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4 years ago

Which of the following ordered pairs represents the solution to the system of equations below?

1 answer:

5 0

<span>-3x + 2y = 2 eq (1)

7x - 3y = 7 eq (2)

Mutiply eq (1) by 3 and eq (2) by 2:

-9x + 6y = 6

14x - 6y = 14

Add th two new equations:

-9x + 14x = 6 + 14

=> 5x = 20

=> x = 20/5

=> x = 4

Now from eq (1)

-3(4) + 2y = 2

=> 2y = 2 + 12

=> 2y = 14

=> y = 14 / 2

=> y = 7

So the ordered pair is (4,7)

Answer: option B. (4,7)
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$\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)$

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You can extract two balls of the same colour in two different way: either you pick two black balls or two red balls. Let's write the probabilities of each pick in each case.

Case 1: two black balls

The probability of picking the first black ball is 2/5, because there are two black balls, and 5 balls in total in the urn.

The probability of picking the second black ball is 1/4, because there is one black ball remaining in the urn, and 4 balls in total (we just picked the other black one!)

So, the probability of picking two black balls is

$P(\text{two blacks}) = \dfrac{2}{5} \cdot \dfrac{1}{4} = \dfrac{2}{20} = \dfrac{1}{10}$

Case 2: two red balls

The probability of picking the first black ball is 3/5, because there are three red balls, and 5 balls in total in the urn.

The probability of picking the second red ball is 2/4=1/2, because there are two red balls remaining in the urn, and 4 balls in total (we just picked the other red one!)

So, the probability of picking two red balls is

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Finally, the probability of picking two balls of the same colour is

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