Answer:
- see the attachment for a graph
- yes, you can carry 5 math books in one load (along with 0–2 science books)
Step-by-step explanation:
If x and y represent the number of math and science books you're carrying, respectively, then 3x and 4y represent their weights in pounds.
The total weight of the carry will be 3x+4y, and you want that to be at most 24 pounds. The expression modeling this is ...
... 3x +4y ≤ 24
A graph of this inequality is shown in the attachment. (We have added the constraints that the number of books not be negative.)
___
5 math books will weigh 5·3 = 15 pounds, so will be within the limit you can carry.
I will base the figure from the one you attached in the other problem you posted. (Also attached here).
Statements [Reasons]
1. Angle X is congruent to Angle Z [Given]
2. Line XY is congruent to Line YZ [Given]
3. Angle Y is congruent to Angle Y [Congruence is reflexive]
4. Triangle XBY is congruent to Triangle ZAY [ASA]
ANSWER: The ASA (Angle Side Angle) Congruency Theorem could be applied.
Answer:
Step-by-step explanation:
![-2\dfrac{1}{3}-1\dfrac{1}{3}=-2\dfrac{1}{3}+\left(-1\dfrac{1}{3}\right)=-\left(2\dfrac{1}{3}+1\dfrac{1}{3}\right)\\\\=-\bigg[(2+1)+\left(\dfrac{1}{3}+\dfrac{1}{3}\right)\bigg]=-\left(3+\dfrac{1+1}{3}\right)\\\\=-3\dfrac{2}{3}](https://tex.z-dn.net/?f=-2%5Cdfrac%7B1%7D%7B3%7D-1%5Cdfrac%7B1%7D%7B3%7D%3D-2%5Cdfrac%7B1%7D%7B3%7D%2B%5Cleft%28-1%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%3D-%5Cleft%282%5Cdfrac%7B1%7D%7B3%7D%2B1%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%5C%5C%5C%5C%3D-%5Cbigg%5B%282%2B1%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%2B%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%5Cbigg%5D%3D-%5Cleft%283%2B%5Cdfrac%7B1%2B1%7D%7B3%7D%5Cright%29%5C%5C%5C%5C%3D-3%5Cdfrac%7B2%7D%7B3%7D)
Answer:
Option C
Step-by-step explanation:
You forgot to attach the expression that models the cost of the camping trip during the three days. However, by analyzing the units, the answer can be reached.
The total cost has to be in units of $.
There are two types of costs in the problem:
Those that depend on the number of days ($/day
)
Those that depend on the number of students and the number of days ($/(student * day))
If there are 3 days of camping and b students, then you have to multiply the costs that depend on the days by the number of days (3), and the costs that depend on the number of students you have to multiply them by 'b'
So, if the costs that must be multiplied by 'b' are only those that depend on the number of students, the coefficient of b must be:
3 days (Cost of training + Cost of food Miscellaneous expenses :).
Therefore the correct answer is option C:
C. It is the total cost of 3 days per student of Mr. Brown, with training, food and miscellaneous expenses.
The expression that represents the total expense should have a formula similar to this:
![y = (3 days) *([\frac{20.dollars}{(day * student)} + \frac{30.dollars}{(student * day)} + \frac{50.dollars}{(student * day)}] b + \frac{200}{day}) + 1050.dollars](https://tex.z-dn.net/?f=y%20%3D%20%283%20days%29%20%2A%28%5B%5Cfrac%7B20.dollars%7D%7B%28day%20%2A%20student%29%7D%20%2B%20%5Cfrac%7B30.dollars%7D%7B%28student%20%2A%20day%29%7D%20%2B%20%5Cfrac%7B50.dollars%7D%7B%28student%20%2A%20day%29%7D%5D%20b%20%2B%20%5Cfrac%7B200%7D%7Bday%7D%29%20%2B%201050.dollars)
y = 3 ($ 100b + $ 200) + $ 1050
