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3 years ago

Liam says - (5/12) is equivalent to 5/-12. Explain if Liam is correct by completing the statements below.

2 answers:

6 0

A positive number divided by a positive number results in a negative quotient, and it's opposite is always positive. The fraction 5/-12 can be described as a positive number divided by a negative number, which always results in a negative quotient. The fractions have different value, so Liam is not correct.

Vinil7 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Hope this helps!!

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Write an inequality that represents 10 less than a 4 times a number more than 10,000

OLga [1]

10-4×>10,000??? I'm sorry Im not good at math

4 0

3 years ago

Read 2 more answers

Solve For Each Variable.

bixtya [17]

The first equation given $6=1-2n+5$

We have to add 1 and 5 to the right side first. We will get,

$6= 6 - 2n$

To get rid of 6 from the right side we have to subtract 6 from both sides.

$6-6 = 6-6-2n$

$6-6 = -2n$

$0=-2n$

To find n we have to move -2 to the other side by dividing both side by -2.

$0/-2 = -2n/-2$

$0= n$

$n=0$

So we have got the required answer for the first question.

The solution is n = 0.

The second equation given,

$8x-2 = -9+7x$

First we have to move 7x to the left side by subtracting it from both sides.

$8x-7x-2 = -9+7x-7x$

$8x-7x-2 = -9$

$x-2 = -9$

Now we have to move -2 to the right side by adding 2 to both sides.

$x-2+2 = -9+2$

$x = -9+2$

$x = -7$

We have got the required answer for the second question.

The solution is x = -7.

The third equation given,

$-8 = -(x+4)$

We have to get rid of that negative sign from both sides. As we have negative sign to both sides we can cancel it out. We will get,

$8=x+4$

Now we have to move 4 to left side by subtracting it from both sides.

$8-4 = x+4-4$

$8-4 = x$

$4=x$

$x = 4$

So we have got the required answer .

The solution is x = 4.

3 0

3 years ago

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monitta

The correct answer is A

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3 years ago

Read 2 more answers

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avanturin [10]

Answer:

2nd one.

Step-by-step explanation:

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2 years ago

Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b

Korolek [52]

Hello,

Very nice as problem.

2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies

since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0

4 0

3 years ago