9ax 6(-4ax 5 - 10x 4

In a recipe, the ratio of fluid ounces of water to fluid ounces of tomato paste is 3:4. You plan to make 35 fluid ounces of sauc

WITCHER [35]

3:4......added = 7

3/7(35) = 105/7 = 15 oz water
4/7(35) = 140/7 = 20 oz tomato paste <==

4 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 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

HELP MUST BE DONE BY 4:00

Leya [2.2K]

Answer:

the antelope is 5280 feet the cougar is 4224 feet the hare is 4136 feet the kangaroo is 3520 feet and the coyote is 3773 feet the ostrich is 3773 feet hope this helps the answer is the antelope

8 0

3 years ago

Plz help it’s due tomorrow

alexira [117]

Answer:

ano bayan? hindi ko makita at ma zoom. dahil sa pc haha nasa loptop kase ako eh. baiiiiiiiiii

Step-by-step explanation:

ti7r7i6ruy;98h6ty79t554e d ty0ty4ytbyyy

8 0

3 years ago

A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain

AysviL [449]

Answer:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

Step-by-step explanation:

For this case we have the following value:

$R = 0.0498 \frac{Kpa}{Km}$

We can convert this first to $\frac{Kpa}{m}$ like this:

$R=0.0498 \frac{Kpa}{Km} *\frac{1km}{1000m}=0.0000498 \frac{Kpa}{m}$

Now we use the fact the the pressure is defined as $P =\frac{F}{A}$ , whre P is the pressure, F the force and A the area, so then $Kpa= \frac{KN}{m^2}$ and then we can replace this:

$R=0.0000498 \frac{KN}{m^3}$

Now from definition of work we know that $W= Fd$ where W is the work, F the force and d the distance, so then is equivalent $KN =\frac{KJ}{m}$

And if we replace this into the equation we got:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

4 0

4 years ago

9ax 6(-4ax 5 - 10x 4 - 9)