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Crank
3 years ago
13

In what ways is it useful to develop an algorithm for adding and subtracting rational numbers?

Mathematics
2 answers:
Arada [10]3 years ago
7 0

Answer:

To add or subtract some rational numbers, you have to change the rational numbers so they have a common denominator. When you have a common denominator, you can then go on with the addition or subtraction on the numerator part.

Oksi-84 [34.3K]3 years ago
4 0

Answer:

To add or subtract these rational numbers, you first need to change your rational numbers so that they share a common denominator. Once you have a common denominator, you can then go ahead with the addition or subtraction of just the numerator part

Step-by-step explanation:

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One hypotenuse and one leg of a right triangle have lengths 61 and 11. Find the length of the other leg
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If the length is the same as one leg, than all the other legs should be the same exact dimensions, legs can not be different measurements by height. Hope that this help you!
5 0
3 years ago
Need Assistance<br>Show Work​
Evgesh-ka [11]

Answer:

-52, and the opposite of this is 52.

Step-by-step explanation:

If we are losing 52 pounds, then our number is -52 since we are losing 52 pounds (adding a negative number to a positive number is the same as subtracting that number).

The opposite of a number is when we negate the number, or multiply it by -1.

A negative number times a negative number is a positive number.

-52\cdot-1 = 52\cdot1 = 52

Hope this helped!

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3 years ago
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Solve the system using combination method to get a solution (x,y)<br> 4x+7y=10<br> -4x-6y=28
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3 years ago
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Leo's bank balances at the end of months 1, 2, and 3 are $1,500.00, $1,530.00, and $1,560.60, respectively. The balances form a
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Rachel has 60 scones. She sells them to Robbie, Cameron, Louis, Tom and Charlie in that order. Each customer buys more scones th
Korvikt [17]
Robbie bought the smallest amount. Let's use x for that amount.
Let's use n for amount that each following customer increases.

We have:
Robbie=x
Cameron=x+n
Louis=x+n+n
Tom=x+n+n+n
Charlie=x+n+n+n+n

We know that they bought total of 60 scones.
Robbie + Cameron + Louis + Tom + Charlie = 60
x + x+n + x+n+n + x+n+n+n + x+n+n+n+n = 60
5x + 10n = 60 /:5
x + 2n = 12

We are also given this information:
(Robbie + Cameron) = 3/7 * (Louis + Tom + Charlie)
We insert the equations from above:
(x + x+n) = 3/7 * (x+n+n + x+n+n+n + x+n+n+n+n) 
2x + n = 3/7 * (3x + 9n) /*7
14x + 7n = 3* (3x + 9n)
14x +7n = 9x + 27n
We take everything on the left side.
14x + 7n - 9x - 27n = 0
5x - 20n = 0/:5
x - 4n = 0

Now we have two equations:
x + 2n = 12
x - 4n = 0
We solve second one for x and insert it into first one.
x + 2n = 12
x = 4n

4n + 2n =12
6n = 12 /:6
n = 2
x=4*2
x=8


Now we can solve for the amount for each customer.
Robbie=x = 8
Cameron=x+n = 8 + 2 = 10
Louis=x+n+n = 8 + 2 + 2 = 12
Tom=x+n+n+n = 8 + 2 + 2 + 2 = 14
Charlie=x+n+n+n+n = 8 + 2 + 2 + 2 + 2 = 16
7 0
3 years ago
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