Given,

Area of rectangle = 25x2 – 35x + 12

We know, area of rectangle = length × breadth

So, by factoring 25x2 – 35x + 12, the length and breadth can be obtained.

25x2 – 35x + 12 = 25x2 – 15x – 20x + 12

=> 25x2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x2 – 35x + 12 = (5x – 3)(5x – 4)

So, the length and breadth are (5x – 3)(5x – 4).

Now, perimeter = 2(length + breadth)

So, perimeter of the rectangle = 2[(5x – 3)+(5x – 4)]

= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14

So, the perimeter = 20x – 14

Let the polynomial be f(x) = 5x – 4x2 + 3

Now, for x = 2,

f(2) = 5(2) – 4(2)2 + 3

=> f(2) = 10 – 16 + 3 = –3

Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1)2 + 3

=> f(–1) = –5 –4 + 3 = -6

The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Consider the equation 3x + 2y = 12

Now, square both sides:

(3x + 2y)2 = 122

=> 9x2 + 12xy + 4y2 = 144

=>9x2 + 4y2 = 144 – 12xy

From the questions, xy = 6

So,

9x2 + 4y2 = 144 – 72

Thus, the value of 9x2 + 4y2 = 72

Solution:

The given two rational numbers are 5/7 and 9/11.

5/7 = 0.714285714…..

9/11 = 0.81818181……

Hence, the three irrational numbers between 5/7 and 9/11 can be:

0.720720072000…

0.730730073000…

0.808008000…

Let x = 0.3333…. 

Multiply with 10,

10x = 3.3333…

Now, 3.3333… = 3 + x (as we assumed x = 0.3333…)

Thus, 10x = 3 + x

10x – x = 3

9x = 3

x = 1/3

Therefore, 0.3333… = 1/3. Here, 1/3 is in the form of p/q and q ? 0.

No, since the square root of a positive integer 16 is equal to 4. Here, 4 is a rational number.

We have to find five rational numbers between 3/5 and 4/5.

So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1)

Now,

3/5 = (3/5) × (6/6) = 18/30

4/5 = (4/5) × (6/6) = 24/30

Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30

Solution:

Solution: