4T = 25 what does T equal<br> ?????????

Does a trend line always pass through every point on a scatter plot?

forsale [732]

No, it does not and it's not supposed to. A trend line is drawn to go through in the best way possible, to go through the "center" of the points as best as it can. It may hit some of the points, but not all of them. Generally, the idea is to try to get the trend line to have as many points "above" it as it does "below" it to say you've gone through them evenly.

6 0

3 years ago

Read 2 more answers

What is 3y^2-y-2 in factored form?

GaryK [48]

BY USING MIDDLE TERM SPLITTING,

3y^2-(3y-2y)-2

(3y^2-3y)+(2y-2)

3y(y-1)+2(y-1)

(3y+2)(y-1)

HOPE THIS WILL HELP U

7 0

3 years ago

Can anyone decompose it into multipliers?

Alex787 [66]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Zora has bag of potting

Xelga [282]

The question is missing the figure which is attached below.

Answer:

The last box that has dimensions 8 in × 8 in × 10 in

Step-by-step explanation:

Given:

Volume of the soil = 924 cubic inches.

There are four different types of boxes that need to be checked whether they can accommodate all of the soil or not.

The volume of the box must be at least 924 cubic inches to accommodate all of the soil.

Now, volume of the first box is given as:

$V_1=l\times b\times h=20\times 5\times 10=1000\ in^3>924\ in^3$

Volume of the second box is given as:

$V_2=l\times b\times h=10\times 10\times 10=1000\ in^3>924\ in^3$

Volume of the third box is given as:

$V_3=l\times b\times h=10\times 20\times 6=1200\ in^3>924\ in^3$

Volume of the fourth box is given as:

$V_4=l\times b\times h=8\times 8\times 10=640\ in^3$

Therefore, only the volume of the fourth box is less than the total volume of the soil. So, last box is the correct option.

6 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 $y=2$ .