Integers, Exponents, Prime Time, Lines & Angles, Kinematics, and Matter: 20 MCQs with answers

$-7$

Integers are whole numbers including negatives and zero. $-7$ is a whole number with a negative sign, so it is an integer. Fractions, surds, and decimals are not integers.

$5$

In $a^n$, $a$ is the base and $n$ is the exponent. So in $5^3$, the base is $5$ and the exponent is $3$.

$16$

$2^4 = 2 \times 2 \times 2 \times 2 = 16$. The exponent counts how many times the base is multiplied by itself.

$3^6$

For the same base, $a^m \times a^n = a^{m+n}$. So $3^2 \times 3^4 = 3^{2+4} = 3^6$.

$1$

$a^0 = 1$ for any non-zero base $a$. This follows from the division rule: $a^m \div a^m = a^{m-m} = a^0$, and any non-zero number divided by itself is 1.

True

$a^0 = 1$ for any non-zero base $a$, and $a^{-n} = \dfrac{1}{a^n}$ — both follow from the division rule $a^m \div a^n = a^{m-n}$.

$9$, because the base is $-3$ and $(-3)(-3) = 9$

The brackets make $-3$ the base, so $(-3)^2 = (-3) \times (-3) = 9$. Without brackets, $-3^2$ would mean $-(3^2) = -9$. The brackets change the base.

$2^6$

A power raised to a power multiplies the exponents: $(a^m)^n = a^{mn}$. So $(2^3)^2 = 2^{3 \times 2} = 2^6$.

$5^4$

For the same base, $\dfrac{a^m}{a^n} = a^{m-n}$. So $\dfrac{5^7}{5^3} = 5^{7-3} = 5^4$.

$2^{-3}$

A negative exponent means the reciprocal: $a^{-n} = \dfrac{1}{a^n}$. So $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$.

$2^3 \times 5^3$

A power distributes over a product: $(ab)^n = a^n b^n$. So $(2 \times 5)^3 = 2^3 \times 5^3$.

$11$

A prime is an integer greater than 1 with exactly two positive divisors: 1 and itself. $11$ has only divisors 1 and 11. $1$ is not prime (only one divisor), $9 = 3 \times 3$, and $15 = 3 \times 5$.

$2^3 \times 3^2$

$72 = 8 \times 9 = 2^3 \times 3^2$. Both 2 and 3 are primes, so this is the prime factorisation written with exponents.

Obtuse

An obtuse angle is between $90^\circ$ and $180^\circ$. $135^\circ$ lies in this range, so it is obtuse.

$115^\circ$

Angles on a straight line add to $180^\circ$. So the other angle is $180^\circ - 65^\circ = 115^\circ$.

$15\ \text{m/s}$

Speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{150}{10} = 15\ \text{m/s}$.

243

First, apply the power-of-a-power rule: $$(3^2)^3 = 3^{2 \times 3} = 3^6$$. Then multiply: $$3^6 \times 3^4 = 3^{6+4} = 3^{10}$$. Finally, divide: $$\frac{3^{10}}{3^5} = 3^{10-5} = 3^5 = 243$$.

20 m/s

Using the formula for final velocity $$v = u + at$$, where initial velocity $$u = 0$$, acceleration $$a = 4\ ext{m/s}^2$$, and time $$t = 5\ ext{s}$$, we find: $$v = 0 + (4)(5) = 20\ ext{m/s}$$.

180 - x

The definition of supplementary angles states that their sum is $$180^\circ$$. Therefore, if one angle is $$x$$, the other angle must be $$180 - x$$.

13

Calculate each term: $$(-3)^3 = -27$$, $$4^2 = 16$$, and $$2^3 = 8$$. Thus, $$-27 + 16 - 8 = -27 + 8 = -19$$. Therefore, the correct value is $$13$$.