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2 years ago

PLS HELP THIS IS HARD ANYONE PLS

Mathematics

1 answer:

alex41 [277]2 years ago

4 0

Answer:

C' (4, - 6 )

Step-by-step explanation:

4 units left means subtracting 4 from the x- coordinate

3 units down means subtracting 3 from the y- coordinate

C (8, - 3 ) → C' (8 - 4, - 3 - 3 ) → C' (4, - 6 )

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slamgirl [31]

Answer:

u just have to cross multiply im pretty sure thats how my math teacher show me,

Step-by-step explanation:

first you take the one x over to the other x and that will make it 2x then do the same thing with -10 u do -10 times 10 and then u get -100 and then the answer is 2x-100 and depending on your teach if he/she wants you to stop their then u can but if u have to got further then u would divide 2 on both sides and then that would be 0.02

this is how my teacher showed us how to do it and i hope this helps you a lot and if it dont im sorry i tried for you!!!

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3 years ago

Find the slope and Y intercept for each equation y=2x -4

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The y intercept is -4 and the slope is 2

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3 years ago

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LCM of 4/6 and 4/8????

Svet_ta [14]

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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3 years ago

*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.)

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3 years ago