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Let after t hours the distances D_{1} traveled by car A

=> D_{1} = 30 t

Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it.

After (t - 1) hours, distance D_{2} traveled by car B

=> D_{2} = 40 (t - 1)

When car B passes car A, they are at the same distance from the starting point and therefore** D**_{1} = D_{2}

=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=>** Car B passes car A at = 9 + 4 = 13 pm.**

=> D

Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it.

After (t - 1) hours, distance D

=> D

When car B passes car A, they are at the same distance from the starting point and therefore

=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=>

First integer = x

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

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(D

Given differential equation, (D^{4 }- 5D^{3} + 5D^{2} + 5D - 6)y = 0

=> For general solution of equation,

Solve D^{4 }- 5D^{3} + 5D^{2} + 5D - 6 = 0

=> D^{4 }- 5D^{3} + 6D^{2} - D^{2 }+ 5D - 6 = 0

=> D^{2} (D^{2} - 5D + 6) - (D^{2} - 5D + 6) = 0

=> (D^{2} - 5D + 6)(D^{2} - 1) = 0 ................................(1)

Now

D^{2 }- 1 = (D - 1)(D + 1) and

Factors of D^{2} - 5D + 6

D^{2} - 5D + 6 = D^{2} - 2D - 3D + 6

= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D^{2} - 5D + 6)(D^{2} - 1) = (D - 3)(D - 2)(D - 1)(D + 1) = 0

=> D = 3, 2, 1, -1 or D = -1, 1,, 2, 3

=> General solution of differential equation is,

=>** y = C**_{1} e^{-x} + C_{2} e^{x} + C_{3} e^{2x} + C_{4} e^{3x} .

=> For general solution of equation,

Solve D

=> D

=> D

=> (D

Now

D

Factors of D

D

= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D

=> D = 3, 2, 1, -1 or D = -1, 1,, 2, 3

=> General solution of differential equation is,

=>

Given progression, 2, 5, 8, 11,......... is a arithmetic progression

Here First term (a) = 2

and Common difference (d) = 5 - 2 = 3

We know, n^{th} term can be represented as, a_{n} = a + (n - 1)d

and sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a_{20} = a + 19 d

= 2 + 19 * 3

= 2 + 57

= 59

=>** 20th term = 59**

Sum of first 14 terms = $\frac{14}{2}${2 * 2 + (14 - 1) * 3}

= 7{4 + 39}

= 7 * 43

= 301

=>**Sum of first 14 terms = 301.**

Here First term (a) = 2

and Common difference (d) = 5 - 2 = 3

We know, n

and sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a

= 2 + 19 * 3

= 2 + 57

= 59

=>

Sum of first 14 terms = $\frac{14}{2}${2 * 2 + (14 - 1) * 3}

= 7{4 + 39}

= 7 * 43

= 301

=>