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3 years ago

What does the 2 represent in 12.789

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

8 0

It is a whole number in the one digits place
:)

11111nata11111 [884]3 years ago

7 0

(Tens digit)(Ones digit) . (tenths digit)(hundredths digit)(thousandths digit)
12.789
2 = (ones digit)

Hope this helps!

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2(x + 4) − 1 = 2x + 7

Margarita [4]

Answer:

All real numbers are solutions to this problem.

Step-by-step explanation:

I) Simplify:

2(x + 4) - 1 = 2x + 7

x + 4 - 0.5 = x + 3.5 (Divide by 2)

x + 3.5 = x + 3.5 (Simplify x + 4 - 0.5)

x = x

II) Conclusion:

Any real number will work as a solution.

7 0

3 years ago

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Given the system of equations, what is the value of the system determinant? x + y = 8 x - y = 10 0 -1 -2

kkurt [141]

Each X and Y in the equations don't have a number in front of them so they are all considered 1.

D = 1*(-1) - 1*(-1) = -1 -1 = -2

The answer is -2

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3 years ago

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Diego had $21.32 and gave $9.50 to Elena. How much did he have left?

Charra [1.4K]

Answer:

11.82

Step-by-step explanation: I know this because 21.32-9.50= 11.82

7 0

3 years ago

The answer is is 26, -44, but i dont understand how to solve it. can you explain?

Anna [14]

║a+9║/7=5
you multiply 7 on both sides which cancel out 7.
now it ║a=9║=35
and now it would be a+9=35 and a+9=-35 then solve it.
35-9 and 9-(-35)= 26 and -44
hope this helped!

4 0

3 years ago

*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