Pleaseeeee helppp!!!!!!!!

Tyler ate x fruit snacks, and Han 3/4 less than that

Harman [31]

The amount Tyler ate would have to be divided by four. After you do that you multiply the number you got to get 3/4. 20/4 = 5 5x3=15

4 0

3 years ago

How to find the area of a circle<br>

Keith_Richards [23]

Answer:

Area = π r^2.

Step-by-step explanation:

Given the radius r of a circle, the area = π r^2.

So for example the area of a circle of radius 2 = π 2^2

= 4π.

8 0

3 years ago

Read 2 more answers

There are 28 students in a class. Two of the students are named Emily. What is the approximate probability that either Emily is

anyanavicka [17]

Answer:

answer is 7

Step-by-step explanation:

they are saying 28 students that means we have to divide the name because their are two students name Emily

28/2= 14 then

14/2=7

5 0

2 years ago

The lowest temperature on May 15 is 2 thirds as warm as the warmest temperature on May 15. The lowest temperature on May 15 is 6

marin [14]

40 degrees because you do 60 times two thirds

8 0

3 years ago

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#24, i am getting <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%26%5Cfrac%7B18%7D%7B7%7D%20

sveticcg [70]

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 1 & -1 & 2 & 2 \\ 1 & 2 & -3 & 4 \end{array} \right]$

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & -7 & 7 & -6 \end{array} \right]$

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]$

Multiply the last row by 1/28 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Eliminate z from the second row by combining 3 (row 3) and row 2 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Multiply the second row by -1 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

$\left[ \begin{array}{ccc|c} 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Multiply the first row by 1/2 :

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

8 0

3 years ago