This polynomial is in STANDARD FORM.<br>4-18x<br>True<br>False

How do you solve this absolute value equation? ∣x−2∣=∣4+x∣

snow_tiger [21]

L x - 2 l = l 4 + x l

Remember : Absolute is just the distance from zero.

Ex : l -4 l turns into 4, as it is 4 spaces away from zero on a number line :)

l x - 2 l turns into x + 2

l 4 + x l stays the same.

So, our equation now looks like this :

x + 2 = 4 + x

So, we're basically just trying to get x by itself.

So, subtract 2 from both sides.

x = 4 + x - 2

Then, subtract x from both sides.

x - x = 4 -2

0 = 2

Therefore, there are no solutions to this equation :)

~Hope I helped!~

6 0

3 years ago

Will someone please help me with these questions. The table shows the average annual cost of tuition at 4-year institutions from

erastovalidia [21]

Step-by-step explanation:

1.) The equation for the line of least squares regression is given by

$Y=mX+c$

where Y = Output Variable = Average Cost (in thousands of dollars)

X = Input Variable = Number of years after 2000

Thus, the given data table transforms as follows:

No. of years after 3 4 5 6 7 8 9 10

after 2000 (X)

Average Cost (Y) 15.505 16.510 17.451 18.471 19.363 20.409 21.093 22.092

To determine the equation precisely, we need to calculate the values of 'm' (Slope) and 'c' (Y-intercept)

Let us take any two random values of X and Y from the above table, so that we have

X₁ = 3; Y₁ = 15.505 and X₂ = 8; Y₂ = 20.409

Plugging in these two sets of values, we get the following two equations:

15.505 = (3)m + c and

20.409 = (8)m + c

Let us subtract the upper one from the lower one so that we have

20.409 - 15.505 = 8m + c - 3m - c

4.904 = 5m

Hence, m = 0.9808

Using this value and plugging it in one of the above equations yields the value of c as

20.409 = 8(0.9808) + c

Therefore, c = 12.563

So, the required equation for the line of least squares regression is

Y = (0.9808)X + 12.563

2.) To estimate the cost of tuition at a 4-year institution in 2000, we can put the value of X to be 0.

We have Y = 0 + 12.563

Thus, the cost of tuition in 2000 was $12,563.

3.) To estimate the cost of tuition in 2020, we put the value of X as 20 so that we have

Y = (0.9808)(20) + 12.563 = 32.179

Therefore, we can expect the cost of tuition to be as high as $32,179 by 2020.

4.) The slope tells us the factor by which the tuition increases each year, starting with the base of 12,563 dollars in 2000.

One way to lower the cost of college tuition is by applying for a waiver based on income level. Alternatively, one can also work at the school.

One can join a part-time graduation program while working for the rest of the day so that one may support oneself as well as pay for their courses. Alternatively, one can opt for Online Courses and get some Diploma in their area of interest while working part-times so that they may pursue a degree later on in their life.

3 0

4 years ago

I will give brainliest to the best answer to this question: Each year, the Department of Water Works measures the chloride conce

Maslowich

Answer:

The total change that will likely occur in 2 years is $=-3\frac{2}{3}$ mg/L

Step-by-step explanation:

Given : The first year, they found the chloride concentration changed by $-1\frac{3}{5}$ mg/L .

It is estimated that the chloride concentration will change $-2\frac{1}{15}$ mg/L the next year.

To find : What is the total change that will likely occur in 2 years?

Solution :

The chloride concentration changed by $-1\frac{3}{5}=-\farc{8}{5}$ mg/L in first year.

The chloride concentration will change $-2\frac{1}{15}=-\frac{31}{15}$ mg/L the next year.

The total change that will likely occur in 2 years is

$=-\frac{8}{5}+(-\frac{31}{15})$

$=\frac{-24-31}{15}$

$=\frac{-55}{15}$

$=\frac{-11}{3}$

In a mixed fraction - $=-3\frac{2}{3}$ mg/L

4 0

3 years ago

Read 2 more answers

You and a friend start riding bikes toward

Setler [38]

Answer:

1.5 hours

I traveled farther

2 miles

Step-by-step explanation:

$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$

My speed

$\dfrac{2\dfrac{1}{6}}{\dfrac{1}{4}}=\dfrac{\dfrac{13}{6}}{\dfrac{1}{4}}=\dfrac{26}{3}\ \text{mph}$

Speed of my friend

$7\dfrac{1}{3}=\dfrac{22}{3}\ \text{mph}$

Time passed for both me and my friend will be equal if we start at the same time and the combined distance covered by both of us will be 24 miles. Let $t$ be the time taken to meet