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damaskus [11]
2 years ago
10

The sum of three numbers is 11 . The sum of twice the first​ number, 5 times the second​ number, and 6 times the third number is

44. The difference between 4 times the first number and the second number is 10. Find the three numbers.
Mathematics
1 answer:
Assoli18 [71]2 years ago
4 0

Answer:

4, 6, 1

Step-by-step explanation:

We can solve this problem using a system of equations:

1) a + b + c = 11

2) 2a + 5b + 6c = 44

3) 4a - b = 10

First, we can substitute the value of b from equation #3 into equation #1:

b = 4a - 10

a + (4a - 10) + c = 11

5a - 10 + c = 11

5a + c = 21

c = 21 - 5a

Now, we can plug the values of b and c into equation #2, as b and c are represented in terms of a:

2a + 5(4a - 10) + 6(21 - 5a) = 44

2a + 20a - 50 + 126 - 30a = 44

-8a + 76 = 44

-8a = -32

a = 4

b = 4a - 10 = 4(4) - 10 = 6

c = 21 - 5a = 21 - 5(4) = 1

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*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.)

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