I'll use the same notation as the other one:
3(y-6)<4(y+5)
I'm going to clean this up a bit by distributing:
3y-18<4y+20
Now, subtract 3y from both sides:
-18And now, subtract 20:
-38So y is greater than or equal to -38
Hope this helps!
When you round you have to check the number to the right of the number you are rounding. If the number to the right is 0-4 then you keep the number the same. 5-9 the number you are rounding goes up by one. Now the 4 before the decimal point is in the ones spot. The 5 to the right of the 4 is the tenth spot and the 6 to the right of that is in the hundredth spot. so this is where we need to round. To the right of the 6 is a 4 so the number would stay the same. Now when you round you drop all the numbers after it so the number is 3184.56
First of all, recall that division by zero is undefined; it's nonsensical; it's just not allowed.So zero certainly needs to be excluded when dividing.
But what about multiplying by zero?
The problem is that multiplying by zero can change the truth of an equation:
It can take a false equation to a true equation.
To see this, consider the false equation ‘
2 = 3
Multiplying both sides by zero results in the new equation
2 ⋅ 0 = 3 ⋅0 (that is, ‘0 = 0’), which is true.
1/4 is 5/20 and 11/10 is 22/20. You have to multiply the numerator the same amount of times you do to the denominator. 4x5 is 20, and you have to do 1x5 as well. At the end, you get 5/20. Hope this helps!