X-3y, x=3, y=-2
(3)-3(-2)
(3)+(3*2)
3+6=9
Answer:
60 minutes
Step-by-step explanation:
We solve this question using Lowest Common Factor method
Find and list multiples of each number of minutes( 12 and 10 minutes) until the first common multiple is found. This is the lowest common multiple.
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80
Multiples of 12:
12, 24, 36, 48, 60, 72, 84
Therefore,
LCM(10, 12) = 60
The number if minutes until they both drip again is 60 minutes
Answer:
Dori has $48 left.
Step-by-step explanation:
The boat costs twice as much as the goggles, divide 8 by 2 to find the price of the goggles. The price of the goggles is $4.
Add $8 and $4 to get $12. This is the total money Dori spent.
Since this is 1/5 of her money, multiply $12 by 5 to get $60. This is the money Dori had.
Subtract the $12 from the $60 to get $48. This is the money Dori has left.
Properties of equality have nothing to do with it. The associative and commutative properties of multiplication are used (along with the distributive property and the fact of arithmetic: 9 = 10 - 1).
All of these problems make use of the strategy, "look at what you have before you start work."
1. = (4·5)·(-3) = 20·(-3) = -60 . . . . if you know factors of 60, you can do this any way you like. It is convenient to ignore the sign until the final result.
2. = (2.25·4)·23 = 9·23 = 23·10 -23 = 230 -23 = 207 . . . . multiplication by 4 can clear the fraction in 2 1/4, so we choose to do that first. Multiplication by 9 can be done with a subtraction that is often easier than using ×9 facts.
4. = (2·5)·12·(-1) = 10·12·(-1) = (-1)·120 = -120 . . . . multiplying by 10 is about the easiest, so it is convenient to identify the factors of 10 and use them first. Again, it is convenient to ignore the sign until the end.
5. = 0 . . . . when a factor is zero, the product is zero
<span>Jill has $30 and Maddie has $60. Jill saves $4 per week and Maddie saves $2 per week. How long will it be before Jill and Maddie have the same amount of money?</span>