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scoundrel [369]
3 years ago
11

How to do 6x times 1/x

Mathematics
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

6

Step-by-step explanation:

6x * 1/x = 6x/1 * 1/x the x on top cancels the x on bottom so

= 6/1 * 1/1 = 6

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Avery had $25.19 in her wallet. If she bought lunch with 7 & dollars from her wallet, how much money did she have in her wal
ivolga24 [154]

Answer:

B) $18.19

Step-by-step explanation:

Avery has $25.19 in her wallet, and she spends $7 on lunch. Subtract 7 from the original amount she had to get the new total:

25.19 - 7 = 18.19

B) $18.19 is your answer.

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2 years ago
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Simplify 8^3 divided by 4^2. A.) 2 B.) 3 C.) 33
N76 [4]

Answer:

32

Step-by-step explanation:

\frac{8^{3}}{4^{2}} = \frac{512}{16} = 32

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3 years ago
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When two parallel lines are intersected by a transversal, alternate exterior angles are?
3241004551 [841]

Answer:

I think it was option 1

Step-by-step explanation:

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2 years ago
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Solve the system. If there is more than one solution, write the general solution. x + y - 2z = 9 3x + y + 2z = 15 x - 5y + 22z =
Margaret [11]

Answer:

<h2>x = (12-k)/2, y = k, z = (k-6)/4 </h2>

Step-by-step explanation:

Given the system of equation

x + y - 2z = 9 ... 1

3x + y + 2z = 15 ...2

x - 5y + 22z = -27... 3

First let us reduce the system of equation into two with two unknowns.

Subtracting 1 from 3

y-(-5y) + (-2z-22z) = 9-(-27)

y+5y + (-24z) = 9+27

6y-24z = 36 ... 4

Multiplying equation 1 by 3 and subtracting from equation 2

3x + 3y - 6z = 27

3x + y + 2z = 15

On subtracting both;

(3y-y)+(-6z-2z) = 27-15

2y-8z = 12 ... 5

Equating 4 and 5

6y-24z = 36 ... 4

2y-8z = 12 ... 5

Multiplying equation 5 by 3 the equation becomes;

6y-24z = 36 ... 6

6y-24z = 36 ... 7

We can see that equation 6 and 7 are the same;

let y = k

6k - 24z = 36

k - 4z = 6

4z = k-6

z = k-6/4

Substituting y = k and z = k-6/4 into equation 1 to get x

From 1; x + y - 2z = 9 ... 1

x + k -2( k-6/4) = 9

x + k - (k-6)/2 = 9

x = 9+(k-6)/2-k

x = {18+(k-6)-2k}/2

x = (12-k)/2

The solutions to the system of equations are x = (12-k)/2, y = k, z = (k-6)/4 where k is any constant. This shows that the system of equation has infinite solutions.

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