Answer:
total is 16.50
Step-by-step explanation:
add 8.50+8.00
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>.
no u can only buy 6, or u dont have enough money for 7
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3
Answer:
3/2 or 1.5
Step-by-step explanation:
0.75 times 4 is 3
0.5 times 4 is 2
Your answer is 3/2 or 1.5
