Math, asked by azadgond4560, 2 days ago

for what value of k will the pain of lineal equation have infinite mang solutions. kx+3y=k-3 12x+ky=k​

Answers

Answered by anupamsgpgi
0

Answer:

Total no. of coins in the piggy is 180

Solution(i):

No. of 50p coins in piggy is 100

Let E be the event of drawing a coin from piggy

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

180

100

=

9

5

Therefore,Probability that a 50p coin wins =

9

5

Solution(ii):

No. of coins with value >Rs. 1 in piggy is 30

Let E be the event of drawing a coin from piggy

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

180

30

=

6

1

Therefore,Probability that a coin with value >Rs.1 falls =

6

1

Solution(iii):

No. of coins with value <Rs 5 in piggy is 170

Let E be the event of drawing a coin from piggy

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

180

170

=

18

17

Therefore,Probability that a coin with value <Rs.5 falls =

18

17

Solution(iv):

No. of Rs. 1 and Rs. 2 coins in piggy is 70

Let E be the event of drawing a coin from piggy

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

180

70

=

18

7

Therefore,Probability that a Rs. 1 and Rs. 2 coin falls =

18

7

Answered by shxb123457964wuzbz
0

Answer:

Given

kx+3y−(k−3)=0

Comparing with a

1

x+b

1

y+c1=0

∴a

1

=k, b

1

=3, c=−(k−3)

12x+ky−k=0

Comparing with a

1

x+b

1

y+c1=0

∴a

1

=12, b

1

=k, c=−k

Since equation has infinite number of solutions

So,

a

2

a

1

=

b

2

b

1

=

c

2

c

1

12

k

=

k

3

=

k

k−3

12

k

=

k

3

k

2

=12×3

k

2

=36

k=±6

k

3

=

k

k−3

3k=k(k−3)

3k=k

2

−3k

k

2

−3k−3k=0

k

2

−6k=0

k(k−6)=0

k=0,6

Therefore, k=6 satisfies both equations

Hence, k=6

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