1

Subtract 1 on both sides:

-1=2x

Divide both sides by 2:

\frac{-1}{2}=x

The x-intercept is (\frac{-1}{2},0).

The y-intercept is when x is 0.

Replace x with 0.

g(0)=\frac{2(0)+1}{0-3}

y=\frac{2(0)+1}{0-3}  

y=\frac{0+1}{-3}

y=\frac{1}{-3}

y=-\frac{1}{3}.

The y-intercept is (0,\frac{-1}{3}).

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is y=\frac{2}{1}.  After simplifying you could just say the horizontal asymptote is y=2.

Or!

I could do some division to make it more clear.  The way I'm going to do this certain division is rewriting the top in terms of (x-3).

y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}

y=2+\frac{7}{x-3}

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0
3 years ago
Rationalise the denominator <br>​
Margaret [11]

Step-by-step explanation:

1. \frac{2 \sqrt{3} }{ \sqrt{5} }

multiply

\sqrt{5}

do both the numerator and denominater

\frac{2 \sqrt{3} }{ \sqrt{5} }   \times  \frac{ \sqrt{5} }{ \sqrt{5} }  =  \frac{2 \sqrt{15} }{ \sqrt{25} }  =  \frac{2 \sqrt{15} }{5}

8 0
3 years ago
What is the approximate circumference of the circle shown below?​
aniked [119]

Answer:

D: 109 cm

Step-by-step explanation:

The formula of circumference is

\pi d

Diameter =

17.4 \times 2 = 34.8

Hence, circumference =

34.8 \times \pi = 109.32742...

rounded of to 109.

8 0
3 years ago
Read 2 more answers
What is the value of x ​
GarryVolchara [31]
<h2><u>We can solve it using trigonometric values.</u></h2>

<h3>So , <u> </u><u>sin</u><u> </u><u>θ</u><u> </u> = ( Perpendicular / Hypotenuese )</h3>

<h2><u>Given</u><u> </u></h2>

Perpendicular = x

Hypotenuese = 30

{ \sf{sin \:  \: 45 {}^{ \degree}   = \dfrac{1}{ \sqrt{2}  }  =  \dfrac{x}{ 30} }}

{ \sf{x =  \dfrac{30}{ \sqrt{2} } }}

<h2>∴ x ( Perpendicular)= 15√2 <u>Ans</u><u>.</u></h2>
3 0
3 years ago
Read 2 more answers
A low-wattage radio station can be heard only within a certain distance from the station. On the graph above, the circular regio
makkiz [27]
I would be happy to help, is there anything else like the options or the graphs you could add?
6 0
3 years ago
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