What is the slope of the line with the equation y=6

From the set {7, 15, 18), which of the values can be substituted for c to make the

Aneli [31]

Answer:

D. 15

Step-by-step explanation:

90 - 30 = 60

2 • 15 = 30

the answer is 15

8 0

3 years ago

A firm manufactured 1000 Tv sets during its first yeae the total firns production during 10 years of operation is 29035 sets If

Alexus [3.1K]

Answer: the increase each year is 423 tv sets

Step-by-step explanation:

The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the arithmetic sequence.

d represents the common difference of the terms in the arithmetic sequence.

a represents the first term of the arithmetic sequence.

From the information given,

n = 10 years

a = 1000

S10 = 29035

We want to determine d which is the amount by which the production increased each year. Therefore, the sum of the first 10 years would be

29035 = 10/2[2 × 1000 + (10 - 1)d]

29035 = 5[2000 + 9d]

29035/5 = [2000 + 9d]

9d = 5807 - 2000 = 3807

d = 3807/9 = 423

3 0

3 years ago

Please help !!!!!!!!

Vsevolod [243]

Answer:

x=3 y=11

Step-by-step explanation:

2x+3=9

2x=6

x=3

y-4=7

y=11

8 0

3 years ago

What is the value of a? Blank units

Damm [24]

Answer:

289

a^2 x b^2 = c^2

a = 8 b= 15 c = s

64 + 225 = 289

square root 289 = 17

4 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