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3 years ago

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Adding and Subtracting Functions (20 POINTS and MARK BRAINLIEST)

1 answer:

Marina86 [1]3 years ago

3 0

Hi I wanna is a good day to you 568

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Please help!!! Here is 25 point please helpp

Ket [755]

Answer:

-3

Step-by-step explanation:

6 0

3 years ago

Three lines of crops in a garden form a triangular shape. Strawberries are planted in a 10 ft line and green beans are planted i

Marat540 [252]

Answer:

$p = 19.18$

Step-by-step explanation:

This question is illustrated using the attachment and will be solved using cosine formula

$a^2 = b^2 + c^2 - 2bcCosA$

<em>Let the strawberry side be s, the Green beans be b and the pumpkins be p.</em>

<em />

The cosine formula in this case is:

$g^2 = s^2 + p^2 - 2spCosG$

Where

$s = 10$

$g = 18$

The equation becomes

$18^2 = 10^2 + p^2 - 2 * 10 * p * Cos\ 68$

$324 = 100 + p^2 - 20 * p * 0.375$

$324 = 100 + p^2 - 7.5p$

Collect Like Terms

$p^2 - 7.5p + 100 - 324 = 0$

$p^2 - 7.5p - 224 = 0$

Using quadratic formula:

$p = \frac{-b\±\sqrt{b^2-4ac}}{2a}$

Where

$a = 1$

$b = -7.5$

$c = -224$

$p = \frac{-(-7.5)\±\sqrt{(-7.5)^2-4*1*-224}}{2*1}$

$p = \frac{7.5\±\sqrt{56.25+896}}{2}$

$p = \frac{7.5\±\sqrt{952.25}}{2}$

$p = \frac{7.5\± 30.86}{2}$

$p = \frac{7.5 + 30.86}{2}$ or $p = \frac{7.5 - 30.86}{2}$

$p = \frac{38.36}{2}$ or $p = \frac{-23.36}{2}$

$p = 19.18$ or $p = -11.68$

But length can not be negative.

So:

$p = 19.18$

6 0

3 years ago

Approximate the value of 13√ to the nearest tenths place. Plot the approximation on the number line.

aalyn [17]

The square root of 13 rounded to the nearest tenth is 3.6, so you would plot this somewhere in between 3 and 4 on a number line.

4 0

3 years ago

Read 2 more answers

*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

Which equation can be used to represent the statement?

serg [7]

Answer:

of the number is X, 16 = 1 + (2/10)×X - 8

7 0

3 years ago

Read 2 more answers