Answering:
188
Explaining:
To solve this problem, we must divide the total amount of money raised by the cost of the stuffed animals. Each stuffed animal costs $17. The club raised $3,207 to buy said stuffed animals. By dividing the money earned, which is also the money the club is able to spend, by the cost of a single/one stuffed animal, we will get how many stuffed animals the club can purchase with the money they currently possess. Our equation will look like this: 3,207 ÷ 17.
After dividing 3,207 by 17, we have the number 188.64705882. This can be rounded to the nearest tenth to create the simpler yet still accurate number 188.6.
Our final step is to round 188.6 down to the whole number it already has. (That is to say, simply cut off the fraction and remove it to get our answer.) This step must be done because we are buying stuffed animals in a real-world situation. The club would not be able to purchase part of a stuffed animal for a fraction of the cost, and the cost of the stuffed animals in the problem is a fixed value. This means that the fraction is irrelevant since we cannot purchase anything with it, effectively making it totally irrelevant to the answer. After removing the fraction from 188.6, we are left with 188.
Therefore, the maximum number of stuffed animals the club can buy is <em>188 stuffed animals</em>.
Answer:
17. xy + z = 51
18 yz - x = 18
21. 5z + ( y - x ) = 20
22. 5x - ( y2 - 4x ) = -10
23 . x2 + y2 - 10z = 70
24. z3 + ( y2 - 4x ) = 1
25. y + xz / 2
26 .3y + x2 / z = 30
Step-by-step explanation:
17.(6 x 8)+3=
48+3=51
18.(8 x 3)- 6 =
24 - 6 = 18
21. 15 + ( 8 - 3 )
15 + 5=
20
22 .30 - (64 - 24)=
30 - 40=
-10
23. 36 + 64 - 30
100 - 30
70
24. 9 + (16 - 24 )=
9 + -8 =
1
25. 8 + 48 / 2
56 / 2
28
26. 24+ 36/3
60/3=
30
Answer:
The answer is...
Step-by-step explanation:
No, because the bike order does not meet the restrictions of 4c+6a (less than or equal to) 120 and 4c+4a (less than or equal to) 100
Answer:
3200 ft-lb
Step-by-step explanation:
To answer this question, we need to find the force applied by the rope on the bucket at time 
At 
After seconds, the weight of the bucket is 
Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.
If the upward direction is positive, the displacement after seconds is 
Since the well is 80 ft deep, the time to pull out the bucket is 
We are now ready to calculate the work done by the rope on the bucket.
Since the displacement and the force are in the same direction, we can write
Use and 
![=\left[63 t-0.2 t^{2}\right]_{0}^{36}](https://tex.z-dn.net/?f=%3D%5Cleft%5B63%20t-0.2%20t%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B36%7D)
