The solution to the algebraic expression is: 23e - 21g - 14j + 32
What are algebraic expressions?
Algebraic expressions are mathematical expressions that contain variables, coefficients, and arithmetic operations such as addition, subtraction, division, and multiplication.
Solving algebraic expressions are an important part of mathematics as it helps to improve the aptitude and solving skills of the students.
From the given information, we have;
23j - 21g + 20e - 13 + 52e - 37j + 45 - 49e
let's rearrange by taking the like terms to the same sides;
= 23j -37j - 21g + 20e + 52e - 49e - 13 + 45
= -14j - 21g + 23e + 32
= 23e - 21g - 14j + 32
Therefore, we can conclude that the solution to the algebraic expression is: 23e - 21g - 14j + 32
Learn more about solving algebraic expressions here:
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Six hundred seven and four hundred nine one-thousandths.
Answer: See attached table
Step-by-step explanation:
![\begin{table}[]\begin{tabular}{lllll} & & & & \\ & & & & \\ & & & & \\ & & & & \end{tabular}\end{table}](https://tex.z-dn.net/?f=%5Cbegin%7Btable%7D%5B%5D%5Cbegin%7Btabular%7D%7Blllll%7D%20%26%20%20%26%20%20%26%20%20%26%20%20%5C%5C%20%26%20%20%26%20%20%26%20%20%26%20%20%5C%5C%20%26%20%20%26%20%20%26%20%20%26%20%20%5C%5C%20%26%20%20%26%20%20%26%20%20%26%20%5Cend%7Btabular%7D%5Cend%7Btable%7D)
+---+---+---+
| 1 | 8 | 2 |
+---+---+---+
| 6 | 4 | 2 |
+---+---+---+
| 5 | 0 | 7 |
+---+---+---+
<u>Proofs:</u>
First row: 1+6+5 = 12
Second row: 8+4+0 = 12
Third row: 2+2+7 = 12
First column: 1+8+2 = 12
Second column: 6+4+2 = 12
Third column: 5+0+7 = 12
Diagonal starting from top left to bottom right: 1+4+7 = 12
Diagonal staring from top right to bottom left: 2+4+5 = 12
Answer:
Tyler is correct. The temperature dropped at a rate of about 4° per hour between 4 and 6, while the temperature dropped at about 2.25° per hour between 6 and 10.
Edit: Explanation
The question is asking about which window of time had a <em>faster</em> decline in temperature, not a larger total change in temperature.
In a 2 hour timeframe, the temperature dropped 8°. (4-6 PM)
In a separate 4 hour timeframe, the temperature dropped 9°. (6-10 PM)
To find which window had a faster change in temp, I took the total temperature drop for each timeframe, then divided it by the number of hours each drop took.
8° / 2 = 4° per hour for 4-6 PM
9° / 4 = 2.25° per hour from 6-10 PM
Since the speed at which the temperature dropped per hour was greater from 4-6 PM than 6-10 PM, Tyler was correct.
