If George completed 3/5 of work in 9 days, it means that he needs 3 days to finish 1/5th of the work.
The remaining part is 2/5 because 1-3/5=2/5. 2/5 is also 6/15
From this, how much did George do? in the first 3 days he did 1/5th and then one more day was left, during which he did 1/5/3=1/15th.
So he did in total 1/15+1/5=1/15+3/15=4/15.
this means that Paul did the rest of 6/15, so Paul did 6/15-4/15=2/15.
So we know that he does 2/15 in 4 days, which means that every two days he can do 1/15 of the work
so he would need 15 times 2 days to finish the work - 30 days!
Answer:
8428
Step-by-step explanation:
According to the Question,
- Given, A theater has 56 rows of seats. If there are 13 seats in the first row, 18 in the 2nd row, 23 in the 3rd row.
We have a number sequence 13, 18, 23, ...
- We can say that this is an arithmetic sequence because of the common difference 'd', is equal to 5, and the first term 'a1' is equal to 13.
- The formula for the sum of an arithmetic sequence with n terms is given is
.
Substitute the given values into the equation to solve for the sum of the 56 rows of seats.
![S_56= \frac{56}{2}[2(13)+(56-1)(5)]\\S_{56}=28\left[26+275\right]\\S_{56}=28\left[301\right]\\S_{56}=8428](https://tex.z-dn.net/?f=S_56%3D%20%5Cfrac%7B56%7D%7B2%7D%5B2%2813%29%2B%2856-1%29%285%29%5D%5C%5CS_%7B56%7D%3D28%5Cleft%5B26%2B275%5Cright%5D%5C%5CS_%7B56%7D%3D28%5Cleft%5B301%5Cright%5D%5C%5CS_%7B56%7D%3D8428)
Therefore, there are 8428 seats in all.
Answer: He can completely decorate 11 cars
Explanation: since he puts 2 snowflakes on each of the 4 windows on one car, in total Jack uses 8 snowflakes per car. If he makes 90 snowflakes, 90 divided by 8 gives you the amount of cars he can decorate. 90/8 = 11.25, but since you can’t have .25 cars the answer is 11.
Answer:
- b/a
- 16a²b²
- n¹⁰/(16m⁶)
- y⁸/x¹⁰
- m⁷n³n/m
Step-by-step explanation:
These problems make use of three rules of exponents:

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)
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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

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4. This works the same way the previous problem does.

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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.
