I would have to say the answer is=<span>C) 16x3 - 8x2y - 16xy2 - 64y3</span>
Answer:
1. $15.80
2. $12
3. $11.06
Step-by-step explanation:
Step-by-step explanation:
1. Buying 20 single candy bars at regular price
The regular price of one is 79 cents.
(# of bars) X (price per bar) = regular price
20 X 79 cents = 1580 cents
1580 cents = $15.80
2. Buying 20 bars in packs of 5
Each pack is $3
(# of packs) X (price per pack) = pack price
(20/5) X ($3) = pack price
4 X $3 = $12
3. Buying 20 bars at special price
For every 2 bought, 1 is free.
(Number of bars paid for) X (price per bar) = special price
(2(20/3)) X (0.79) = special price
14 X 0.79 = $11.06
Because for every 2 bars that are paid for, 1 is free, he pays for 14 bars and he gets the 21st free.
Andy's 20 regular priced candy bars are $15.80, bars in packs are $12, and bars at the special price is $11.06.
Answer:
34.5 miles
Step-by-step explanation:
Answer:
It depends on the equation.
If the bases are equal and the variables are only in the exponents, set the exponents equal.
If there are variables in the exponents, but you cannot set the bases equal, then use logarithms.
Example 1:
Here you have the same base on both sides. The variables are in the exponents. Set the exponents equal and solve for x.
2x + 5 = 9
2x = 4
x = 2
Example 2:
The bases are different, but you can make the bases equal using laws of exponents. Remember that 9 = 3^2.
Now you have equal bases, so the exponents must be equal.
2x = 12
x = 6
Example 3:
Here you can't make the bases equal, so you take the log of both sides and use laws of logs.
Recall that:
Answer:
Option B.
Step-by-step explanation:
Two events are said to be independent of each other, if the probability of one event ocurrin in not way affects the probability of the other event occurring.
The interception of two independent events P(A ∩ B) = P(A) × P(B), where:
P(A) = 0.70
P(B) = 0.20
P(A ∩ B) = P(A) × P(B) = 0.70x0.20 = 0.14
The two events are independent if the probability of buying Bread AND cheese equals: 0.14, which is Option B.