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aksik [14]
2 years ago
15

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Mathematics
1 answer:
Kaylis [27]2 years ago
3 0

Answer:

Adults $ 17

Students: $8

Step-by-step explanation:

Adults: x

Students: y

a) 25x + 31y = 673

b) 27x +31y = 707

Elimination:

a) -1(25x + 31y = 673) = -25 -31y = -673

a + b

= (-25 -31y = -673) + (27x + 31y = 707)

= 2x = 34

x = 34/2

x = 17

Substitution a or b:

a) 25(17) + 31y = 673

425 + 31y = 673

31y = 673 - 425

31y = 248

y = 248/31

y = 8

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Answer:

12

Step-by-step explanation:

For the top triangle, he has 3 options

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2C1 = 2

For the ones on sides of the middle one, 2 options each

2C1 × 2C1 = 4

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These include all three outer ones same, so subtract 3

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24 - 3 - 9 = 12

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3 years ago
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If a+b+c=0 for a,b,c then prove a²/bc+b²/ca+c²/ab=3.
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According to the identity if a+b+c=0
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3 years ago
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Step-by-step explanation:

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andrew-mc [135]
\dfrac{ \dfrac{x+3}{4x^2-16} }{ \dfrac{2x^2+10x+12}{2x-4} }

-----------------------------------------------------------------------------------
Write the divide fraction horizontally:
-----------------------------------------------------------------------------------

= \dfrac{x+3}{4x^2-16} \div \dfrac{2x^2+10x+12}{2x-4}

-----------------------------------------------------------------------------------
Factorise the numerators and denominators when possible:
-----------------------------------------------------------------------------------

= \dfrac{x+3}{4(x+2)(x-2)} \div \dfrac{ 2(x + 3) (x + 2)}{2(x-2)}

-----------------------------------------------------------------------------------
Convert the divide fraction to multiplication fraction
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= \dfrac{x+3}{4(x+2)(x-2)} \times \dfrac{2(x-2)}{2(x+3)(x+2)}

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Cancel the factors
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= \dfrac{1}{4(x+2)} \times \dfrac{1}{(x+2)}

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Combine to single fraction
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= \dfrac{1}{4(x+2)^2}



4 0
3 years ago
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