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CTJan27 Online JMSS - Linear Graphs - Distance,Midpoint, Gradient Point Formula, Parallel Lines and Perpendicular Lines

Student Name

Student Email

Complete All The Questions

$5$ $2$ $5\sqrt{2x}$ $2x$

$81$ $54$ $144$ $121$

$4\sqrt{3}$ $2\sqrt{12}$ $16\sqrt{3}$ $6\sqrt{8}$

$10\sqrt{12.5}$ $5\sqrt{25}$ $5\sqrt{5}$ $25\sqrt{5}$

$-28\sqrt{7}$ $-14\sqrt{2}$ $-2\sqrt{49}$ $-49\sqrt{2}$

$x^2\sqrt{x^5}$ $x^{10}$ $\sqrt{x^{10}}$ $x^5$

$y^7$ $y^4\sqrt{y^3}$ $y^{3.5}$ $y^3\sqrt{y}$

$a^2b\sqrt{b}$ $a^4b\sqrt{b}$ $a^2\sqrt{b^3}$ $a^2b^2\sqrt{b}$

$25m\sqrt{3}$ $3m\sqrt{5}$ $5m\sqrt{3}$ $5\sqrt{3m^2}$

$5\sqrt{6}$ $12\sqrt{5}$ $6\sqrt{5}$ $3\sqrt{10}$

$3x^5\sqrt{2}$ $3x^2\sqrt{2x}$ $9x^2\sqrt{2x}$ $6x^2\sqrt{3x}$

$20x^2\sqrt{2x}$ $20x\sqrt{2x}$ $40x^2\sqrt{x}$ $4x\sqrt{8x^3}$

$-18y^2\sqrt{2y^3}$ $-12y^2\sqrt{6y}$ $-18y^3\sqrt{2y}$ $-216y^3\sqrt{y}$

$20a^4b^5\sqrt{b}$ $20a^3b^4\sqrt{ab}$ $20a^7b^{10}$ $10a^4b^5\sqrt{b}$

$12k^3\sqrt{2k}$ $12k^4\sqrt{2k}$ $24k^3\sqrt{2k}$ $4k^2\sqrt{18k}$

$-10\sqrt{15}$ $-5\sqrt{6}$ $-6\sqrt{5}$ $-25\sqrt{6}$

$x^4\sqrt{3}$ $\frac{3x^4\sqrt{3}}{3}$ $3x^4\sqrt{3}$ $x^4\sqrt{9}$

$-20x^3y^4\sqrt{5x}$ $-10x^4y^4\sqrt{2x}$ $-10x^5y^3\sqrt{2x}$ $-10x^5y^4\sqrt{2x}$

The student extracted $\sqrt{12}$ instead of $\sqrt{36}$. The student failed to simplify the radical completely. The student multiplied the coefficient by the radicand. The student used 36 as the factor but incorrectly identified the remaining radicand as 6 instead of 2.

Ensuring the coefficient $5x^2$ is included in the final answer. Identifying $16$ as the largest perfect square factor of $48$. Multiplying the extracted $4$ by $5x^2$. Dividing the variable exponent $7$ by $2$.

$4\sqrt{63}$ $6\sqrt{7}$ $18\sqrt{2}$ $3\sqrt{28}$

$-12\sqrt{2}$ $-24\sqrt{3}$ $-72\sqrt{2}$ $-36\sqrt{2}$

$9a^5b^6$ $81a^5b^6$ $9a^8b^{10}$ $9a^{\sqrt{10}}b^{\sqrt{12}}$

$x^{16}\sqrt{x}$ $x^8\sqrt{x}$ $x^9\sqrt{x}$ $x^8$

$20x^{10}y^7\sqrt{xy}$ $20x^{11}y^8\sqrt{y}$ $20x^{11}y^{7.5}$ $20x^{11}y^7\sqrt{y}$

$4a^2b^3\sqrt{8a}$ $12a^3b^6\sqrt{2a}$ $6ab^3\sqrt{8a^3}$ $12a^2b^3\sqrt{2a}$

$-30x^6y^6\sqrt{2xy}$ $-10x^6y^6\sqrt{18xy}$ $-30x^5y^6\sqrt{2xy}$ $-30x^6y^7\sqrt{2x}$

$5q$ $q$ $p^2q^6$ $5$

$5m^5n^5\sqrt{2mn}$ $25m^5n^5\sqrt{2mn}$ $5m^4n^5\sqrt{2mn}$ $10m^5n^4\sqrt{2mn}$

$-56k^{51}j^{100}\sqrt{j}$ $-28k^{51}j^{101}\sqrt{2}$ $-28k^{50}j^{100}\sqrt{2kj}$ $-28k^{51}j^{100}\sqrt{2j}$

$18a^4b^5c$ $6a^4b^5c$ $3a^4b^5c$ $6a^2b^4c$

$-14x^2y\sqrt{3}$ $-7x^2y\sqrt{6}$ $-28x^2y\sqrt{3}$ $-14xy\sqrt{3x^2y}$

$x^2\sqrt{10}$ $\sqrt{10x}$ $\sqrt{10x^2}$ $5x\sqrt{10}$

$x^2y\sqrt{x^3y}$ $5p\sqrt{2q}$ $10\sqrt{15}$ $-3a^3b^2\sqrt{7}$

The remaining radicand $60$ still contains the perfect square factor $4$. The extracted coefficient $2$ was not multiplied by the existing coefficient $6$. The numerical factor $4$ was missed in the initial extraction. The student failed to identify the largest perfect square factor of $240$ (which is $16$).

$16x^2y^4\sqrt{3x}$ $4x^2y^4\sqrt{3}$ $4x^{2.5}y^4$ $4x^2y^4\sqrt{3x}$

$63a^6b^6\sqrt{3a}$ $63a^7b^6\sqrt{3a}$ $21a^6b^6\sqrt{3a}$ $9a^6b^6\sqrt{21a}$