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2 years ago

Alicia says that 6÷2+5 is the same as 5+6÷2. is Alicia correct or incorrect

2 answers:

6 0

She is correct, as long as you don't move the symbols around, you will always get the same answer. Because of Order of Operations. Try it.

6/2+5

3+5=8

then,

5+6/2

5+3=8

This means that Alicia is correct.

As long as you don't switch the symbols around.
Example: 6+2/5 or switch numbers around 6/5+2.

liq [111]2 years ago

3 0

It is not correct because of the order the problem is organized in.

You might be interested in

Simplify the following math expression: 5+4(6-3)+4/2

lesantik [10]

Answer:

you have to solve the brackets first(6-3=3)

Then solve the division (4/2=2)

lastly the addition (5+4=9)

so we have 9×3=27

27+2=29.

3 0

2 years ago

Read 2 more answers

What is the multiplicative inverse of 5 in z11, z12, and z13? you can do a trial-and-error search using a calculator or a pc?

grin007 [14]

A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.

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