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Vladimir79 [104]
3 years ago
7

margo can purchase tile at a store for $0.79 per tile and rent a tile saw for $24 at another store she can borrow the tile saw f

or free if she buys tiles there for $1.19 per tile how mant tiles must she buy for the cost to be the same at both stores?
Mathematics
1 answer:
dimaraw [331]3 years ago
4 0
She must by 60 tiles for the prices to be even. 24+0.79x=1.19x x=amount of tiles. you subtract 0.79x from both sides. you should get 24=0.4x. then you divide 0.4 from both sides. you should get 60=x
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Step-by-step explanation:

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Answer:

Possible answer: \displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6.

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Rewrite the bounds of c as fractions:

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\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}.

Hence the range for c would be:

\displaystyle \frac{3}{2} < c < \frac{4}{2}.

If the denominator of c is also 2, then the range for its numerator (call it p) would be 3 < p < 4. Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than 2.

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}.

\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}.

At this point, the difference between the numerators is now 2. That allows a number (7 in this case) to fit between the bounds. However, \displaystyle \frac{1}{c} = \frac{4}{7} can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

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\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}.

\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}.

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It is important to note that some expressions for c can be simplified. For example, \displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5} because of the common factor 2.

Apparently \displaystyle c = \frac{16}{10} = \frac{8}{5} works. c = 1.6 while \displaystyle \frac{1}{c} = \frac{5}{8} = 0.625.

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