125 = 5^3 = 25 * 5
The simplification you are trying to do involves finding the prime factorization of a number. Since 125 is not divisible by 2, move on to 3. 125 is also not divisible by 3, so move on to 5. Then 125/5 = 25, and 25/5 = 5, so 125 = 5^3. The prime factorization of 125 is 5^3.
The digit in the ten thousands place is printed slightly to the left
of the one in the thousands place.
If they are the same digit, then the value of the one on the left is
ten times the value of the one on the right.
Direct variation is a relation that has the form
y = kx
where k is the constant of proportionality.
If you are told that a relation is a direct proportion, and you are given one data point, you can find k. The you can write the equation of the direct relation.
Here is an example.
The price of gasoline follows a direct variation.
John bought 5 gallons of gas and paid $15.
a) Write an equation for the relation.
b) Using the relation you found, how much do 13.8 gallons cost?
Solution:
Since the relation is a direct variation, it follows the general equation of a direct variation:
y = kx
We are given one data point, 5 gallons cost $15.
We plug in 5 for x and 15 for y and we find k.
y = kx
15 = k * 5
k = 3
Now that we know that k = 3, we rewrite the relation using our value of k.
y = 3x
This is the answer to part a).
Part b)
We use our relation, y = 3x, and we plug in 13.8 into x and find y.
y = 3x
y = 3 * 13.8
y = 41.4
The price of 15 gallons of gas is $41.40.
Answer:
the prices were $0.05 and $1.05
Step-by-step explanation:
Let 'a' and 'b' represent the costs of the two sodas. The given relations are ...
a + b = 1.10 . . . . the total cost of the sodas was $1.10
a - b = 1.00 . . . . one soda costs $1.00 more than the other one
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Adding these two equations, we get ...
2a = 2.10
a = 1.05 . . . . . divide by 2
1.05 -b = 1.00 . . . . . substitute for a in the second equation
1.05 -1.00 = b = 0.05 . . . add b-1 to both sides
The prices of the two sodas were $0.05 and $1.05.
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<em>Additional comment</em>
This is a "sum and difference" problem, in which you are given the sum and the difference of two values. As we have seen here, <em>the larger value is half the sum of the sum and difference</em>: a = (1+1.10)/2 = 1.05. If we were to subtract one equation from the other, we would find <em>the smaller value is half the difference of the sum and difference</em>: b = (1.05 -1.00)/2 = 0.05.
This result is the general solution to sum and difference problems.
