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Question Number 13888 Answers: 1 Comments: 3

Prove that if A′, B′ and C′ are the midpoints of the sides BC, CA and AB, respectively, then AA′ + BB′ + CC′ < AB + BC + CA

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}, \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$${AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$

Question Number 13886 Answers: 1 Comments: 0

Question Number 13294 Answers: 0 Comments: 3

If a, b and c are the sides of a triangle and a + b + c = 2, then prove that a^2 + b^2 + c^2 + 2abc < 2

$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{abc}\:<\:\mathrm{2} \\ $$

Question Number 13283 Answers: 1 Comments: 0

If f : A → B given by 3^(f(x)) + 2^(−x) = 4 is a bijection, then find A and B if possible.

$$\mathrm{If}\:{f}\::\:{A}\:\rightarrow\:{B}\:\mathrm{given}\:\mathrm{by}\:\mathrm{3}^{{f}\left({x}\right)} \:+\:\mathrm{2}^{−{x}} \:=\:\mathrm{4}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{bijection},\:\mathrm{then}\:\mathrm{find}\:{A}\:\mathrm{and}\:{B}\:\mathrm{if}\:\mathrm{possible}. \\ $$

Question Number 13281 Answers: 1 Comments: 0

Let f(x) = { ((1 + x, 0 ≤ x ≤ 2)),((3 − x, 2 < x ≤ 3)) :} . Find fof.

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{1}\:+\:{x},\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{2}}\\{\mathrm{3}\:−\:{x},\:\mathrm{2}\:<\:{x}\:\leqslant\:\mathrm{3}}\end{cases}\:.\:\mathrm{Find}\:{fof}. \\ $$

Question Number 13276 Answers: 0 Comments: 2

Question Number 17715 Answers: 1 Comments: 0

Two forces 5N and 6N acting at a point are inclined at 60° to each other. Calculate their resultant.

$$\mathrm{Two}\:\mathrm{forces}\:\mathrm{5N}\:\mathrm{and}\:\mathrm{6N}\:\mathrm{acting}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{are}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{60}°\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}.\: \\ $$$$\mathrm{Calculate}\:\mathrm{their}\:\mathrm{resultant}.\: \\ $$

Question Number 13261 Answers: 2 Comments: 0

Balance the equation K_2 CrO_4 + HCl → K_2 Cr_2 O_7 + KCl + H_2 O

$$\mathrm{Balance}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{K}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{HCl}\:\rightarrow\:\mathrm{K}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{KCl}\:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$

Question Number 13260 Answers: 2 Comments: 0

Number of moles of KMnO_4 required to oxidise one mole of Fe(C_2 O_4 ) in acidic medium is (a) 0.6 (b) 1.67 (c) 0.2 (d) 0.4

$$\mathrm{Number}\:\mathrm{of}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{required} \\ $$$$\mathrm{to}\:\mathrm{oxidise}\:\mathrm{one}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{Fe}\left(\mathrm{C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} \right)\:\mathrm{in} \\ $$$$\mathrm{acidic}\:\mathrm{medium}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{1}.\mathrm{67} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{2} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{0}.\mathrm{4} \\ $$

Question Number 13295 Answers: 1 Comments: 3

For n ∈ N, n > 1, show that (1/n) + (1/(n + 1)) + (1/(n + 2)) + ... + (1/n^2 ) > 1

$$\mathrm{For}\:{n}\:\in\:\mathbb{N},\:{n}\:>\:\mathrm{1},\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{n}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{2}}\:+\:...\:+\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:>\:\mathrm{1} \\ $$

Question Number 13257 Answers: 1 Comments: 0

Question Number 13258 Answers: 1 Comments: 0

Question Number 13252 Answers: 1 Comments: 0

lim_(x→0) sin x

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}sin}\:\mathrm{x} \\ $$

Question Number 17714 Answers: 0 Comments: 0

Two boys pull a cart with two a rope along the horizontal. if the two boy exact a force of 50N each. Calculate the resultant force on the cart.

$$\mathrm{Two}\:\mathrm{boys}\:\mathrm{pull}\:\mathrm{a}\:\mathrm{cart}\:\mathrm{with}\:\mathrm{two}\:\mathrm{a}\:\mathrm{rope}\:\mathrm{along}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{two}\:\mathrm{boy} \\ $$$$\mathrm{exact}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{50N}\:\mathrm{each}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{resultant}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the}\:\mathrm{cart}. \\ $$

Question Number 17713 Answers: 0 Comments: 2

Evaluate: (−(√3))^((−(√2)))

$$\mathrm{Evaluate}:\:\:\:\:\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$

Question Number 13249 Answers: 0 Comments: 2

If msinθ = nsin(θ + 2α), then tan(θ + α)cotα equal to [Answer given in my book is ((1 − n)/(1 + n))]

$$\mathrm{If}\:{m}\mathrm{sin}\theta\:=\:{n}\mathrm{sin}\left(\theta\:+\:\mathrm{2}\alpha\right),\:\mathrm{then} \\ $$$$\mathrm{tan}\left(\theta\:+\:\alpha\right)\mathrm{cot}\alpha\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left[\mathrm{Answer}\:\mathrm{given}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{is}\:\frac{\mathrm{1}\:−\:{n}}{\mathrm{1}\:+\:{n}}\right] \\ $$

Question Number 13237 Answers: 1 Comments: 2

Question Number 13236 Answers: 0 Comments: 16

For positive a,b,c such that a b c=1 show that a^(b+c) b^(c+a) c^(a+b) ≤1 solution: a^(b+c) b^(c+a) c^(a+b) =(a^b a^c b^c b^a c^a c^b ) =(b×c)^a (a×c)^b (a×b)^c =(a^0 ×b×c)^a (a×b^0 ×c)^b (a×b×c^0 )^c ≤(a×b×c)^(a ) (a×b^ × c)^b (a×b×c)^c ≤(1)^a (1)^b (1)^c ;since a b c=1 ≤1

$${For}\:{positive}\:\:{a},{b},{c}\:\:{such}\:{that}\:\:{a}\:{b}\:{c}=\mathrm{1} \\ $$$${show}\:{that}\:\:{a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:\leqslant\mathrm{1} \\ $$$${solution}: \\ $$$${a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:=\left({a}^{{b}} {a}^{{c}} \:{b}^{{c}} {b}^{{a}} \:\:{c}^{{a}} {c}^{{b}} \right) \\ $$$$\:\:\:\:\:\:=\left({b}×{c}\right)^{{a}} \:\left({a}×{c}\right)^{{b}} \:\left({a}×{b}\right)^{{c}} \\ $$$$\:\:\:\:\:\:\:=\left({a}^{\mathrm{0}} ×{b}×{c}\right)^{{a}} \:\left({a}×{b}^{\mathrm{0}} ×{c}\right)^{{b}} \left({a}×{b}×{c}^{\mathrm{0}} \right)^{{c}} \\ $$$$\:\:\:\:\:\:\:\leqslant\left({a}×{b}×{c}\right)^{{a}\:\:\:\:} \left({a}×{b}^{} ×\:{c}\right)^{{b}} \:\left({a}×{b}×{c}\right)^{{c}} \\ $$$$\:\:\:\:\:\:\:\leqslant\left(\mathrm{1}\right)^{{a}} \:\left(\mathrm{1}\right)^{{b}} \:\left(\mathrm{1}\right)^{{c}} \:\:\:\:;{since}\:\:{a}\:{b}\:{c}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\leqslant\mathrm{1} \\ $$

Question Number 13234 Answers: 2 Comments: 0

The angle of elevation of the sun is 27°. A man is 180 cm tall . How long is his shadow. Give your answer to the nearest 10cm

$$\mathrm{The}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sun}\:\mathrm{is}\:\mathrm{27}°.\:\:\mathrm{A}\:\mathrm{man}\:\mathrm{is}\:\mathrm{180}\:\mathrm{cm}\:\mathrm{tall}\:.\:\mathrm{How}\:\mathrm{long}\:\mathrm{is} \\ $$$$\mathrm{his}\:\mathrm{shadow}.\:\mathrm{Give}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{10cm} \\ $$

Question Number 13228 Answers: 0 Comments: 5