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2 years ago

Somebody HELP ME PLS IM TRYING TO GET MY GRADES UP FROM AN F! IF ITS CORRECT AND HELPFUL U WILL BE BRAINLIEST

Mathematics

1 answer:

kramer2 years ago

5 0

Answer:

144

Step-by-step explanation:

A=BH

your base is 9 and your height is 16. 16 times 9= 144. Hope this helps :)

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Write an equivalent equation that does not contain fractions 2/3 x + 9 equals 3/4

AlekseyPX

Answer:

8x + 108 = 9

Step-by-step explanation:

Given

$\frac{2}{3}$ x + 9 = $\frac{3}{4}$

Multiply through by the lowest common multiple of 3 and 4, that is 12 to clear the fractions.

8x + 108 = 9 ← equivalent equation with no fractions

6 0

3 years ago

Which one is bigger 95.580 or 95.58

mezya [45]

They're both equal. The zero in 95.580 doesn't matter in this comparison.

5 0

3 years ago

Read 2 more answers

i have been on this test like- forever trying to figure this out im horrible at math someone please help me-

ss7ja [257]

Answer:

Step-by-step explanation:

total cost = c

students = s

cost = 7

c = s*7

independent is s

dependent is c

7 0

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.)

6 0

3 years ago

Bob will rent a car for the weekend. He can choose one of two plans. The first plan has an Initial fee of $60 and costs an addit

shepuryov [24]

Answer:

Step-by-step explanation:

+55+%2B+.5x+=+.7x+

+.2x+=+55+

+x+=+275+

3 0

3 years ago