Mathematics Questions and Answers

\(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\)

\(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\)

\(2(2 - 12) + 5(2x) + 3(3x) = 132\)

\(-20 + 10x + 9x = 132\)

\(19x = 152\)

\(x = 8\)

When you divide a polynomial p(x) by (x - a), the remainder = p(a)

i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2).

= 2(2)\(^2\) + 2 - 3

= 8 + 2 - 3

= 7.

Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder.

i.e In this case, the polynomial = (x\(^2\) - x - 5)(3x - 1) + 7.

= (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7

= (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7

= 3x\(^3\) - 4x\(^2\) - 14x + 12

Explanation

\(x^2 - 7x + 10 \leq 0\)

Solve for \(x^2 - 7x + 10 = 0\)

We have, (x - 5)(x - 2) \(\leq\) 0.

Conditions:

Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0.

\(\implies\) x \(\leq\) 5; x \(\geq\) 2.

2 \(\leq\) x \(\leq\) 5.

Choosing x = 3,

3\(^2\) - 7(3) + 10 = 9 - 21 + 10

= -2 \(\leq\) 0.

\(\therefore\) 2 \(\leq\) x \(\leq\) 5.

\(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\)

\(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\)

\(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\)

\(5(2x - 4) \geq 3(4x - 6)\)

\(10x - 20 \geq 12x - 18\)

\(10x - 12x \geq -18 + 20\)

\(-2x \geq 2\)

\(x \leq -1\)