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3 years ago

Can y’all answer questions 1,3 and 5? Or just 1 and 5 pls

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

I believe 5 is 3 inches per year because 57 minus 40 and 1/2 is 16 and 1/2, so you divide 16 and 1/2 by 5 and 1/2 which equals 3, so 3 inches per year

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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

Please please help no links!!

dedylja [7]

Answer:

Step-by-step explanation:

Brainliest please!

8 0

3 years ago

Will ran four miles on his first day of training.The next hint he ran one-third that distance.How far did he run the second day?

zhenek [66]

4 miles = 1 day

4miles divided by 1/3 distance on 2nd day = 1.33 miles ran on second day
4/1 x 1/3=4/3 4 divided by 3= 1.33 miles ran on the second day
If Will ran 1/3 of the distance that he ran on the first day (4 miles) the second day, he ran 1.33 miles

3 0

3 years ago

If 2(4x-7)=22, then 8x-14=22

Travka [436]

If this is a true or false statement then it is true

4 0

3 years ago

What is 1 4/5 + 2 3/20 +5/ 3

sveticcg [70]

$1 \frac{4}{5} +2 \frac{3}{20} + \frac{5}{3} \\ \\ 3 + \frac{4}{5} + \frac{3}{20} + \frac{5}{3} \\ \\ LCD = 60 \\ \\ 3 + \frac{4 \times 12}{5 \times 12} + \frac{3 \times 3}{20 \times 3} + \frac{5 \times 20}{3 \times 20} \\ \\ 3 + \frac{48}{60} + \frac{9}{60} + \frac{100}{60} \\ \\ 3 + \frac{48 + 9 + 100}{60} \\ \\ 3 + \frac{157}{60} \\ \\ 3 + 2 \frac{37}{60} \\ \\ 5 \frac{37}{60} \\ \\ Answer: \fbox {5 37/60} \ or \ \fbox {5.6167}$

5 0

3 years ago