# Test: Algebra III - Ambitious

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Question 1:   Simplify: $4x÷2\sqrt{x}$ , $x>0$
$\frac{1}{2}\sqrt{x}$
$2\sqrt{x}$
$\frac{1}{2}x\sqrt{2}$
$x\sqrt{2}$
Question 2:   Solve: $\frac{x}{7}-5=2$
$x=7$
$x=49$
$x=21$
$x=14$
Question 3:   Multiply out: $\left({t}^{2}-3k+3\right)\left(k+9\right)$
${t}^{2}k-9{t}^{2}+{k}^{2}-24k+27$
${t}^{2}k+9{t}^{2}-3{k}^{2}-24k+27$
${t}^{2}k+6{t}^{2}-3{k}^{2}+12k+27$
${t}^{2}k-9{t}^{2}+{k}^{2}-16k+27$
Question 4:   Expand: $\left(4xy-6\right)\left(4-6xy\right)$
$-16{x}^{2}{y}^{2}-38xy+24$
$-16{x}^{2}{y}^{2}+26xy+10$
$-24{x}^{2}{y}^{2}+52xy-24$
$-24{x}^{2}{y}^{2}+26xy-10$
Question 5:   Expand: $\left(3{x}^{4}+y\right)\left({x}^{4}-2y\right)$
$3{x}^{8}-7{x}^{4}y+2{y}^{2}$
$4{x}^{8}-7{x}^{4}y-2{y}^{2}$
$3{x}^{8}-5{x}^{4}y-2{y}^{2}$
$4{x}^{8}-y$
Question 6:   Expand: $\left(x\sqrt{3}+1\right)\left(3\sqrt{3}-1\right)$
$6x-x\sqrt{3}-2\sqrt{3}$
$9x-x\sqrt{3}+3\sqrt{3}-1$
$3\sqrt{3}$
$-2x\sqrt{3}+3\sqrt{3}$
Question 7:   Expand: $\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{3}+\sqrt{6}\right)$
$\sqrt{6}+2\sqrt{3}-3\sqrt{2}-3$
$\sqrt{6}-\sqrt{2}-3$
$2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-3$
$\sqrt{6}-3\sqrt{2}+3$
Question 8:   Expand: $\left(a-1\right)\left(a+2\right)+\left(a-3\right)\left(a+2\right)$
$2{a}^{2}-6a+4$
${a}^{2}-8a-6$
$2{a}^{2}-5a-6$
$2{a}^{2}-8$
Question 9:   Expand: ${\left(x+y\right)}^{2}-\left({x}^{2}+{y}^{2}\right)$
$2{x}^{2}+2xy$
$2{x}^{2}+2{y}^{2}-2xy$
$2xy$
$2{y}^{2}+2xy$
Question 10:   Formula for the area of a trapezium is: $A=\frac{1}{2}\left(a+b\right)\cdot h$ Rearrange the formula to make $h$ the subject
$h=\frac{1}{2}\cdot \frac{A}{a+b}$
$h=\frac{a+b}{2A}$
$h=2A\left(a+b\right)$
$h=\frac{2A}{a+b}$