A Algebra — coverage pack

Answer

• $m^2n^3$

Method: There are two factors of $m$ and three factors of $n$, so $m\times m=m^2$ and $n\times n\times n=n^3$. Therefore the product is $m^2n^3$.

Answer

• $\frac{p^3q}{5}$
• Coefficient: $\frac15$

Method: The three factors of $p$ give $p^3$, and division by $5$ is shown by a fraction bar. Hence the expression is $p^3q/5=\frac15p^3q$, so its coefficient is $\frac15$.

Answer

• Area: $\frac{3xy}{2}$
• Perimeter: $3x+2y$

Method: Area is length multiplied by width, so $A=(3x/2)y=3xy/2$. Perimeter is twice the length plus twice the width: $P=2(3x/2)+2y=3x+2y$.

Answer

• $13$

Method: Substitute using brackets: $2(-3)^2-5=2\times9-5=18-5=13$.

Answer

• $40\,\text{J}$

Method: Substitute both values: $E=\frac12\times3.2\times5^2$. Since $5^2=25$, $E=0.5\times3.2\times25=40\,\text{J}$.

Answer

• $3.78\times10^9\,\text{J}$

Method: Substitute before evaluating: $E=(4.2\times10^{-8})(3\times10^8)^2$. Squaring gives $9\times10^{16}$, so $E=4.2\times9\times10^8=37.8\times10^8=3.78\times10^9\,\text{J}$.

Answer

• Expression

Method: The algebra has no equality or inequality sign, so it is an expression.

Answer

• Two terms
• Factors $4x$ and $(2x-3)$

Method: The addition or subtraction signs separate $8x^2$ and $-12x$, so there are two terms. On the right, $4x$ is multiplied by the bracket $(2x-3)$, so these are the two factors.

Answer

• $3(2x-1)=9$ is an equation
• $3(2x-1)\equiv6x-3$ is an identity
• $3(2x-1)<9$ is an inequality

Method: The first statement is true only for a particular value of $x$, so it is an equation. Expanding the second gives $6x-3$ for every $x$, so it is an identity. The final statement compares two quantities using $<$, so it is an inequality.

Answer

• $5a+8b$

Method: Collect the like $a$-terms and the like $b$-terms separately: $(7a-2a)+(3b+5b)=5a+8b$.

Answer

• $(3x+2)(2x-1)$

Method: The product of the leading and constant coefficients is $6\times(-2)=-12$. Split the middle term using $4x-3x$: $6x^2+4x-3x-2=2x(3x+2)-(3x+2)=(3x+2)(2x-1)$.

Answer

• $\frac{x-3}{x-2}$
• $x\ne-3$ and $x\ne2$

Method: Factorise both parts: $x^2-9=(x-3)(x+3)$ and $x^2+x-6=(x+3)(x-2)$. Cancelling the common factor gives $(x-3)/(x-2)$. The original denominator is zero at $x=-3$ or $x=2$, so both values remain excluded.

Answer

• $w=\frac{A}{l}$

Method: Divide both sides by $l$ to isolate $w$: $A/l=w$, so $w=A/l$.

Answer

• $t=\frac{v-u}{a}$

Method: Subtract $u$ from both sides to obtain $v-u=at$. Divide both sides by $a$, giving $t=(v-u)/a$.

Answer

• $a=\frac{b(P+1)}{P-1}$

Method: Multiply by $a-b$: $P(a-b)=a+b$. Expanding gives $Pa-Pb=a+b$. Collect the $a$-terms: $Pa-a=Pb+b$, so $a(P-1)=b(P+1)$. Dividing by $P-1$ gives $a=b(P+1)/(P-1)$.

Answer

• $4(n+2)-3(n-1)\equiv n+11$

Method: Expand both brackets, remembering that the subtraction acts on both terms: $4n+8-3n+3=n+11$. Therefore the two expressions are equivalent.

Answer

• $n+(n+1)=2n+1$, so the sum is odd

Method: Let the first integer be $n$, so the next is $n+1$. Their sum is $n+(n+1)=2n+1$. Since $2n$ is even for every integer $n$, $2n+1$ is odd, proving the claim.

Answer

• $(2n+1)^2=8k+1$ for an integer $k$

Method: Write an odd integer as $2n+1$. Then $(2n+1)^2=4n^2+4n+1=4n(n+1)+1$. One of the consecutive integers $n$ and $n+1$ is even, so $n(n+1)=2k$ for some integer $k$. Therefore the square is $4(2k)+1=8k+1$.

Answer

• $f(6)=14$

Method: Substitute $x=6$: $f(6)=3(6)-4=18-4=14$.

Answer

• $f^{-1}(x)=\frac{x-2}{5}$
• $f^{-1}(27)=5$

Method: Write $y=5x+2$ and rearrange: $x=(y-2)/5$. Hence $f^{-1}(x)=(x-2)/5$. Substituting $27$ gives $f^{-1}(27)=(27-2)/5=5$.

Answer

• $x=-3$ or $x=3$

Method: Form the composite by substituting $g(x)$ into $f$: $fg(x)=2(x^2-1)+3=2x^2+1$. Hence $2x^2+1=19$, so $2x^2=18$ and $x^2=9$. Therefore $x=-3$ or $x=3$.

Answer

• Quadrant II

Method: The $x$-coordinate is negative and the $y$-coordinate is positive, which places the point in quadrant II.

Answer

• $(-1,-1)$

Method: Average the $x$-coordinates and the $y$-coordinates: $((-5+3)/2,(2-4)/2)=(-2/2,-2/2)=(-1,-1)$.

Answer

• $P=(0,-1)$

Method: The vector from $A$ to $B$ is $(10,-10)$. Since $AP$ is $3/5$ of the whole segment, $AP=(6,-6)$. Adding this to $A$ gives $P=(-6,5)+(6,-6)=(0,-1)$.

Answer

• $y=4x-7$

Method: In $y=mx+c$, use $m=4$ and $c=-7$. This gives $y=4x-7$.

Answer

• $y=-x+3$

Method: The gradient is $m=(-1-5)/(4-(-2))=-6/6=-1$. Using $(-2,5)$ in $y=-x+c$ gives $5=2+c$, so $c=3$. Therefore the equation is $y=-x+3$.

Answer

• $y=-\frac32x-\frac12$

Method: Rearrange the given line: $-3y=6-2x$, so $y=(2/3)x-2$ and its gradient is $2/3$. A perpendicular line has gradient $-3/2$. Write $y=-(3/2)x+c$ and substitute $(3,-5)$: $-5=-9/2+c$, so $c=-1/2$. Hence $y=-(3/2)x-1/2$.

Answer

• Gradient $-3$
• $y$-intercept $8$

Method: Compare with $y=mx+c$. The coefficient of $x$ is $m=-3$, and the constant is $c=8$.

Answer

• Gradient $-2$
• $y$-intercept $12$
• $x$-intercept $6$

Method: Using the two points, $m=(0-12)/(6-0)=-12/6=-2$. The point with $x=0$ gives the $y$-intercept $12$, and the point with $y=0$ gives the $x$-intercept $6$.

Answer

• Gradient $-6$ litres per minute
• $V$-intercept $120$ litres, the initial volume
• $V=120-6t$

Method: The gradient is $(72-120)/(8-0)=-48/8=-6$, meaning the volume decreases by $6$ litres each minute. At $t=0$, $V=120$, so the initial volume is $120$ litres. Therefore $V=-6t+120$.

Answer

• Roots $x=2$ and $x=3$
• $y$-intercept $6$

Method: Factorise: $x^2-5x+6=(x-2)(x-3)$, so $y=0$ at $x=2$ and $x=3$. Setting $x=0$ gives $y=6$, so the $y$-intercept is $6$.

Answer

• Turning point $(4,-9)$
• Axis $x=4$
• Roots $x=1$ and $x=7$

Method: The completed-square form gives turning point $(4,-9)$ and axis $x=4$. For the roots, set $y=0$: $(x-4)^2=9$, so $x-4=\pm3$ and $x=1$ or $x=7$.

Answer

• $y=2(x+2)^2-18$
• Turning point $(-2,-18)$
• Roots $x=-5$ and $x=1$

Method: Factor $2$ from the quadratic and linear terms: $y=2(x^2+4x)-10=2((x+2)^2-4)-10=2(x+2)^2-18$. The turning point is therefore $(-2,-18)$. Setting $y=0$ gives $(x+2)^2=9$, so $x+2=\pm3$ and the roots are $-5$ and $1$.

Answer

• Exponential

Method: Equal increases in $x$ multiply the output by a constant factor, so $y=2^x$ is an exponential function.

Answer

• Asymptotes $x=0$ and $y=0$
• Branches in quadrants I and III

Method: The expression is undefined at $x=0$, giving vertical asymptote $x=0$. As $|x|$ grows, $6/x$ approaches $0$, giving horizontal asymptote $y=0$. Since $6/x$ has the same sign as $x$, the branches lie in quadrants I and III.

Answer

• $x$-intercepts at $x=-90^\circ,90^\circ,270^\circ$
• Maxima $(0^\circ,1)$ and $(360^\circ,1)$
• Minima $(-180^\circ,-1)$ and $(180^\circ,-1)$

Method: Cosine is zero at odd multiples of $90^\circ$, giving $-90^\circ$, $90^\circ$ and $270^\circ$ in the interval. It reaches $1$ at multiples of $360^\circ$, here $0^\circ$ and $360^\circ$, and reaches $-1$ at odd multiples of $180^\circ$, here $-180^\circ$ and $180^\circ$.

Answer

• Translation by vector $\begin{pmatrix}0\\4\end{pmatrix}$

Method: Adding $4$ outside the function increases every $y$-coordinate by $4$ and leaves every $x$-coordinate unchanged. This is translation by vector $(0,4)$.

Answer

• $(-3,-2)$
• Reflection in the $y$-axis

Method: Replacing $x$ by $-x$ reverses every $x$-coordinate and leaves every $y$-coordinate unchanged. Thus $(3,-2)$ maps to $(-3,-2)$, a reflection in the $y$-axis.

Answer

• Corresponding point $(3,2)$
• Reflect in the $y$-axis, translate $2$ units right, then translate $3$ units down

Method: If $(a,b)$ lies on $y=f(x)$, then $f(2-x)=b$ when $2-x=a$, so the new $x$-coordinate is $2-a$ and the new $y$-coordinate is $b-3$. With $(a,b)=(-1,5)$ this gives $(3,2)$. Since $f(2-x)=f(-(x-2))$, the graph is reflected in the $y$-axis, moved $2$ units right, and then moved $3$ units down.

Answer

• $(16,1.5)$
• At an average speed of $16$ km/h, the journey takes $1.5$ hours.

Method: Substitute $v=16$: $t=24/16=1.5$. The coordinates are speed then time, so the graph contains $(16,1.5)$ and this represents a $1.5$-hour journey at $16$ km/h.

Answer

• $x\approx2.2$

Method: Draw the increasing exponential curve $y=2^x$ and the decreasing straight line $y=9-2x$ on the same axes. Their intersection has $x$-coordinate about $2.2$ (the numerical value is about $2.201$), so the graphical estimate to one decimal place is $2.2$.

Answer

• $t\approx4.5$ seconds

Method: Plot the quadratic $d=3t^2+2$ and the line $d=14t$. Equal distances occur at intersections. The later intersection has $t\approx4.52$, so an appropriate graph gives about $4.5$ seconds; the earlier intersection near $0.15$ seconds is not requested.