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Kazeer [188]
2 years ago
5

Gavin needs to buy candy for the 11 students in his class plus himself. Gavin and each student will get one bag of candy. He use

s $20 to buy two different types of candy. Gavin buys x bags of gummy bears that cost $1.50 per bag and y bags of chocolates that cost $2.00 per bag. The system of equations used to represent this situation is graphed, as shown.
Mathematics
2 answers:
Bingel [31]2 years ago
6 0

Answer:

20=1.50(6)+2.00(4)

Step-by-step explanation:

faltersainse [42]2 years ago
6 0

Answer:

4

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

6 0
2 years ago
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<h3>Answer: D) 5</h3>

=======================================================

Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, \frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2

Allowing us to say,

\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\

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Explanation:

I hope this helped!

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7 0
2 years ago
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