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andreev551 [17]
3 years ago
12

Identify any solutions to the system given below.

Mathematics
2 answers:
fiasKO [112]3 years ago
3 0
Answer:
• (6, -7)
• (-4, 13)

Explanation:

• Option One

2x + y = 5
2(6) + (-7) = 5
12 - 7 = 5
5 = 5 {true}

3y = 15 - 6x
3(-7) = 15 - 6(6)
-21 = 15 - 36
-21 = -21 {true}

• Option Two

2x + y = 5
2(2) + 1 = 5
4 + 1 = 5
5 = 5 {true}

3y = 15 - 6x
3(1) = 15 - 6(2)
3 = 15 - 12
3 = 13 {not true}

• Option Three

2x + y = 5
2(-2) + (-9) = 5
-4 - 9 = 5
-13 = 5 {not true}

3y = 15 - 6x
3(-9) = 15 - 6(-2)
-27 = 15 + 12
-27 = 27 {not true}

• Option Four

2x + y = 5
2(-4) + 13 = 5
-8 + 13 = 5
5 = 5 {true}

3y = 15 - 6x
3(13) = 15 - 6(-4)
39 = 15 + 24
39 = 39 {true}
telo118 [61]3 years ago
3 0

Answer:

(6, -7)

(2, 1)

(-4, 13)

Step-by-step explanation:

2х + y = 5  ⇒ y= 5-2x

Зу = 15 – 6х ⇒ y= 5-2x

Both equations are same so any point satisfying one of them is the solution.

(6, -7)

  • -7= 5-12 - yes

(2, 1)

  • 1= 5- 4 - yes

(-2, -9)

  • -9= 5 + 4 - no

(-4, 13)

  • 13= 5+8- yes

So 3 of given points satisfy the equation

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Answer:

\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)}  \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

Step-by-step explanation:

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                    \binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756

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