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kotykmax [81]
3 years ago
5

Bryce spent $5.26 on some apples prices at $0.64 each and some pears priced at $0.45 each. at another store he could have bought

the same number of apples at $0.32 each and the same number of pears at $0.39 each, for a total cost of $3.62. how many apples and how many pears did bryce buy?
Mathematics
1 answer:
Reil [10]3 years ago
3 0
Let the number of apples be x and that of pears be y, then:
0.64x + 0.45y = 5.26 . . . (1)
0.32x + 0.39y = 3.62 . . . (2)

(2) x 2 => 0.64x + 0.78y = 7.24 . . . (3)
(1) - (3) => -0.33y = -1.98
y = -1.98 / -0.33 = 6

From (2), 0.32x + 0.39(6) = 3.62
0.32x = 3.62 - 2.34 = 1.28
x = 1.28 / 0.32 = 4

Therefore, he bought 4 apples and 6 pears.
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3 years ago
Find the equivalent fraction for 10/20
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1.find the sum. (3^3+5x^2+3x-7)+(8x-6x^2+6)<br> 2. Find the Difference: (8x-4x^2+3\)-(x^3+7x^2+3x-8)
vlada-n [284]

1.

(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1

2.

(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10

3 0
3 years ago
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lisov135 [29]
He spent $9.45
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6 0
3 years ago
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