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Question Number 15497    Answers: 1   Comments: 5

P=Σ_(n∈P) ^∞ n Q=Σ_(n∉P) ^∞ n P=2+3+5+7+... Q=1+4+6+8+... Is P>Q? Is Q>P?

$${P}=\underset{{n}\in\mathbb{P}} {\overset{\infty} {\sum}}{n}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}=\underset{{n}\notin\mathbb{P}} {\overset{\infty} {\sum}}{n} \\ $$$${P}=\mathrm{2}+\mathrm{3}+\mathrm{5}+\mathrm{7}+... \\ $$$${Q}=\mathrm{1}+\mathrm{4}+\mathrm{6}+\mathrm{8}+... \\ $$$$\: \\ $$$$\mathrm{Is}\:{P}>{Q}?\:\:\:\mathrm{Is}\:{Q}>{P}? \\ $$

Question Number 15495    Answers: 0   Comments: 2

Question Number 15485    Answers: 1   Comments: 0

Question Number 15480    Answers: 1   Comments: 1

Question Number 15471    Answers: 1   Comments: 0

Question Number 15470    Answers: 0   Comments: 0

Question Number 15474    Answers: 0   Comments: 0

Question Number 15473    Answers: 0   Comments: 0

Question Number 16613    Answers: 1   Comments: 2

Question Number 15463    Answers: 1   Comments: 0

Find the domain and range of a function for which f(x)=((1+2x)/x).

$${Find}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{a}\:{function}\:{for}\:{which}\:{f}\left({x}\right)=\frac{\mathrm{1}+\mathrm{2}{x}}{{x}}. \\ $$

Question Number 15452    Answers: 1   Comments: 0

Question Number 15448    Answers: 0   Comments: 2

A vector A^→ of magnitude A is turned through an angle θ. Calculate the change in the magnitude of vector.

$$\mathrm{A}\:\mathrm{vector}\:\overset{\rightarrow} {{A}}\:\mathrm{of}\:\mathrm{magnitude}\:{A}\:\mathrm{is}\:\mathrm{turned} \\ $$$$\mathrm{through}\:\mathrm{an}\:\mathrm{angle}\:\theta.\:\mathrm{Calculate}\:\mathrm{the} \\ $$$$\mathrm{change}\:\mathrm{in}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{vector}. \\ $$

Question Number 15440    Answers: 3   Comments: 9

Question Number 15434    Answers: 1   Comments: 8

Question Number 15407    Answers: 1   Comments: 0

A ball is thrown vertically upward with velocity 20 m/s from a rail road car moving with a velocity 5 m/s horizontally. A person standing on the ground observes its motion as projectile. Find maximum height attained by the ball if point of projection is at a height 3 m from the ground.

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward} \\ $$$$\mathrm{with}\:\mathrm{velocity}\:\mathrm{20}\:\mathrm{m}/\mathrm{s}\:\mathrm{from}\:\mathrm{a}\:\mathrm{rail}\:\mathrm{road} \\ $$$$\mathrm{car}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{horizontally}.\:\mathrm{A}\:\mathrm{person}\:\mathrm{standing}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{observes}\:\mathrm{its}\:\mathrm{motion}\:\mathrm{as}\:\mathrm{projectile}. \\ $$$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{attained}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{if}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{3}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{ground}. \\ $$

Question Number 15405    Answers: 1   Comments: 0

A body is projected at time t = 0 from a certain point on a planet surface with a certain velocity at a certain angle with the planet′s surface (assumed horizontal). The horizontal and vertical displacement x and y in metre are related to time as x = 10(√3)t and y = 10t − 4t^2 . Find vertical component of velocity of the particle when it is at a height half of the maximum height attained.

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{certain}\:\mathrm{point}\:\mathrm{on}\:\mathrm{a}\:\mathrm{planet}\:\mathrm{surface}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{certain}\:\mathrm{velocity}\:\mathrm{at}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{angle} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{planet}'\mathrm{s}\:\mathrm{surface}\:\left(\mathrm{assumed}\right. \\ $$$$\left.\mathrm{horizontal}\right).\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{and}\:\mathrm{vertical} \\ $$$$\mathrm{displacement}\:{x}\:\mathrm{and}\:{y}\:\mathrm{in}\:\mathrm{metre}\:\mathrm{are} \\ $$$$\mathrm{related}\:\mathrm{to}\:\mathrm{time}\:\mathrm{as}\:{x}\:=\:\mathrm{10}\sqrt{\mathrm{3}}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{10}{t}\:−\:\mathrm{4}{t}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{vertical}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{height}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{attained}. \\ $$

Question Number 15393    Answers: 1   Comments: 0

A man observes that when he moves up a distance c metres on a slope, the angle of depression of a point on the horizontal plane from the base of the slope is 30°, and when he moves up further a distance c metres, the angle of depression of that point is 45°. The angle of inclination of the slope with the horizontal is?

$$\mathrm{A}\:\mathrm{man}\:\mathrm{observes}\:\mathrm{that}\:\mathrm{when}\:\mathrm{he}\:\mathrm{moves}\:\mathrm{up} \\ $$$$\mathrm{a}\:\mathrm{distance}\:{c}\:\mathrm{metres}\:\mathrm{on}\:\mathrm{a}\:\mathrm{slope},\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{depression}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{from}\:\mathrm{the}\:\mathrm{base}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{slope}\:\mathrm{is}\:\mathrm{30}°,\:\mathrm{and}\:\mathrm{when}\:\mathrm{he}\:\mathrm{moves}\:\mathrm{up} \\ $$$$\mathrm{further}\:\mathrm{a}\:\mathrm{distance}\:{c}\:\mathrm{metres},\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of} \\ $$$$\mathrm{depression}\:\mathrm{of}\:\mathrm{that}\:\mathrm{point}\:\mathrm{is}\:\mathrm{45}°.\:\mathrm{The} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{inclination}\:\mathrm{of}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{is}? \\ $$

Question Number 15392    Answers: 1   Comments: 0

Each side of an equilateral triangle subtends angle of 60° at the top of a tower of height h standing at the centre of the triangle. If 2a be the length of the side of the triangle, then (a^2 /h^2 ) = ?

$$\mathrm{Each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{subtends}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}°\:\mathrm{at}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{h}\:\mathrm{standing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{centre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}.\:\mathrm{If}\:\mathrm{2}{a}\:\mathrm{be}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle},\:\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\:=\:? \\ $$

Question Number 15377    Answers: 1   Comments: 0

−39 mod 4 = ?

$$−\mathrm{39}\:\mathrm{mod}\:\mathrm{4}\:=\:? \\ $$

Question Number 15384    Answers: 1   Comments: 0

If a flagstaff subtends equal angles at 4 points A, B, C and D on the horizontal plane through the foot of the flagstaff, then A, B, C and D must be the vertices of (1) Square (2) Cyclic quadrilateral (3) Rectangle (4) Parallelogram

$$\mathrm{If}\:\mathrm{a}\:\mathrm{flagstaff}\:\mathrm{subtends}\:\mathrm{equal}\:\mathrm{angles}\:\mathrm{at}\:\mathrm{4} \\ $$$$\mathrm{points}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{on}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{plane}\:\mathrm{through}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{flagstaff}, \\ $$$$\mathrm{then}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{must}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{vertices}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Square} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Cyclic}\:\mathrm{quadrilateral} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Rectangle} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Parallelogram} \\ $$

Question Number 15418    Answers: 2   Comments: 0

A grasshopper can jump a maximum horizontal distance of 40 cm. If it spends negligible time on the ground then in this case its speed along the horizontal road will be?

$$\mathrm{A}\:\mathrm{grasshopper}\:\mathrm{can}\:\mathrm{jump}\:\mathrm{a}\:\mathrm{maximum} \\ $$$$\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{40}\:\mathrm{cm}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{spends}\:\mathrm{negligible}\:\mathrm{time}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\mathrm{then}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{its}\:\mathrm{speed}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{road}\:\mathrm{will}\:\mathrm{be}? \\ $$

Question Number 15358    Answers: 1   Comments: 0

If a^→ , b^→ , c^→ are mutually perpendicular vectors of equal magnitudes, show that the vector a^→ + b^→ + c^→ is equally inclined to a^→ , b^→ and c^→ .

$$\mathrm{If}\:\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}},\:\overset{\rightarrow} {{c}}\:\mathrm{are}\:\mathrm{mutually}\:\mathrm{perpendicular} \\ $$$$\mathrm{vectors}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitudes},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:+\:\overset{\rightarrow} {{b}}\:+\:\overset{\rightarrow} {{c}}\:\mathrm{is}\:\mathrm{equally}\:\mathrm{inclined} \\ $$$$\mathrm{to}\:\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}}\:\mathrm{and}\:\overset{\rightarrow} {{c}}\:. \\ $$

Question Number 15352    Answers: 0   Comments: 4

Let a^→ = i^∧ + 4j^∧ + 2k^∧ , b^→ = 3i^∧ − 2j^∧ + 7k^∧ and c^→ = 2i^∧ − j^∧ + 4k^∧ . Find a vector d^→ which is perpendicular to both a^→ and b^→ , and c^→ ∙ d^→ = 15.

$$\mathrm{Let}\:\overset{\rightarrow} {{a}}\:=\:\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}\:+\:\mathrm{2}\overset{\wedge} {{k}},\:\overset{\rightarrow} {{b}}\:=\:\mathrm{3}\overset{\wedge} {{i}}\:−\:\mathrm{2}\overset{\wedge} {{j}}\:+\:\mathrm{7}\overset{\wedge} {{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:=\:\mathrm{2}\overset{\wedge} {{i}}\:−\:\overset{\wedge} {{j}}\:+\:\mathrm{4}\overset{\wedge} {{k}}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{vector}\:\overset{\rightarrow} {{d}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{both}\:\overset{\rightarrow} {{a}}\:\mathrm{and}\:\overset{\rightarrow} {{b}}, \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:\centerdot\:\overset{\rightarrow} {{d}}\:=\:\mathrm{15}. \\ $$

Question Number 15348    Answers: 2   Comments: 0

Solve the equation z^2 + 2(1 + j)z + 2 = 0, Givin each result in form a + jb with a and b correct to 2dp.

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{j}\right)\mathrm{z}\:+\:\mathrm{2}\:=\:\mathrm{0},\:\mathrm{Givin}\:\mathrm{each}\:\mathrm{result}\:\mathrm{in}\:\mathrm{form}\:\:\mathrm{a}\:+\:\mathrm{jb} \\ $$$$\mathrm{with}\:\:\mathrm{a}\:\:\mathrm{and}\:\:\mathrm{b}\:\:\mathrm{correct}\:\mathrm{to}\:\:\mathrm{2dp}. \\ $$

Question Number 15343    Answers: 0   Comments: 0

Question Number 15344    Answers: 0   Comments: 0

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