1. The Associative Property of Addition.
Both equations equal 28.
2. The equations are not equal.
The volume of a rectangular prism is equal to length*width*height. In this case, we have V = 243 and H = 3.
243 = LW(3), so LW = 81, where L & W are whole numbers greater than 1. There are only 3 possible pairs of values for this: 27*3, 9*9, and 3*27 (we cannot use 1*81 or 81*1, since we need dimensions > 1).
Part A. This could be true if Length = 9 in and Width = 9 in, as 9*9 = 81.
Part B. The claim could be false if Length = 27 in and Width = 3 in, as 27*3 = 81, but 27 is not equal to 3.
*Note that the width cannot be longer than the length.
Answer:
To calculate the mean of a set of data, you have to work out the sum of the data (in this case, you need to work out the sum of all the points) and divide it by the number of data points there are (in this case, number of games played).
So you have to do:
(67 + 45 + 84 + 55 + 73 + 36 + 80 + 62 + 38)/9 = 60
The mean number of points is 60.
Step-by-step explanation:
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Answer: The salon charges $16 for a manicure and $25 for a pedicure.
Step-by-step explanation:
Let x represent the amount charged by the salon for a manicure.
Let y represent the amount charged by the salon for a pedicure.
Over the weekend, they performed 49 manicures and 16 pedicures, bringing in a total of $1,184 in receipts. It means that
49x + 16y = 1184- - - - - - - - - - 1
So far this week, they have administered 31 manicures and 33 pedicures, with receipts totalling $1,321. It means that
31x + 33y = 1321- - - - - - - - - - 2
Multiplying equation 1 by 31 and equation 2 by 49, it becomes
1519x + 496y = 36704
1519x + 1617y = 64729
Subtracting, it becomes
- 1121y = - 28025
y = - 28025/- 1121
y = $25
Substituting y = 25 into equation 1, it becomes
49x + 16 × 25 = 1184
49x + 400 = 1184
49x = 1184 - 400
49x = 784
x = 784/49
x = $16
If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.
So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.
(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.
(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)
= x^4 + 29x^2 + 100
So the equation would be x^4 + 29x^2 + 100 = 0.
Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.
