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Question Number 18722 Answers: 0 Comments: 0

Question Number 18715 Answers: 1 Comments: 0

∣x + 1∣

$$\mid\mathrm{x}\:+\:\mathrm{1}\mid \\ $$

Question Number 18712 Answers: 0 Comments: 0

Question Number 18704 Answers: 1 Comments: 0

Question Number 18702 Answers: 1 Comments: 1

If a sin^2 x+b cos^2 x = c, b sin^2 y+a cos^2 y = d and a tan^2 x = b tan y then (a^2 /b^2 ) is equal to

$$\mathrm{If}\:\:\:{a}\:\mathrm{sin}^{\mathrm{2}} {x}+{b}\:\mathrm{cos}^{\mathrm{2}} {x}\:=\:{c},\: \\ $$$$\:\:\:\:\:\:\:{b}\:\mathrm{sin}^{\mathrm{2}} {y}+{a}\:\mathrm{cos}^{\mathrm{2}} {y}\:=\:{d}\:\:\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:{a}\:\mathrm{tan}^{\mathrm{2}} {x}\:=\:{b}\:\mathrm{tan}\:{y}\:\: \\ $$$$\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18701 Answers: 1 Comments: 1

The value of tan 6° tan 42° tan 66° tan 78° is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\mathrm{tan}\:\mathrm{6}°\:\mathrm{tan}\:\mathrm{42}°\:\mathrm{tan}\:\mathrm{66}°\:\mathrm{tan}\:\mathrm{78}°\:\:\mathrm{is} \\ $$

Question Number 18695 Answers: 1 Comments: 0

if ((n tanθ)/(cos^2 (α−θ)))=((m tan(α−θ))/(cos^2 θ)) then show that 2θ=α−tan^(−1) (((n−m)/(n+m))tanα)

$$\mathrm{if}\:\:\frac{\mathrm{n}\:\mathrm{tan}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\theta\right)}=\frac{\mathrm{m}\:\mathrm{tan}\left(\alpha−\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}\theta=\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\mathrm{tan}\alpha\right) \\ $$

Question Number 18881 Answers: 0 Comments: 0

If (6 (√6) +14)^(2n+1) = m and if f is the fractional part of m, then f m is equal to

$$\mathrm{If}\:\left(\mathrm{6}\:\sqrt{\mathrm{6}}\:+\mathrm{14}\right)^{\mathrm{2}{n}+\mathrm{1}} =\:{m}\:\mathrm{and}\:\mathrm{if}\:{f}\:\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{m},\:\mathrm{then}\:{f}\:{m}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18693 Answers: 1 Comments: 0

prove that 2tan^(−1) [tan(α/2)tan((π/4)−(β/2))] =tan^(−1) (((sinα cosβ)/(cosα +sinβ)))

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$

Question Number 18967 Answers: 0 Comments: 3

Let PQRS be a rectangle such that PQ = a and QR = b. Suppose r_1 is the radius of the circle passing through P and Q and touching RS and r_2 is the radius of the circle passing through Q and R and touching PS. Show that : 5(a + b) ≤ 8(r_1 + r_2 )

$$\mathrm{Let}\:\mathrm{PQRS}\:\mathrm{be}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{PQ}\:=\:{a}\:\mathrm{and}\:\mathrm{QR}\:=\:{b}.\:\mathrm{Suppose}\:\mathrm{r}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{P} \\ $$$$\mathrm{and}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{touching}\:\mathrm{RS}\:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{Q} \\ $$$$\mathrm{and}\:\mathrm{R}\:\mathrm{and}\:\mathrm{touching}\:\mathrm{PS}.\:\mathrm{Show}\:\mathrm{that}\:: \\ $$$$\mathrm{5}\left({a}\:+\:{b}\right)\:\leqslant\:\mathrm{8}\left(\mathrm{r}_{\mathrm{1}} \:+\:\mathrm{r}_{\mathrm{2}} \right) \\ $$

Question Number 18681 Answers: 1 Comments: 0

y = ∣sin x∣ + 2 y = ∣x∣ + 2 −π −π ≤ x ≤ π Find the area that have created from the equations above

$${y}\:=\:\mid\mathrm{sin}\:{x}\mid\:+\:\mathrm{2} \\ $$$${y}\:=\:\mid{x}\mid\:+\:\mathrm{2}\:−\pi \\ $$$$−\pi\:\leqslant\:{x}\:\leqslant\:\pi \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{that}\:\mathrm{have}\:\mathrm{created} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{above} \\ $$

Question Number 18678 Answers: 2 Comments: 0

lim_(x→∞) ((4^(x + 1) + 2^(x +1) − 3^(x + 1) )/(4^(x − 1) + 2^(x − 1 ) + 3^(x + 1) ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4}^{{x}\:+\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:+\mathrm{1}} \:−\:\mathrm{3}^{{x}\:+\:\mathrm{1}} }{\mathrm{4}^{{x}\:−\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:−\:\mathrm{1}\:} +\:\mathrm{3}^{{x}\:+\:\mathrm{1}} \:} \\ $$

Question Number 18667 Answers: 2 Comments: 0

If x=2^(1/3) − 2^(−1/3) , find the value of 2x^3 +6x.

$$\mathrm{If}\:{x}=\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\:\mathrm{2}^{−\mathrm{1}/\mathrm{3}} ,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\mathrm{2}{x}^{\mathrm{3}} +\mathrm{6}{x}. \\ $$

Question Number 18666 Answers: 1 Comments: 0

An elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. what is the Original length of the string ?

$$\mathrm{An}\:\mathrm{elastic}\:\mathrm{material}\:\mathrm{has}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\:\mathrm{36cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{40N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}\:\mathrm{and} \\ $$$$\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{45cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{60N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Original}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:? \\ $$

Question Number 18884 Answers: 1 Comments: 0

Determine the smallest positive integer x, whose last digit is 6 and if we erase this 6 and put it in left most of the number so obtained, the number becomes 4x.

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive}\:\mathrm{integer} \\ $$$$\mathrm{x},\:\mathrm{whose}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{6}\:\mathrm{and}\:\mathrm{if}\:\mathrm{we}\:\mathrm{erase} \\ $$$$\mathrm{this}\:\mathrm{6}\:\mathrm{and}\:\mathrm{put}\:\mathrm{it}\:\mathrm{in}\:\mathrm{left}\:\mathrm{most}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{so}\:\mathrm{obtained},\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{becomes}\:\mathrm{4x}. \\ $$

Question Number 18663 Answers: 1 Comments: 2

Find the product of 101 × 10001 × 100000001 × ... × (1000...01) where the last factor has 2^7 − 1 zeros between the ones. Find the number of ones in the product.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{101}\:×\:\mathrm{10001}\:× \\ $$$$\mathrm{100000001}\:×\:...\:×\:\left(\mathrm{1000}...\mathrm{01}\right)\:\mathrm{where}\:\mathrm{the} \\ $$$$\mathrm{last}\:\mathrm{factor}\:\mathrm{has}\:\mathrm{2}^{\mathrm{7}} \:−\:\mathrm{1}\:\mathrm{zeros}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{ones}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ones}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$

Question Number 18968 Answers: 1 Comments: 1

Find the side lengths of a triangle if side lengths are consecutive integers,and one of whose angles is twice as large as another.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{if}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{are}\:\mathrm{consecutive}\: \\ $$$$\mathrm{integers},\mathrm{and}\:\mathrm{one}\:\mathrm{of}\:\mathrm{whose}\:\mathrm{angles} \\ $$$$\mathrm{is}\:\mathrm{twice}\:\mathrm{as}\:\mathrm{large}\:\mathrm{as}\:\mathrm{another}. \\ $$

Question Number 18655 Answers: 0 Comments: 3

Find the number of odd integers between 30,000 and 80,000 in which no digit is repeated.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{integers} \\ $$$$\mathrm{between}\:\mathrm{30},\mathrm{000}\:\mathrm{and}\:\mathrm{80},\mathrm{000}\:\mathrm{in}\:\mathrm{which}\:\mathrm{no} \\ $$$$\mathrm{digit}\:\mathrm{is}\:\mathrm{repeated}. \\ $$

Question Number 18653 Answers: 1 Comments: 0

Solve the inequality, ∣x − 1∣ + ∣x + 1∣ < 4

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{inequality},\:\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{4} \\ $$

Question Number 18652 Answers: 1 Comments: 0

Show that for any natural number n, the fraction ((21n + 4)/(14n + 3)) is in its lowest term.