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Naya [18.7K]
2 years ago
9

There is a ratio of 5 to 3, dogs to cats. There are 15 cats; how many dogs would you have?

Mathematics
2 answers:
LenaWriter [7]2 years ago
8 0
Since we have 5 dogs to 3 cats, we can find a scale factor. Our new proportion is x:15.
To solve for this, we need to divide 15 by 3 to get our scale factor to find x.
15/3 = 5, so our scale factor is 5.
Multiply 5 dogs by 5 to get the new ratio.
Our new ratio is 25:15, meaning that we have 25 dogs total.
Your answer is 25 dogs.
I hope this helps!
denpristay [2]2 years ago
3 0
15×5/3=25. As a result, I would have 25 dogs. Hope it help!
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 Sandra sells necklaces at a school craft fair. She uses the equation P= 7.5n - (2.25n + 15) to determine her total profit at th
gregori [183]
The way you find profit is to subtract the revenue and the cost
Profit = Revenue - Cost
The revenue is the amount of money coming in, the cost is the amount of money going out. The goal of course is to have the revenue larger than the cost so that the profit is positive. 

So the equation given is
P = 7.5n - (2.25n+15)
and its in the form
P = R - C
where...
R = 7.5n is the revenue equation
C = 2.25n+15 is the cost equation

Focus on the revenue equation
R = 7.5n
which is the same as 
R = 7.50*n

This tells us that Sandra pulls in a total of 7.50*n dollars where n is some positive whole number. It represents the number of necklaces sold. For example, if she sold n = 10 necklaces, then
R = 7.50*n
R = 7.50*10
R = 750
meaning that Sandra has made $750 in revenue

As you can see above, the revenue is computed by multiplying the price per necklace ($7.50) by the number of necklaces sold (n) to get R = 7.50*n

So that's why the answer is $7.50

Note: The 2.25 is part of the cost equation. This is known as the variable cost. It is the cost to make one necklace ignoring the fixed cost (eg: rent). The variable cost often doesn't stay the same, but algebra textbooks often simplify this aspect. 
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olga nikolaevna [1]
The answer is b because are congruent
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Answer:

Step-by-step explanation:

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