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amid [387]
2 years ago
14

tressa's home is 4/5 mile from school. anton's home is 3/5 mile from school. how many times the distance from anton's home to sc

hool is the distance from tressa's home to school?
Mathematics
1 answer:
WINSTONCH [101]2 years ago
7 0

The distance from Tressa's home to school is 1\frac{1}{3} times the distance from Anton's home to school

From the question,

Tressa's home is 4/5 mile from school

and Anton's home is 3/5 mile from school.

To determine how many times the distance from Anton's home to school is the distance from Tressa's home to school, we will divide the distance from Tressa's home to school by the distance from Anton's home to school

That is, 4/5 mile ÷ 3/5 mile

∴ we get

\frac{4}{5} \div \frac{3}{5}

= \frac{4}{5} \times \frac{5}{3}

= \frac{4}{3}

=1 \frac{1}{3}

Hence, the distance from Tressa's home to school is 1\frac{1}{3} times the distance from Anton's home to school

Learn more here: brainly.com/question/17420260

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Marat540 [252]

Answer:

x/2x+3 + 2x +3/x = 184 /65

x(2x + 3)–1 + (2x+3)x-1 = 184/65

(2x + 3)x / (2x +3)x = 184/65

2x² + 3x / 2x² + 3x = 184/65

1x= 184/64

x = 184/64

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I'll give you brainliest! Just need help with 5-10! (Least common factors)
umka2103 [35]

Answer:

  • 60
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Step-by-step explanation:

5.

12,\:30\\\\\mathrm{Prime\:factorization\:of\:}12:\quad 2\times\:2\times\:3\\\mathrm{Prime\:factorization\:of\:}30:\quad 2\times\:3\times\:5\\\\=2\times\:2\times\:3\times\:5\\\\=60

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6,\:10\\\\\mathrm{Prime\:factorization\:of\:}6:\quad 2\times\:3\\\mathrm{Prime\:factorization\:of\:}10:\quad 2\times \:5\\\\=2\times \:3\times \:5\\\\=30

7.

16,\:24\\\\\mathrm{Prime\:factorization\:of\:}16:\quad 2\times \:2\times\:2\times\:2\\\mathrm{Prime\:factorization\:of\:}24:\quad 2\times\:2\times\:2\times\:3\\\\=2\times\:2\times\:2\times\:2\times\:3\\\\=48

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14,21\\\\\mathrm{Prime\:factorization\:of\:}14:\quad 2\times\:7\\\mathrm{Prime\:factorization\:of\:}21:\quad 3\times\:7\\\\=2\times\:7\times\:3\\\\=42

9.

9,15\\\\\mathrm{Prime\:factorization\:of\:}9:\quad 3\times\:3\\\mathrm{Prime\:factorization\:of\:}15:\quad 3\times\:5\\\\=3\times\:3\times\:5\\\\=45

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5,\:11\\\\\mathrm{Prime\:factorization\:of\:}5:\quad 5\\\mathrm{Prime\:factorization\:of\:}11:\quad 11\\\\=5\times \:11\\\\=55

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2 years ago
Solve the following equation simultaneously 1/x-5/y=7, 2/x+1/y=3​
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Using the second equation, we can find x:

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The solution is (x, y) = (1/2, -1).

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<em>Additional comment</em>

If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.

A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).

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