Mathematics Questions and Answers

m² + n² = 29 .......(1)
m + n = 7 ............(2)

From (2),
m = 7 - n
but m² + n² = 29, substituting;
(7-n)² + n² = 29
49 - 14n + n² + n² = 29
=> 2n² -14n + 20 = 0
Thus n² -7n + 10 = 0

Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.

When n = 5,
m + n = 7, => m = 2,

When n = 2,
m + n = 7, => m = 5.

Thus (m,n) = (5,2) and (2,5)

Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.

\(y = px^{2} + q ... (i)\)

\(y = 2x^{2} - 1 ... (ii)\)

At x = 2, 

(i): \(y = p(2^{2}) + q = 4p + q\)

(ii): \(y = 2(2^{2}) - 1 = 7\)

\(\therefore \text{The coordinates of the point of intersection = (2, 7)}\)

(i): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)

\(\frac{a^{3x} - 26a^{2x} + 156a^{x} - 216}{a^{2x} - 24a^{x} + 108}\)

Let \(a^{x} = z\)

\(\therefore = \frac{z^{3} - 26z^{2} + 156z - 216}{z^{2} - 24y + 108} ... (i)\)

Dividing (i) above, we get \(z - 2\)

= \(a^{x} - 2\)

\(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\)

= \(0.14 \times 0.275\)

= \(0.0385 \approxeq 0.039\)

Note: Embedded zeroes are counted just the same as non-zero digits and For the number of decimal places stated, count that number of digits to the right of the decimal and underline it.

The next number to its right is called the 'rounder decider'. If the 'rounder decider' is 5 or more, then round the previous digit up by 1