Further Mathematics Questions and Answers

\(\frac{3}{3 - √3}\)

Rationalize

\(= \frac{3}{3 - √6} \times \frac{3 + √6}{3 + √6}\)

\(=\frac{3(3 + √6)}{(3 - √6)(3 + √6)}\)

\(=\frac{3(3 + √6)}{9 - 6}=\frac{3(3 + √6)}{3}\)

∴ 3 + √6

\(y = \frac{1} {(1 - x^2)^5} = (1-x^2)^{-5}\)

Let u = \(1 - x^2; y = u ^{-5}\)

\(\frac{du}{ dx}=-2x;\frac{dy}{ du}=-5u^{-6}\)

Using chain rule:

\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{ dx}\)

\(\frac{dy}{dx}=-2x\times-5u^{-6}\)

\(\frac{dy}{dx}=-2x\times-5(1 - x^2)^{-6}\)

\(\therefore\frac{dy}{dx} = 10x(1-x^2)^{-6}=\frac{10x}{(1-x^2)^6}\)

The statement "y ⇒ x" represents a logical implication, where y implies x. This means that if y is true, then x must also be true. In other words, if the condition represented by y is satisfied, the condition represented by x must also be satisfied.

Out of the provided options:

"Emmanuella did not go to school because the school bus arrived late."
In this case, "Emmanuella did not go to school" (not y) is the outcome of the late school bus (x). However, the implication y ⇒ x would mean that if the student walked down to school (y), then the school bus arrived late (x). So this case does not match the y ⇒ x structure.

"The school bus arrived early and Kate ran to school" does not match the y ⇒ x structure, as there's no clear implication between the two events.

"Mary walked to school because the school bus arrived late" can be represented by y ⇒ x because if Mary walked to school (y), it implies that the school bus arrived late (x).

"Either the school bus arrived late or Maryam walked to school" does not match the y ⇒ x structure, as it represents a logical OR relationship rather than an implication.

So, the correct statement that can be represented by y ⇒ x is:

Mary walked to school because the school bus arrived late

Explanation

P = {x : 2 ≤ x ≤ 8} = {2, 3, 4, 5, 6, 7, 8}

Q = {x : 4 < x ≤ 12} = {5, 6, 7, 8, 9, 10, 11, 12}

Q1 = R - {5, 6, 7, 8, 9, 10, 11, 12} = {..., 2, 3, 4, 13, 14, 15, ...}

∴ P ⋂ Q\(^1\) = {2, 3, 4} = {x : 2 ≤ x ≤ 4}

rth term of a binomial expansion = \(^nC_r - _1a^{n - (r - 1)} b^{r - 1}\)

n = 7, r = 5 ∴ r - 1 = 4

5th term = \(^7C_4 q^{7 - 4} x^4\)

= \(^7C_4 q^3x^4\)

\(\therefore 35q^3x^4\)