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3 years ago

5

sue wrote a doubles fact. it has a sum less than 10 and greater than 4. the addends are each less than 5 what facts might she ha

ve written?

1 answer:

Sergeu [11.5K]3 years ago

7 0

6, 7, 8, and 9 welcome :)

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Which statement is true

timofeeve [1]

Answer:

The third one.

Step-by-step explanation:

Hope this helps.

4 0

3 years ago

F(x)=2x^2-5;find when f(x)=-7

Daniel [21]

Answer:

x=i or no solution exists depending on the user's grade

Step-by-step explanation:

f(x) = 2x^2-5

-7 = 2x^2-5

-2 = 2x^2, x^2=-1, x=i

6 0

2 years ago

Read 2 more answers

*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

What is the value of x in the following equation? 5x + 12 = 6(x + 1) <br> A. 6 <br> B. 5 C. 24 D. 7

Maurinko [17]

Answer:

A: 6

Step-by-step explanation:

Our equation is 5x+12=6(x+1)

Our possible numbers are 6, 5, 24, or 7.

To figure this out you start putting the possible numbers in place of x so we'll start off using 6:

5*6+12=6(6+1).

5*6=30. 30+12=42. 6+1=7, 6*7=42.

This equals the same on both sides.

Next we'll go to 5:

5*5+12=6(5+1)

5*5=25, 25+12=37. 5+1=6, 6*6=36.

This doesn't equal the same amount on both sides.

Next is 24:

5*24+12=6(24+1)

5*24=120, 120+12=132. 24+1=25, 6*25=150

This doesn't equal the same amount.

Next is 7:

5*7+12=6(7+1)

5*7=35, 35+12=47. 7+1=8, 6*8=48.

This doesn't equal.

The only number that equals the same amount on both sides is A:6

4 0

3 years ago

Which among the four choices gives the difference of (x/4x²) and (3/4x²)?

Sergeeva-Olga [200]

Answer:

2

Step-by-step explanation:

the difference is that

6 0

3 years ago