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baherus [9]
3 years ago
11

Solve the system of equations -5y+4x= 49 7y+2x= -23 Find X and Y Will mark Brainlest

Mathematics
1 answer:
grigory [225]3 years ago
8 0

<em>Answer:</em>

<em>y = - 5 ; x = 6</em>

<em>Step-by-step explanation:</em>

<em>-5y+4x= 49</em>

<em>7y+2x= -23 | × - 2</em>

<em />

<em>-5y+4x= 49</em>

<em>-14y - 4x = 46</em>

<em>___________ +</em>

<em>- 5y - 14y + 4x - 4x = 49 + 46</em>

<em>- 19y = 95 | × - 1</em>

<em>19y = - 95</em>

<em>y = - 95 : 19</em>

<em>y = - 5</em>

<em />

<em>- 5(-5) + 4x = 49</em>

<em>25 + 4x = 49</em>

<em>4x = 49 - 25</em>

<em>4x = 24</em>

<em>x = 24 : 4</em>

<em>x = 6</em>

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Answer: The correct number of balls is (b) 4.

Step-by-step explanation:  Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

^{10}C_r=210.

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10

If r = 2, then

^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45

If r = 3, then

^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120

If r = 4, then

^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

Thus, (b) is the correct option.

4 0
3 years ago
Read 2 more answers
Tim said that to solve the equation x - 2 = 9, he would
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