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3 years ago

Jackie cut a 2 yard spool into 5 equal lengths of ribbon?

Mathematics

1 answer:

lidiya [134]3 years ago

6 0

5 units = 2 yd.
1 unit = 2/5
Each piece of ribbon is 2/5 yd. long.

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⎧

MrRissso [65]

Answer:

f(n)=-5-3n

Step-by-step explanation:

Given the recursive formula of a sequence

f(1)=−8

f(n)=f(n−1)−3

We are to determine an explicit formula for the sequence.

f(2)=f(2-1)-3

=f(1)-3

=-8-3

f(2)=-11

f(3)=f(3-1)-3

=f(2)-3

=-11-3

f(3)=-14

We write the first few terms of the sequence.

-8, -11, -14, ...

This is an arithmetic sequence where the:

First term, a= -8

Common difference, d=-11-(-8)=-11+8

d=-3

The nth term of an arithmetic sequence is determined using the formula:

T(n)=a+(n-1)d

Substituting the derived values, we have:

T(n)=-8-3(n-1)

=-8-3n+3

T(n)=-5-3n

Therefore, the explicit formula for f(n) can be written as:

f(n)=-5-3n

7 0

3 years ago

Ratio between 80% and 20%

musickatia [10]

4:1 is the answer............

3 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

Verify that the given value is the solustion of the equation x=6 for 2z-4=8

klemol [59]

Answer:

8-4=2z.. 2z+4=6 so x=6

6 0

3 years ago

Put the steps of solving the equation in order from first to last.<br> 2x + 1 = -1

Alborosie

Answer:

x = 1

Step-by-step explanation:

2x + 1 = -1

2x = -1 - 1

2x = -2

x = -2 ÷ -2

x = 1

3 0

3 years ago