All you have to do is multiply 18.3 x 5.1
18.3×5.1=93.33
It will take 93.33 minutes to repair the entire section. Hope that helps!
p= 1/7
Alright, so first, you need to flip the equation.<span><span>It will turn into p+<span>4/7</span></span>=<span>5/7
Next, you want to get the variable by itself so you need to subtract 4/7 from both sides. 5/7-4/7=1/7
p=1/7</span></span>
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
If Deliah does jumping jacks at a constant rate, this means that she does them at the same pace or you could say that she does the same amount of jumping jacks in a specified amount of time, ie. if you counted how many jumping jacks she did in one minute, it would be same as how many she would complete in the next minute, and the next, and so on.
Now given that she does 184 jumping jacks in four minutes, and she has kept a constant pace throughout, to find out how many she does each minute, we simply need to divide the number of jumping jacks she does in 4 minutes by 4. Thus:
Jumping jacks in 1 minute = Jumping jacks in 4 minutes / 4
= 184 / 4
= 46
Thus, Deliah can do 46 jumping jacks per minute.
All the numbers in this range can be written as

with

and

. Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)
so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.
For each number that occupies an entire diagonal in the table, it's easy to see that that number

shows up

times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.
So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.