Haybalor's Five Bales Of Hay?

($372 + $135 + 500) / ($2.21 - ($0.83 + .40)) = 1,028 [+/- 31]

Bales A+C=82, A+D=83, A+E=86, B+C=84, B+D=85, B+E=88, C+D=87, C+E=91, D+E=91. That is just one way to do the problem. My proof that this is the correct answer is because the numbers for each letter added up, equals the weights as given in the book. There are 10 different ways to this problem. I only could figure out one of the ten different ways to do the problem.

The solution is 0.25(15) + 0.25(35). The order in which the products are added does not matter. This is an example of using the commutative property.

K 13. C 14. C 15. E 16. D 17.

seven. Then I tried 21, but that didn’t fit either. I kept going up by multiples of seven trying to find

(Note: You can use tables or a financial calculator. If you use a calculator, please provide the inputs you used to solve the problems.) (5 points each = total 20 points)

MATH 533 Week 2

It’s hard to keep all the math rules straight. You may have been thinking of addition. With addition we would

We have 5 bales of hay. We don’t know how much they weigh. When they are weighed in pairs, their weights are 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. Your initial task is to find is to find out how much each bale weighs. Is your answer the only possible set of weights? Explain how you know

To start the discussion, I wrote the following open sentence for Madison: 5 + 6 = ____. Madison quickly responded “11”. When I asked how she arrived at her answer, she indicated that this number fact was something that she had known since kindergarten.

Since m = 7, I immediately know that (8m - 18) = 38, and (7m + 3) = 52.

Running with this information can now write out the equation AB2 + BC2 = AC2. One important thing is that we must note that AB is equal to “X” and the line segment of BC is equal to that of 2x+4, and that AC will be equal to that of 2x+6. So we will now input this information to create (x)2 + (2x + 4)2 = (2x + 6)2 and begin factoring each term into two sections. These two sections will be as x*x + (2x + 4)(2x + 4) = (2x + 6)(2x + 6). x times x is x2. An important tool to use now would be the FOIL method, so we will take (2x + 4)(2x + 4) and create 4x2 + 16x + 16. Right off the bat we notice that we have like terms. So we will add x2 to 4x2 to get 5x2. This will create 5x2 + 16x + 16 = 4x2 + 24x+ 36. Now we will use the subtraction property to get 5x2 – 4x2 + 16x – 24x + 16 – 36 = 0, however we still have like terms, so because 5x2 is a like term with -4x2 we will add them together to get x2. We will also combine 16x and –24x and also 16 and –36 which are also like terms and create –8x and –20, our equation should now look like x2 – 8x -- 20 = 0.We will now factor the equation from left to right, first factoring x2 which has 1 coefficient so the fact will be 1 and -1. The other term will be 20 which have no coefficient so we will do 5x4 and then 4 still can be divided so 2x2. This will create 20=225.

Explain to Anna, Blake, and Christine who is correct, and identify any errors that you find. Provide the correct manner to fix those problems, and identify the correct answer. Use complete sentences.

(0+6 and 4+2). C and F could be 0, 2, 4, or 6. The fifth clue, “C is 2 less than D”, I knew that C and D were both even 1-digit whole numbers. C could only be 0, 2, 4, or 6. Adding two to 0, 2, 4, and 6 would give me D’s possibilities (2, 4, 6, or 8). The last clue, “E is greater than C”, I knew that E could be 3 or 5, so C could not be 6 because that is greater than 3 and

As a class, we got our solution. The weights for the individual hay bales are 39, 41, 43, 44, 47. Here are some