The length of each side of the equilateral triangle is 10. what is the length of altitude h?

PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!

max2010maxim [7]

Answer:

Step-by-step explanation:

y intercept is -6

the slope is already in it's fraction form but if you want it back to whole number it's y=1.33x-6

the slope is positive

Some plots you can put are (6, 2) and (12, 10)

Just start at (0, -6) and counts 4 up and 3 right

4 0

2 years ago

Which of the following expressions represents “the quotient of nine times a number and seven ?

blsea [12.9K]

Answer:

9n÷7

Step-by-step explanation:

A quotient means a ratio or one quantity being divided by another, or in algebraic terms (a/b). In this case, the first quantity is "nine times a number" represented in terms of n as (n). The other quantity is just the number 7.

Dividing the 9 times n by 9 gives us the following:

9n÷7

Hope this helps :>

7 0

3 years ago

5a+5b=25 and -5a+5b=35

Alex_Xolod [135]

add the two equations together

5a+5b=25

-5a+5b=35

------------------

0 + 10b =60

divide by 10

b=6

5a + 5b = 25

5a +5(6) = 25

5a +30 =25

subtract 30 from each side

5a =-5

divide by 5

a = -1

Answer (-1,6)

or a=-1 b=6

3 0

3 years ago

Three times two less than a number is greater than or equal to five times the number. Find all of the numbers that satisfy the g

Sholpan [36]

It would be 3(n-2) is greater than or equal to 5n.

3 0

3 years ago

Read 2 more answers

*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