Question Number 66856 by John Kaloki Musau last updated on 20/Aug/19 $${Two}\:{towns}\:{T}\:{and}\:{S}\:{are}\:\mathrm{300}\:{km}\:{apart}. \\ $$$${Two}\:{buses}\:{A}\:{and}\:{B}\:{started}\:{from} \\ $$$${T}\:{at}\:{the}\:{same}\:{time}\:{travelling}\:{towards} \\ $$$${S}.\:{Bus}\:{B},\:{travelling}\:{at}\:{an}\:{average} \\ $$$${speed}\:{of}\:\mathrm{10}{km}/{h}\:{greater}\:{than}\:{that} \\ $$$${of}\:{A}\:{reached}\:{S}\:\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}}\:{hours}\:{earlier}. \\ $$$$\left({a}\right)\:{Find}\:{the}\:{average}\:{speed}\:{of}\:{A} \\…
Question Number 1322 by 123456 last updated on 22/Jul/15 $${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{{n}} −{a}_{\mathrm{0}} ={a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ,{n}\geqslant\mathrm{2} \\ $$$${a}_{\mathrm{10}} =? \\ $$$$\underset{{n}\rightarrow+\infty}…
Question Number 66852 by John Kaloki Musau last updated on 20/Aug/19 Commented by John Kaloki Musau last updated on 20/Aug/19 The cross-section of a head of a bolt is the form of a regular hexagon as shown in the figure below. Determine the area of the cross-section. Commented by John Kaloki…
Question Number 1318 by 123456 last updated on 22/Jul/15 $$\eta:\mathbb{C}^{\mathrm{2}} \rightarrow\mathbb{R}_{+} \\ $$$$\eta\left({x},{y}\right)=\mid\mid{x}\mid−\mid{y}\mid\mid \\ $$$$\eta\left({x},{y}\right)=\mathrm{0}\overset{?} {\Leftrightarrow}{x}={y} \\ $$$$\eta\left({x},{y}\right)\overset{?} {=}\eta\left({y},{x}\right) \\ $$$$\eta\left({x},{z}\right)\overset{?} {\leqslant}\eta\left({x},{y}\right)+\eta\left({y},{z}\right) \\ $$$${x}\in\mathbb{C} \\…
Question Number 66851 by John Kaloki Musau last updated on 20/Aug/19 Commented by John Kaloki Musau last updated on 20/Aug/19 A piece of wire, pcm long, is bent to form the shape shown in the figure above. The figure consists of a semicircular arc of radius rcm and two perpendicular sides of length xcm each. Express x in terms of p and r, hence show that the area of the figure is given by A=1/2πr^2+1/8(p-πr)^2 Answered by John Kaloki…
Question Number 66846 by John Kaloki Musau last updated on 20/Aug/19 Commented by John Kaloki Musau last updated on 20/Aug/19 The diagram above shows a cross-section of a bottle. The lower part ABC is a hemisphere of radius 5.2cm and the upper part is a frustrum of a cone. The top radius of the frustrum is one third of the radius of the hemisphere. The hemispherical part is conpletely filled with water as shown in the diagram. When the container is inverted, the water now conpletely fills only the frustfum part. (a) Determine the height of the frustrum part. (b) Find the surface area of the frustrum part of the bottle. Commented by John Kaloki…
Question Number 66849 by John Kaloki Musau last updated on 20/Aug/19 Commented by John Kaloki Musau last updated on 20/Aug/19 A girl wanted to make a regular octagon of side 14cm. She made it from a square piece of a card of size ycm by cutting off four isosceles triangles whose equal sides were xcm each, as shown in the figure above. (a) Write down an expression for the area of the octagon in terms of x and y. (b) Find the value of x. (c) Find the area of the octagon. Commented by John Kaloki…
Question Number 66842 by John Kaloki Musau last updated on 20/Aug/19 Commented by John Kaloki Musau last updated on 20/Aug/19 In the figure above,BAD and CBD are right angled triangles. Find the length of AB. Commented by John Kaloki…
Question Number 66845 by John Kaloki Musau last updated on 20/Aug/19 $${A}\:{cylindrical}\:{tank}\:{of}\:{radius}\:\mathrm{2}{m}\: \\ $$$${and}\:{height}\:\mathrm{1}.\mathrm{5}{m}\:{initially}\:{contains} \\ $$$${water}\:{to}\:{a}\:{depth}\:{of}\:\mathrm{50}{cm}.\:{Water} \\ $$$${is}\:{added}\:{to}\:{the}\:{tank}\:{at}\:{the}\:{rate}\:{of}\: \\ $$$$\mathrm{62}.\mathrm{84}{l}\:{per}\:{minute}\:{for}\:\mathrm{15}\:{minutes}. \\ $$$${Find}\:{the}\:{new}\:{height}\:{of}\:{water}\:{in} \\ $$$${the}\:{tank}. \\…
Question Number 66832 by Rio Michael last updated on 20/Aug/19 $${solve}\:{the}\:{system}\:{of}\:{congruence} \\ $$$$\:\:\:\left.\begin{matrix}{{x}\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{5}\right)}\\{{x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{7}\right)}\\{{x}\equiv\:\:\mathrm{3}\left({mod}\:\mathrm{9}\right)}\\{{x}\:\equiv\:\mathrm{4}\left(\:{mod}\:\mathrm{11}\right)}\end{matrix}\right\} \\ $$ Answered by mr W last updated on 26/Aug/19 $${x}=\mathrm{5}{h}+\mathrm{1}\:\:\:\:…\left(\mathrm{1}\right) \\…