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Today · 2026-06-18

Physics P30–31
Chemistry P34–35
Math Exercise 2.2 / 2.3

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Math Exercise 2.1 — Simultaneous Equationsopen ↗

redo2Your factorisation $(2x+2)(x-4)=0$ is correct. Re-check the root from $2x+2=0$ — what sign should $x$ be? You wrote $x=1$.redo3$x^2+y^2=25$ is not the same as $x+y=5$ — you cannot square-root a sum term by term. Substitute $y=x-1$ into $x^2+y^2$, keep the squares and expand; you should get a quadratic with two solutions.redo14The working under Q14 is actually Q15 ($y=3x$). Q14 is a different pair: $x+y=4$ and $x^2+y^2=10$. Redo it against the correct equations.do19Skipped. Find where the line $y=1-2x$ cuts the curve $x^2+y^2=2$ — same substitution method you used elsewhere.fix23You found the two intersection points, but the question asks for the MIDPOINT of $AB$ — take the average of the two points. Also re-check the $x$-coordinate of the first point by substituting back into $2x+5y=1$.

Math Exercise 2.3 — Modulus Functionsopen ↗

fixQ4 (stationary point + 4 solutions)(a) Re-expand $(x-7)(x+1)$: the middle term is $x-7x=-6x$, so it's $x^2-6x-7$ — roots $7,-1$ and stationary point $(3,16)$, not $(-3,16)$. You wrote $+6x$.redoQ5 (two solutions)Two issues: this curve is $|(x+5)(x+1)|$ — its turning point is at $x=-3$ with value $-4$, so the peak of $|f|$ is 4 (not 16). And for TWO solutions you need $k$ ABOVE the peak (or $k=0$); $0<k<\text{peak}$ gives FOUR.fixQ7(e) (|x²−5x+1|=3)You left $x^2-5x-2=0$ unsolved — finish it with the quadratic formula to get the other two roots $\tfrac{5\pm\sqrt{33}}{2}$.

Math Exercise 3.3 — Factor Theorem & Polynomialsopen ↗

fix2cYour setup $-\tfrac{27}{2}+\tfrac{9a}{4}-\tfrac{87}{2}+30=0$ is correct — you just stopped. Combine the number terms, then isolate $\tfrac{9a}{4}$ and solve for $a$.do7bNot answered. A cubic has three linear factors and you already have two of them — think about what the third must be (or test $x=-3$ in the expression and state the factor theorem).

Math Exercise 3.4 — Factorising & Solving Cubicsopen ↗

fix8You've written down the given roots, but the question asks for $f(x)$. Build it: multiply the factors $(x-\text{root})$ together and expand to a cubic with integer coefficients (leading coefficient 1).fix9Same as Q8, but the leading coefficient is 2 — so use a factor like $(2x-1)$ for the root $\tfrac12$ before expanding.fix10The surd pair $1\pm\sqrt2$ comes from $x^2-2x-1$. Multiply that by $(x+3)$ and expand — your final cubic should be monic (leading coefficient 1). Please re-check your result.fix11Method is right: $2\pm\sqrt3$ gives $x^2-4x+1$, times $(2x-1)$ for the root $\tfrac12$. Re-check the middle ($x$) term when you expand — it looks off.

Math Exercise 3.5 — Remainder Theoremopen ↗

fix1For (b), re-add $8-24+22-7$ carefully — group the positives and negatives and re-check that remainder.do16The challenge question — give it a go. With roots $1,k,k+1$ and $f(2)=20$, you can reach $k^2-3k-18=0$; then solve that quadratic for the possible $k$.

Physics Workbook — Ch.4 Forcesopen ↗

fixp26 · Q3 (resultant forces)(b) Looked it up on the clean figure: there are TWO 3 N arrows, both pointing left (negative), and one 5 N pointing right. Add all three with signs ($+5-3-3$). You wrote −2 — did you include BOTH 3 N forces?fixp28 · Q2 (seesaw)(b) You described what happens to the 160 N child but didn't answer the question — which WAY does the plank turn when $d$ decreases? (Which side's moment now wins?)fixp28 · Q3 (beam + rope, Fig 4.11)Checked the clear figure — the distances are easy to misassign, and your $d=0.53$ m is actually plausible under one reading. Write out your full moment equation (each force × its distance about the pivot, including the 8.0 N rope) so it can be confirmed.fixp29 · Q2 (metre rule)(d) Check where the weight $W$ sits relative to the pivot, then use moments: $W \times (\text{its distance}) = 1.5\,\text{N} \times 20\,\text{cm}$. Your 1.5 N looks like the rule's own weight, not the calculated $W$.

Chemistry — Ch.4 Ionic, Covalent & Metallic Bondingopen ↗

fixp29 · Q6Sodium part is good (loses 1e⁻ → Na⁺, more protons than electrons). The question also needs chlorine: it GAINS 1e⁻ → Cl⁻, so more electrons than protons → negative.redop29 · Q7a RbClRbCl is IONIC — electron transfer, not sharing. Don't draw overlapping (shared) circles. Show Rb giving its 1 outer electron to Cl, as separate ions in square brackets with charges $[\text{Rb}]^+\,[\text{Cl}\,]^-$. (You DID draw covalent ones correctly later — same idea, but transfer instead of share.)fixp29 · Q7b Al₂S₃Formula $\text{Al}_2\text{S}_3$ is right — but no diagram. Show 2 Al each losing 3e⁻ to 3 S (each gaining 2e⁻), drawn as ions with their charges.dop30 · Q8b m.p.Left blank. Compare the charges: MgO is $\text{Mg}^{2+}/\text{O}^{2-}$ vs NaCl's $\text{Na}^+/\text{Cl}^-$ — bigger charges → stronger attraction → more energy to break → higher melting point.fixp30 · Q9bYour method is muddled. Give ONE clear way: either melt it (heat until molten) OR dissolve it in water — each frees the ions to move. Don't mix 'heat' with 'dissolve in water'.fixp33 · Q1(b) conductivityYou explained diamond but stopped — now do graphite: each carbon uses only 3 of its 4 outer electrons for bonds, leaving 1 free/delocalised electron per atom that can move and carry charge.fixp33 · Q2 (SiO₂ structure)(c) Re-count: each black (oxygen) atom bridges how many silicon atoms in the structure? You wrote 3 — look again at how many bonds an oxygen forms here.

Chemistry — Ch.4 (pp.34–35: giant structures & metallic bonding)open ↗

fixp34 · Q3(a) describe structuresAlso describe silicon(IV) oxide: each silicon is covalently bonded to 4 oxygen atoms (and each oxygen to 2 silicons), giving the tetrahedral giant structure.fixp35 · Q2 metallic passageThere's one more error — a metallic bond is the electrostatic ATTRACTION (not repulsion) between the positive metal ions and the sea of delocalised electrons.fixp35 · Q3 metallic diagramYou drew the arrangement, but the question asks for a LABELLED diagram — label the positive metal ions and the 'sea' of delocalised electrons.

Today · 2026-06-18

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