\n| \n (f(x) textual content{ is rising when } f'(x) > 0.)<\/p>\n start{align*}
\nf'(x) &= 2x \u2013 6x^2
\n&= 2x(1-3x)
\nfinish{align*}<\/p>\n (2x(1-3x) geq 0)
\n(\u2234 f'(x) geq 0 textual content{ when } x in (0, frac{1}{3}) )
\n(\u2234 f(x) textual content{ is rising when } x in (0, frac{1}{3}))<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 17 a<\/h3>\n(i)<\/strong><\/p>\n\n \n\n\nstart{align*}
\ny &= 29.2 \u2013 0.011(540)
\n&= 23.26^circ textual content{C}
\n&approx 23.3^circ textual content{C}\u00a0 textual content{(1 d.p.)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n (ii)<\/strong><\/p>\nAs the peak of the city above sea degree will increase, the typical most every day temperature decreases. Particularly, for each metre above sea degree, there\u2019s roughly a (0.011^circ textual content{C}) lower within the common most every day temperature. Because the correlation coefficient is 0.494, there\u2019s a average destructive correlation of this knowledge.<\/p>\n <\/p>\n Query 17b<\/h3>\nLatitude can be a greater predictor of the city\u2019s common every day most temperature as there\u2019s a stronger destructive correlation for this knowledge of (r = -0.897) versus solely a average destructive correlation coefficient of (-0.494) given by the comparability of the peak above sea degree and common every day most temperature.<\/p>\n <\/p>\n Query 18<\/h3>\nUtilizing sine rule in (triangle ABC), let (angle ABC = alpha).<\/p>\n \n \n\n\nstart{align*}
\nfrac{sin(alpha)}{25} &= frac{sinleft(28^circright)}{16}
\nsin(alpha) &= frac{25}{16} sin(28^circ)
\n\u2234alpha &= sin^{-1}left(frac{25}{16} sin(28^circ)proper)
\nfinish{align*}The 2 options are,start{align*}
\n\u2234 alpha &= sin^{-1}left(frac{25}{16} sin(28^circ)proper) textual content{, } 180^circ \u2013 sin^{-1}left(frac{25}{16} sin(28^circ)proper).
\nfinish{align*}Since (alpha) is obtuse,start{align*}
\n\u2234 alpha &= 180^circ \u2013 sin^{-1}left(frac{25}{16} sin(28^circ)proper)
\n&approx 133^circ ; textual content{(nearest diploma)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 19<\/h3>\nThe graph of (y = 2 + frac{1}{x+4}) is:<\/p>\n ![\"2021](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2021\/11\/2021-HSC-Maths-Adv-Exam-Paper-Solutions-question.png\") <\/p>\n
Query 20<\/h3>\n\n \n\n\nstart{align*}
\n2 sin(4x) &= 1
\nsin(4x) &= frac{1}{2}
\nfinish{align*}To unravel for (0 leq x leq frac{pi}{4}), we have to clear up the above trigonometric equation for (0 leq 4x leq pi).start{align*}
\n\u2234 4x &= frac{pi}{6}, frac{5pi}{6}.
\n\u2234 \u00a0x &= frac{pi}{24}, frac{5pi}{24}.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 21<\/h3>\nThe turning two turning factors are at (0,0) and (2, -32) and (x)-intercepts at 0 and three.<\/p>\n <\/p>\n ![\"2021](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2021\/11\/2021-HSC-Maths-Adv-Exam-Paper-Solutions-question-21-graph.png\") <\/p>\n
<\/p>\n Query 22a<\/h3>\n\n \n\n\nstart{align*}
\n& 0.1915 \u2013 0.0398 = 0.1517
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 22b<\/h3>\n\n \n\n\nstart{align*}
\n& P(W > 3528)
\n&= P(Z > frac{3528 \u2013 3300}{570})
\n&= P(Z > 0.4)
\n&= 1 \u2013 P(Z leq 0) \u2013 P(0 < Z < 0.4)
\n&= 1 \u2013 frac{1}{2} \u2013 0.1554
\n&= 0.3446.
\nfinish{align*}start{align*}
\n\u2234\u00a0textual content{Anticipated variety of infants out of } 1000 &= 1000 occasions 0.3446
\n&= 345 textual content{(nearest entire child)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 23<\/h3>\nFirst clear up for (b) by substituting (P = 1250) and (t = 20),<\/p>\n \n \n\n\nstart{align*}
\n1250 &= 5000b^{-frac{20}{10}}
\nfrac{1}{4} &= b^{-2}
\nb^2 &= 4
\n\u2234b &= 2\u00a0 textual content{ since } b textual content{ is constructive}.
\nfinish{align*}Differentiate (P(t)) to be able to discover (t) when (frac{dP}{dt} = -30).start{align*}
\nfrac{dP}{dt} &= frac{5000 occasions ln{2} occasions 2^{frac{-t}{10}}}{-10}
\n&= -500 occasions ln{2} occasions 2^{frac{-t}{10}}
\nfinish{align*}Let (frac{dP}{dt} = -30) and clear up for (t),start{align*}
\n-30 &= -500 occasions ln{2} occasions 2^{frac{-t}{10}}
\n2^{frac{-t}{10}} &= frac{30}{500 occasions ln{2}}
\nfrac{-t}{10} &= log_2 left(frac{30}{500 occasions ln{2}}proper)
\n\u2234 t &= -10 log_2 left(frac{30}{500 occasions ln{2}}proper)
\n\u2234 t &approx 35.3 textual content{ years (1 d.p.)}.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 24<\/h3>\n\n \n\n\nstart{align*}
\ntextual content{Shaded space} &= frac{1}{2}(2)(3) + int_{2}^4 frac{3}{x \u2013 1} ; dx
\n&= 3 + 3 left[ln{(x-1)}right]_{2}^{4}
\n&= 3 + 3left[ln{(3)} \u2013 ln{(1)}right]
\n&= 3 + 3ln{(3)}
\n&= 3 + ln{(27)} ; textual content{items}^2
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 25<\/h3>\nUtilizing the desk of future worth curiosity elements for the primary 8 years,<\/p>\n \n \n\n\nstart{align*}
\nA &= 8.2132 occasions 1000
\n&= 8213.2
\nfinish{align*}Leaving the cash in her financial savings account for an additional 2 years at 1.25%,start{align*}
\nA &= 8213.2 occasions (1.0125)^2
\n&approx $ 8419.81 textual content{(2 d.p.)}.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 26a<\/h3>\nTo seek out the utmost top, first differentiate (y(t)) with respect to (t).<\/p>\n \n \n\n\nstart{align*}
\ny(t) &= -5t^2 + 70t + 100
\ny'(t) &= -10t + 70
\ny\u201d(t) &= -10
\nfinish{align*}
\nStationary level is attained when (y'(t) = 0), so substituting in,
\nstart{align*}
\ny'(t) &= 0
\n-10t + 70 &= 0
\n\u2234\u00a0t &= 7
\nfinish{align*}
\nConfirm that (t = 7) is a most as a result of (y\u201d(7) = -10 < 0).Therefore, the utmost top reached by the particle is,start{align*}
\ny(7) &= -5(7)^2 + 70(7) + 100
\ny(7) &= 345 textual content{ m}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 26b<\/h3>\nSo as to discover the speed, i.e. (y'(t)) of the particle earlier than hitting the bottom. First discover (t) when the particle has a top of zero, i.e. (y(t) = 0).<\/p>\n \n \n\n\nstart{align*}
\ny(t) &= 0
\n-5t^2 + 70t + 100 &= 0 textual content{(use quadratic method)}
\n\u2234 t &= 7 pm sqrt{69}
\nfinish{align*}Since (7 \u2013 sqrt{69} < 0), then the particle hits the bottom when (t = 7 + sqrt{69}).
\nThen, we are going to solely discover (y'(7 + sqrt{69})),start{align*}
\ny'(7 + sqrt{69}) &= -10(7 + sqrt{69}) + 70
\n&= -70 \u2013 10sqrt{69} + 70
\n&= \u2013 10sqrt{69}
\nfinish{align*}(\u2234textual content{ velocity } = -10 sqrt{69} ; textual content{m}^2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 27a<\/h3>\nThe turning level under is at (6,400) the place the x-intercepts are at 0 and 12.<\/p>\n ![\"2021](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2021\/11\/2021-HSC-Maths-Adv-Exam-Paper-Solutions-question-27a-graph.png\") <\/p>\n
Query 27b<\/h3>\n\n \n\n\nstart{align*}
\nE &= int_{a}^{b} P(t) ; dt
\n&= int_{a}^{b} 400 sinleft( frac{pi}{12} tright) ; dt
\n&= frac{400 occasions 12}{pi} left[ -cosleft( frac{pi}{12}tright)right]_{a}^{b}; dt
\n&= frac{4800}{pi} left[ cosleft( frac{api}{12}right) \u2013 cosleft( frac{bpi}{12}right)right]
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 27c<\/h3>\nUse half (b) and let (a = 3) and (E=300), then clear up for (b).<\/p>\n \n \n\n\nstart{align*}
\n300 &= frac{4800}{pi} left[ cosleft( frac{3pi}{12}right) \u2013 cosleft( frac{bpi}{12}right)right]
\nfrac{300 pi}{4800} &= frac{1}{sqrt{2}}- cosleft( frac{bpi}{12}proper)
\ncosleft( frac{bpi}{12}proper) &= frac{1}{sqrt{2}} \u2013 frac{pi}{16}
\nb &= frac{12}{pi} cos^{-1} left(frac{1}{sqrt{2}} \u2013 frac{pi}{16}proper)
\nb &approx 3.95238
\nfinish{align*}Therefore, ready time isbegin{align*}
\nb \u2013 a &= 3.95238 \u2013 3
\n&= 0.95238 ; textual content{hours}
\n&approx 57.142 ; textual content{minutes}
\nfinish{align*}Subsequently, the least period of time that Kenzo should wait to the closest minute is 58 minutes.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 27d<\/h3>\nFrom the graph drawn partly (a), we are able to observe that the utmost energy output is attained 6 hours after dawn. Thus it could not require as lengthy for Kenzo to cost his cellphone ranging from 6 hours after dawn as in comparison with ranging from 3 hours after dawn.<\/p>\n <\/h3>\nQuery 28a<\/h3>\nFirst discover the x-intercept of the perform (f(x) = 8 \u2013 2^x).<\/p>\n \n \n\n\nstart{align*}
\n0 &= 8 \u2013 2^x
\n2^x &= 8
\n\u2234 x &= 3
\nfinish{align*}Subsequent, combine (f(x)) from 0 to three to be able to discover the realm below the curve.start{align*}
\ntextual content{Space}
\n&= int_{0}^{3} 8 \u2013 2^x ; dx
\n&= left[ 8x \u2013 frac{2^x}{ln{(2)}}right]_{0}^{3}
\n&= left(8(3) \u2013 frac{2^{(3)}}{ln{(2)}}proper) \u2013 left( 8(0) \u2013 frac{2^{(0)}}{ln{(2)}}proper)
\n&= 24 \u2013 frac{8}{ln(2)} + frac{1}{ln(2)}
\n&= 24 \u2013 frac{7}{ln{(2)}}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 28b<\/h3>\nThe graph under has an x-intercept at 2 and asymptote of (y=-8).<\/p>\n ![\"2021](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2021\/11\/2021-HSC-Maths-Adv-Exam-Paper-Solutions-question-28b-graph.png\") <\/p>\n
<\/p>\n Query 28c<\/h3>\n\n \n\n\nstart{align*}
\nint_{2}^{5} g(x) ; dx
\n&= int_{-3}^{0} -f(-x) ; dx
\n&= int_{0}^{3} -f(x) ; dx
\n&= \u2013 int_{0}^{3} f(x) ; dx
\n&= \u2013 left[ 24 \u2013 frac{7}{ln{(2)}}right]
\n&= frac{7}{ln{(2)}} \u2013 24
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 29a<\/h3>\nLet (A_n) be the quantity within the account on Megan\u2019s n-th birthday.<\/p>\n \n \n\n\nstart{align*}
\nA_1 &= 5000(1.03) + 1000
\nA_2 &= A_1(1.03) + 1000 = 5000(1.03)^2 + 1000(1.03) + 1000
\nA_3 &= A_2(1.03) + 1000 = 5000(1.03)^3 + 1000(1.03)^2 + 1000(1.03) + 1000 = $8554.54
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 29b<\/h3>\nWe\u2019re on condition that (A_{17} = 30 025.83).
\nLet (B_n) be the quantity within the account after the n-th annual withdrawal.<\/p>\n \n \n\n\nstart{align*}
\nB_1 &= A_{17}(1.03) \u2013 2000
\nB_2 &= A_{17}(1.03)^2 \u2013 2000(1.03) \u2013 2000
\n&vdots
\nB_n &= A_{17}(1.03)^n \u2013 2000[(1.03)^{n-1} + (1.03)^{n-2} + dots + 1.03 + 1]
\n&= A_{17}(1.03)^n \u2013 frac{2000(1 \u2013 1.03^n)}{1 \u2013 1.03}
\nfinish{align*}Resolve for (B_n = 0),start{align*}
\nA_{17}(1.03)^n \u2013 frac{2000(1 \u2013 1.03^n)}{1 \u2013 1.03} &= 0
\nA_{17}(1.03)^n \u2013 frac{2000}{-0.03} + frac{2000(1.03)^n}{-0.03} &= 0
\nleft(A_{17} \u2013 frac{2000}{0.03}proper)(1.03)^n &= \u2013 frac{2000}{0.03}
\n(1.03)^n &= \u2013 frac{2000}{0.03} occasions frac{1}{A_{17} \u2013 frac{2000}{0.03}}
\nn &= frac{1}{ln(1.03)} ln left(-frac{2000}{0.03} occasions frac{1}{A_{17} \u2013 frac{2000}{0.03}}proper)
\n&approx 20.249
\nfinish{align*}Therefore, Megan could make a most of 20 withdrawals of ($2,000) earlier than her account stability is inadequate.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 30<\/h3>\nResolve for (x) when (F(x) = 0.99)<\/p>\n \n \n\n\nstart{align*}
\n0.99 &= 1- e^{-0.01x}
\ne^{-0.01x} &= 0.01
\n-0.01x &= frac{ln{(0.01)}}{-0.01}
\n\u2234 x &= 460.52 textual content{ hours (2 d.p.)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 31<\/h3>\n\n \n\n\nstart{align*}
\ny &= x^2 \u2013 1
\nfrac{dy}{dx} &= 2x
\nfinish{align*}Equate the gradient of the tangent when (x = a) with the gradient of the road that passes via ((3, -8)) and ((a, a^2-1)),start{align*}
\nfrac{-8 \u2013 (a^2 \u2013 1)}{3 \u2013 a} &= 2(a)
\n-8 \u2013 a^2 + 1 &= 6a \u2013 2a^2
\na^2 \u2013 6a -7 &= 0
\n(a \u2013 7)(a + 1) &= 0
\n\u2234 a &= 7, -1
\nfinish{align*}Subsequently, the 2 factors on (y = x^2-1) the place their tangents go via ((3,-8)) are (A(7,48)) and (B(-1, 0)).
\nUtilizing the point-gradient method, the equation of the tangent at (A(7,48)) isbegin{align*}
\ny \u2013 48 &= 14(x \u2013 7)
\ny &= 14x \u2013 98 + 48
\n\u2234 y &= 14x \u2013 50
\nfinish{align*}Equally for (B(-1,0)),start{align*}
\ny \u2013 0 &= -2(x + 1)
\n\u2234 y &= -2x \u2013 2
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 32<\/h3>\nLet (X) be a feminine\u2019s top within the inhabitants and (Y) be a male\u2019s top.<\/p>\n \n \n\n\nstart{align*}
\nP(X< 175) &= 97.5%
\nP(X < 160.6) &= 16%
\nP(Y < h) &= 84%
\nfinish{align*}Our goal is the discover (h). Utilizing the empirical rule with the knowledge given above.start{align*}
\nmu + 2sigma &= 175 & textual content{(1)}
\nmu \u2013 sigma &= 160.6 & textual content{(2)}
\nfinish{align*}Subtracting equation (1) by (2) givesbegin{align*}
\n3 sigma &= 14.4
\n\u2234 sigma &= 4.8
\nfinish{align*}Substitute (sigma = 4.8) into (1),start{align*}
\nmu + 2(4.8) &= 175
\nmu = 165.4
\nfinish{align*}Therefore, for the grownup male inhabitants, the imply is (1.05 mu = 173.67) and the usual deviation is (1.1 sigma = 5.28).
\nSince being (84%) taller than the male inhabitants is 1 normal deviation above the imply, then we are able to calculate (h) asbegin{align*}
\nh &= 1.05 mu + 1.1 sigma
\n&= 173.67 + 5.28
\n&= 178.95 ; textual content{cm}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 33a<\/h3>\nUtilizing the truth that the entire likelihood is all the time (1),<\/p>\n \n \n\n\nstart{align*}
\nint_{-infty}^{infty} f(x) ; dx &= 1
\nint_{0}^{6} frac{Ax}{x^2 + 4} ; dx &= 1
\nA left[frac{1}{2} ln{(x^2 + 4)}right]_{0}^{6} &= 1
\nfrac{A}{2}left[ln(40) \u2013 ln(4)right] &= 1
\nfinish{align*}start{align*}
\n\u2234 A &= frac{2}{ln{(40)} \u2013 ln{(4)}}
\n&= frac{2}{ln(frac{40}{4})}
\n&= frac{2}{ln(10)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 33b<\/h3>\nDiscover when the primary spinoff is the same as zero to search out the utmost turning level.
\nThe primary spinoff is<\/p>\n \n \n\n\n(frac{dy}{dx} = frac{2}{ln{(10)}} left[frac{4 \u2013 x^2}{(x^2+4)^2} right])
\nResolve for (frac{dy}{dx} = 0),start{align*}
\nfrac{2}{ln{(10)}} left[frac{4 \u2013 x^2}{(x^2+4)^2} right] &= 0
\n4 \u2013 x^2 &= 0
\nx &= pm 2
\n\u2234x &= 2 textual content{ since } x in [0, 6]
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 33c<\/h3>\n\n \n\n\nstart{align*}
\nP(X < 2) &= int_{0}^{2} frac{Ax}{x^2 + 4} ; dx
\n&= frac{2}{ln{(10)}}left[frac{1}{2} ln{(x^2+4)} right]_{0}^{2}
\n&= frac{1}{ln{(10)}} left[ln{left(2^2 + 4right)} \u2013 ln{left(0^2 + 4right)} right]
\n&= frac{1}{ln{(10)}} left[ln{(8)} \u2013 ln{(4)} right]
\n&= frac{ln{left(frac{8}{4} proper)}}{ln{(10)}}
\n&= frac{ln{(2)}}{ln{(10)}}
\n\u2234 P(X <2) &= log_{10}{(2)} textual content{(utilizing change of base rule)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/h3>\nQuery 33d<\/h3>\nWe are able to write the next information through the use of the empirical rule and outcomes from earlier components:<\/p>\n \n \n\n\nstart{align*}
\n&P(IQ > 130) = 2.5% = 0.025
\n&P(X < 2) = log_{10}{(2)}
\n&P(X < 2 | IQ > 130) = 80% = 0.8
\nfinish{align*}The goal is to search out the likelihood that an individual has an IQ larger than 130, on condition that they\u2019ll full the puzzle in lower than two hours.
\nThat\u2019s, we need to discover (P(IQ > 130 | X < 2)),start{align*}
\nP(IQ > 130 | X < 2) &= frac{P(IQ < 130 textual content{ and } X < 2)}{P(X < 2)}
\n&= frac IQ > 130) occasions P(IQ > 130){P(X < 2)}
\n&= frac{0.8 occasions 0.025}{log_{10}(2)}
\n&approx 0.066 textual content{(3 d.p.)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Query 34<\/h3>\n\n \n\n\nstart{align*}
\nEleft(Xright) &= r occasions r^n + r^2 occasions r^{n \u2013 1} + \u2026 + r^{okay} occasions r^{n \u2013 okay + 1} + .. + r^n occasions r
\nEleft(Xright) &= r^{n + 1} + r^{n + 1} + \u2026 + r^{n + 1} textual content{ (the summation is $r^{n + 1}$ summed $n$ occasions)}
\nEleft(Xright) &= n(r^{n + 1})\u00a0 \u2026 (1)
\nfinish{align*}Additionally word that the summation of the possibilities equate to at least one.start{align*}
\nsum_{x}P(X = x) &= 1
\nr^n + r^{n \u2013 1} + \u2026 + r &= 1
\nfrac{r occasions (1 \u2013 r^n)}{1 \u2013 r} &= 1
\n\u2234\u00a0r^{n+1} &= 2r \u2013 1 \u2026 (2)
\nfinish{align*}By substituting equation (2) into equation (1),start{align*}
\n\u2234 Eleft(Xright) &= n(2r \u2013 1)
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n <\/h2>\n,<\/p>\n <\/div>\n","protected":false},"excerpt":{"rendered":" On this submit, we are going to work our method via the 2021 HSC Maths Superior (2 Unit) paper and provide the options, written by our Head of Arithmetic Oak Ukrit and his staff. (Doing follow papers? See the options for the 2020 HSC Maths Adv Examination right here.) Learn on to see tips on<\/p>\n","protected":false},"author":1,"featured_media":370,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-369","post","type-post","status-publish","format-standard","has-post-thumbnail","category-math"],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/369","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/comments?post=369"}],"version-history":[{"count":1,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/369\/revisions"}],"predecessor-version":[{"id":648,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/369\/revisions\/648"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media\/370"}],"wp:attachment":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media?parent=369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/categories?post=369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/tags?post=369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} | | | | | | | | | | | | | | | | | | | | | | | | |
![\"2021](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2021\/11\/2021-hsc-maths-adv-exam-paper-solutions-question-6-answer-diagram.png\")
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