Question of Exercise 2

# Question Expand : (i) (2a + 3b + 4c)2 (ii) (a + 2b – 5c)2 (iii) (3x + 2y - z)2 (iv) (3x – 2y – 1)2 (v) (vi)

Write the coefficients of x^2 in each of the following

(i) 2 + x^2 + x

(ii) 2 - x^2 + x^3

(iii)  pi/2x^2 + x

(iv) √(2)x - 1

Solution:

Explanation:

A coefficient is a number or quantity that is associated with a variable. It's commonly an integer multiplied by the variable immediately adjacent to it.

(i) 2+x2+x

The x2 coefficient is 1.

(ii) 2-x2+x3

The x2 coefficient is -1.

(iii) (π/2)x2+x

The x2 coefficient is π /2.

(iv)√2x-1

In the provided formulation, there is no x2 term. It can be rewritten as 0x2 +√ 2x - 1

The coefficient of x2 equals 0 because x2 does not exist.

(i) 2+x2+x , the coefficient of  x2 is 1.

(ii) 2-x2+x3, the coefficient of   x2  is -1.

(iii) (π/2)x2+x, here the coefficient of   x2  is  π/2.

(iv) √2x-1, the coefficient of   x is  0.

Which of the following is are correct

A:6 ÷ 3 =2

B:3 ÷ 6 =1/2

C:6 ÷ 3 ≠ 3 ÷ 6

D: None of the above

Solution:

Explanation

How many faces and edges does a triangular prism have

Solution:

Explanation:

• The triangular prism's sides and bases are either congruent or oblique.
• The prism's edges connect to the appropriate sides.
• The two bases of this prism are equilateral triangles, and their edges are parallel to one another.
• To grasp the structure, look at the diagram below.

• It contains 9 edges, 5 faces, and 6 vertices in total (which are joined by the rectangular faces).
• It features three rectangular sides and two triangular bases.
• The triangular prism is considered to be semiregular if the triangular bases are equilateral and the other faces are squares rather than rectangles.

5 faces and 9 edges.

Find the cube root of 125

Solution:

Explanation:

• A number's cube root is a number that, when multiplied three times, returns the original number 125.
• The cube root of 125 is represented as ∛125 using the 3rd root sign.
• Let P be an integer such that its cube is 125, P3 = 125 or .
• Assume that P = 1,2,3... and that its cube equals 125.

1 x 1 x 1 = 1

2 x 2 x 2 = 8

3 x 3 x 3 = 27

4 x 4 x 4 = 64

5 x 5 x 5 = 125

• As a result, ∛125 = ∛5x5x5=5 .

Hence, the cube root of 125 is 5 .

Find the square root of 5

Solution:

From the question, we have to find the square of .

So,