Question

solve 6,7,8 thanks you
...

Answer 1

ques5.)Let the speed of the car be C kmph
Let the speed of the train be T kmph

4 hours = 250 km / T kmph + (370 - 250) km / C kmph

         4 = 250/T + 120/C  -- equation 1

4 hrs 18 minutes = 130 km / T kmph + (370 - 130)km / C kmph
       258/60 = 4.3 = 130 / T + 240 / C      --- equation 2

Multiply equation 1 by 2 and subtract equation 2 from it.

       8 - 4.3 = 500/T - 130/T + 240/C - 240/C

       3.7 = 370 / T

       T = 370/3.7 = 100 kmph

Substitute the value of T in equation 1 to get,

       4 = 250/T + 120/C  -- equation 1
       4 = 250/100 + 120/C
       4 - 2.5 = 120/C

       C = 120/1.5 = 80 kmph

The train runs at 100 kmph and the car runs at 80 kmph

Answer 2

Heya !!!

★ Question 6:-

Solution :- F(X) => KX²+4X+4

Here,

A = K , B = 4 and C = 4

Sum of zeroes = -B/A

Alpha + Beta = (-4/K)---------(1)

Product of zeroes = C/A

Alpha × Beta =( 4/K)---------(2)

(Alpha² + Beta²) = 24

(Alpha+Beta)² - 2 Alpha × Beta = 24

(-4/K)² - 2 × 4/K = 24

16/K² - 8/K = 24

16/K² = 24 + 8/K

16/K² = 24K + 8/K

16K/K² = 24K + 8

16/K = 24K + 8

24K² + 8K = 16

24K²+8K-16 = 0

24K² +24K - 16K -16 = 0

24K(K+1) - 16(K+1) = 0

(K+1) (24K-16) = 0

(K+1) = 0 OR (24K-16) = 0

K = -1 OR K = 16/24 = 8/12 = 4/6 = 2/3

Hence,

K = -1 Or 2/3

★Question 7 :-

Solution :- F(X) =3X⁴+6X³-2X²-10X-5

✓5/3 and -✓5/3 are the two zeroes of the polynomial P(X).

(X-✓5/3) ( X+✓5/3)

(X)²-(✓5/3)² => X²-5/3 = 3X²-5

(3X²-5) is also a factor of the polynomial P(X).

On dividing P(X) by G(X) we get,

3X²-5) 3X⁴+6X³-2X²-10X-5( X²+2X+1

******** 3X⁴ -5X²

____- _______+ _________
+6X³ + 3X² -10X -5

+6X³**** -10X

_-_______+__ ________
* 3X²****** -5

***** **+3X²**** -5

_____-______+______

We get,

Remainder = 0

And,

Quotient = X²+2X+1

Factories the Quotient then we will get two other zeroes of polynomial P(X).

=> X²+2X+1

=> X²+X+X+1

=> X(X+1) +1(X+1)

=> (X+1) (X+1) = 0

=> (X+1) = 0 OR (X+1) = 0

=> X = -1 OR X = -1

Hence,

-1 , ✓5/3 , -✓5/3 , -1 are the four zeroes of the polynomial 3X⁴+6X³-2X²-10X-5

★ Question 7:-

Solution:- F(X) => X³-3X²+X+1

A-B , A , A+B are the zeros of polynomial F(X).

=> X³-3X²+X+1

Here,

A = 1 , B = -3 , C= 1 and D = 1

Sum of zeroes = -B/A

A-B + A + A+B = -(-3)/1

3A = 3

A = 3/3

A = 1.

And,

Product of zeroes = C/A

(A-B) (A) (A+B) = 1/1

(1-B) (1) (1+B) = -1/1

1-B² = -1

B²= 2

B = +- ✓2

HOPE IT WILL HELP YOU...... :-)