For the experiment, you need 2L of cola. Your first option would be to purchase 1 2L bottle of cola for $2.25.
To calculate the second option, let's convert milliliters to liters first. There are 1,000 milliliters in 1 liter. With this, we know that there are 2,000 milliliters in 2 liters. Option 2 comes in 500-milliliter cans, which means that you would need 4 of them (2,000/500 = 4). 4 cans multiplied by $0.50 would cost you $2.00.
Let's check the cost of your answer options.
A. 4 cans - As seen above, this would cost $2.00.
B. 1 bottle - From the question, we know this would cost $2.25.
C. 2 bottles - This would be more soda than you need and would cost $4.50 ($2.25x2)
D. 1 can - This would be .5L and not enough soda for the experiment.
E. 5 cans - This would cost $2.50, but would be an extra 500mL of soda.
F. 2 cans - This would only be 1L of soda and not enough for the experiment.
G. 3 cans - This would be 1.5L of soda and not enough for the experiment either.
For the best price option, you would choose A (four cans of soda). This would give you the amount of soda that you need at the lowest price.
Answer:
<h2>
The answer is B. $8.82</h2>
Step-by-step explanation:
given data
mints $0.96 per pound
chocolates $4.70 per pound
lollipops $0.07 each
cost of each items based on the amount item
mints $0.96*0.75= 0.72
chocolates $4.70*1.5=7.05
lollipops $0.07*15=1.05
total cost= 0.72+7.05+1.05
total cost=$8.82
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
So we have the system of equations:
equation (1)
equation (2)
To use substitution, we are going to solve for one variable in one of our equations, and then we are going to replace that value in the other equation:
Solving for
in equation (2):
equation (3)
Replacing equation (3) in equation (1):
equation (4)
Replacing equation (4) in equation (3):
We can conclude that the solution of our system of equations is <span>
(7/5, 21/10)</span>
