1. $Z_{11}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{4};$
$5\cdot 4=20=\overline{9};$
$5\cdot 5=25=\overline{3};$
$5\cdot 6=30=\overline{8};$
$5\cdot 7=35=\overline{2};$
$5\cdot 8=40=\overline{7};$
$5\cdot 9=45=\overline{1};$
$5\cdot 10=50=\overline{6}.$

The multiplicative inverse of 5 in $Z_{11}$ is 9.

2. $Z_{12}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{3};$
$5\cdot 4=20=\overline{8};$
$5\cdot 5=25=\overline{1};$
$5\cdot 6=30=\overline{6};$
$5\cdot 7=35=\overline{11};$
$5\cdot 8=40=\overline{4};$
$5\cdot 9=45=\overline{9};$
$5\cdot 10=50=\overline{2};$
$5\cdot 11=55=\overline{7}.$

The multiplicative inverse of 5 in $Z_{12}$ is 5.

3. $Z_{13}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{2};$
$5\cdot 4=20=\overline{7};$
$5\cdot 5=25=\overline{12};$
$5\cdot 6=30=\overline{4};$
$5\cdot 7=35=\overline{9};$
$5\cdot 8=40=\overline{1};$
$5\cdot 9=45=\overline{6};$
$5\cdot 10=50=\overline{11};$
$5\cdot 11=55=\overline{3};$
$5\cdot 12=60=\overline{8}.$

The multiplicative inverse of 5 in $Z_{13}$ is 8.

8 0

3 years ago

Solve the following system of equations.

Elan Coil [88]

Answer:

(2,-3)

Step-by-step explanation:

I am not sure if you meant the first equation to be y or -y. I solved it as y.

y = x-5 -x -3y =7

I am going to take the second equation and write it as x =

-x - 3y = 7 Give equation

-x = 3y +7 Add 3y to both sides

x = -3y-7 Multiplied each term in the equation by -1 so that x could be positive

I am going to substitute -3y-7 for x in the first equation up above

y = x - 5

y = -3y -7 - 5 Substitute -3y-7 for x

y = -3y -12 Combined -7-5

4y = -12 Added 3y to both sides

y = -3 Divided both sides by 4.

I now know that y is -3, I will plug that into x = -3y-7 to solve for x

x = -3(-3) -7

x = 9-7 A negative times a negative is a positive

x = 2

5 0

1 year ago

The product of (-5xy2) and (-4x2y)

solong [7]

Combine like variables:

(20x^2)(2y^3)

Combine coefficients:

40(x^2)(y^3)

3 0

2 years ago

In a game, Janeesa started with 0 points. She then earned 50 points, lost 80 points and earned 10 points. Which number line show

umka2103 [35]

Answer:

-20

Step-by-step explanation:

If Jeneesa started with, 0 points and she earned 50, then she would have 50. If she lost 80 points, she would go into the negatives. She would be at -30, and when she earned 10 more points, she would be at -20.

I don't know what your options look like, but look at the one(s) with -20 at the end and figure out what one best fits my explanation (and the numbers I use)!

3 0

3 years ago

give the slope of the line (if it exists) and the y intercept (if it exists) and graph the line 5y=3. what is the slope?​

give the slope of the line (if it exists) and the y intercept (if it exists) and graph the line 5y=3. what is the slope?