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Masteriza [31]
3 years ago
11

What is 5 multiplied by 2/7 as a mixed number

Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0
10/7 which equals 1 and 3/7
Luba_88 [7]3 years ago
4 0

Answer:

the answer is 1 and 3/7 as a mixed number

Step-by-step explanation:


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The first three steps in determining the solution set of the system of equations algebraically are shown.
Ivenika [448]

(2,-1), (-4,17).

Step-by-step explanation:

Equate the equation A and equation B

Convert the quadratic equation in factored form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Complete the square. Remember to balance the equation by adding the same constants to each side.

Rewrite as perfect squares

Square root both sides

Find the values of y

Substitute the value of x in the equation B

3 0
2 years ago
Express the system as AX = B; then solve using matrix inverses
Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]

\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B

Adj.A=Transpose of cofactor of Matrix A

Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

   A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]

\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4

7 0
3 years ago
Does anyone know this?? ​
Evgesh-ka [11]
The answer is letter A
7 0
3 years ago
Help to solve. B. Present in the form of a fraction.
Studentka2010 [4]
I hope this helps you

8 0
2 years ago
Read 2 more answers
Raymond wants to make a box that has a volume of 360 cubic inches. He wants the height to be 10 inches and the other two dimensi
Aleks [24]
V=LWH
h=10
V=360

360=LW10
divide both sides by 10
36=LW
try some values
ok so if we get L=1 and W=36, that is the same box as L=36 and W=1
so
L,W
36,1
18,2
12,3
9,4
6,6

5 of them

answer is 5 different boxes
dunno what the 'compare volume' is
8 0
3 years ago
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