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Question Number 13606    Answers: 1   Comments: 0

Show that 19^(93) − 13^(99) is a positive integer divisible by 162.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{19}^{\mathrm{93}} \:−\:\mathrm{13}^{\mathrm{99}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{162}. \\ $$

Question Number 13732    Answers: 1   Comments: 3

Sum the following: tan x+2tan 2x+2^2 tan 2^2 x+...+2^n tan 2^n x

$$\mathrm{Sum}\:\mathrm{the}\:\mathrm{following}: \\ $$$$\mathrm{tan}\:{x}+\mathrm{2tan}\:\mathrm{2}{x}+\mathrm{2}^{\mathrm{2}} \mathrm{tan}\:\mathrm{2}^{\mathrm{2}} {x}+...+\mathrm{2}^{{n}} \mathrm{tan}\:\mathrm{2}^{{n}} {x} \\ $$

Question Number 13601    Answers: 1   Comments: 0

Let f : R − {(3/5)} → R be defined by f(x) = ((3x + 2)/(5x − 3)) . Then, (a) f^(−1) (x) = x (b) f^(−1) (x) = −f(x) (c) fof(x) = −x (d) f^(−1) (x) = (1/(19))f(x)

$$\mathrm{Let}\:{f}\::\:\mathbb{R}\:−\:\left\{\frac{\mathrm{3}}{\mathrm{5}}\right\}\:\rightarrow\:\mathbb{R}\:\mathrm{be}\:\mathrm{defined}\:\mathrm{by} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{3}{x}\:+\:\mathrm{2}}{\mathrm{5}{x}\:−\:\mathrm{3}}\:.\:\mathrm{Then}, \\ $$$$\left(\mathrm{a}\right)\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:{x} \\ $$$$\left(\mathrm{b}\right)\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:−{f}\left({x}\right) \\ $$$$\left(\mathrm{c}\right)\:{fof}\left({x}\right)\:=\:−{x} \\ $$$$\left(\mathrm{d}\right)\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{19}}{f}\left({x}\right) \\ $$

Question Number 13622    Answers: 1   Comments: 0

Evaluate: ∫_0 ^(2π) e^(x/2) sin ((x/2) + (π/4))dx

$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{\frac{{x}}{\mathrm{2}}} \mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right){dx} \\ $$

Question Number 13623    Answers: 1   Comments: 0

Evaluate: ∫_0 ^(2π) e^x cos ((π/4) + (x/2))dx

$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{{x}} \:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\:+\:\frac{{x}}{\mathrm{2}}\right){dx} \\ $$

Question Number 13598    Answers: 0   Comments: 1

If g(x) = x^2 + x − 2 and (1/2) gof(x) = 2x^2 − 5x + 2, then prove that f(x) = 2x − 3.

$$\mathrm{If}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+\:{x}\:−\:\mathrm{2}\:\mathrm{and} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{gof}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{f}\left({x}\right)\:=\:\mathrm{2}{x}\:−\:\mathrm{3}. \\ $$

Question Number 13595    Answers: 1   Comments: 0

If f : R → (−1, 1) is defined by f(x) = ((−x∣x∣)/(1 + x^2 )) , then prove that f^(−1) (x) = −sgn(x)(√((∣x∣)/(1 − ∣x∣)))

$$\mathrm{If}\:{f}\::\:\mathbb{R}\:\rightarrow\:\left(−\mathrm{1},\:\mathrm{1}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$${f}\left({x}\right)\:=\:\frac{−{x}\mid{x}\mid}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:,\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:−\mathrm{sgn}\left({x}\right)\sqrt{\frac{\mid{x}\mid}{\mathrm{1}\:−\:\mid{x}\mid}} \\ $$

Question Number 13591    Answers: 1   Comments: 1

for: x^2 +(y−1)^2 =1 show that it can be written as: r=2sin(θ)

$$\mathrm{for}: \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}: \\ $$$${r}=\mathrm{2sin}\left(\theta\right) \\ $$

Question Number 13584    Answers: 2   Comments: 10

Calculate 19^(93) (mod 81).

$$\mathrm{Calculate}\:\mathrm{19}^{\mathrm{93}} \:\left(\mathrm{mod}\:\mathrm{81}\right). \\ $$

Question Number 13570    Answers: 1   Comments: 0

A motor boat going downstream overcomes a float at a point A. 60 minutes later it turns and after some time passes the float at a distance of 12 km from the point A. Find the velocity of the stream (assuming constant velocity for the boat in still water)

$$\mathrm{A}\:\mathrm{motor}\:\mathrm{boat}\:\mathrm{going}\:\mathrm{downstream} \\ $$$$\mathrm{overcomes}\:\mathrm{a}\:\mathrm{float}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:{A}.\:\mathrm{60}\:\mathrm{minutes} \\ $$$$\mathrm{later}\:\mathrm{it}\:\mathrm{turns}\:\mathrm{and}\:\mathrm{after}\:\mathrm{some}\:\mathrm{time}\:\mathrm{passes} \\ $$$$\mathrm{the}\:\mathrm{float}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{12}\:\mathrm{km}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{point}\:{A}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{stream} \\ $$$$\left(\mathrm{assuming}\:\mathrm{constant}\:\mathrm{velocity}\:\mathrm{for}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{boat}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}\right) \\ $$

Question Number 13563    Answers: 1   Comments: 3

Question Number 13544    Answers: 0   Comments: 6

Question Number 13529    Answers: 0   Comments: 7

Let p be a prime number > 3. What is the remainder when p^2 is divided by 12?

$$\mathrm{Let}\:{p}\:\mathrm{be}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:>\:\mathrm{3}.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:{p}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\mathrm{12}? \\ $$

Question Number 13578    Answers: 4   Comments: 0

Question Number 13514    Answers: 1   Comments: 21

From Developer We are planning an update to this app in next 4−6 weeks. Please provide your input on problems/suggestion for improvement on this app. You can give feedback as a comment to this post or email us at infoattinkutara.com.

$$\mathrm{From}\:\mathrm{Developer} \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{planning}\:\mathrm{an}\:\mathrm{update}\:\mathrm{to} \\ $$$$\mathrm{this}\:\mathrm{app}\:\mathrm{in}\:\mathrm{next}\:\mathrm{4}−\mathrm{6}\:\mathrm{weeks}. \\ $$$$\mathrm{Please}\:\mathrm{provide}\:\mathrm{your}\:\mathrm{input}\:\mathrm{on} \\ $$$$\mathrm{problems}/\mathrm{suggestion}\:\mathrm{for}\: \\ $$$$\mathrm{improvement}\:\mathrm{on}\:\mathrm{this}\:\mathrm{app}. \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{give}\:\mathrm{feedback}\:\mathrm{as}\:\mathrm{a}\:\mathrm{comment} \\ $$$$\mathrm{to}\:\mathrm{this}\:\mathrm{post}\:\mathrm{or}\:\mathrm{email}\:\mathrm{us}\:\mathrm{at} \\ $$$$\mathrm{infoattinkutara}.\mathrm{com}. \\ $$

Question Number 13508    Answers: 1   Comments: 0

5!! = ? please workings, how is the answer 15

$$\mathrm{5}!!\:=\:? \\ $$$$\mathrm{please}\:\mathrm{workings},\:\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{15} \\ $$

Question Number 13501    Answers: 1   Comments: 0

Find x 10_C_x = 5_C_2

$$\mathrm{Find}\:\mathrm{x} \\ $$$$\mathrm{10}_{\mathrm{C}_{\mathrm{x}} } \:=\:\mathrm{5}_{\mathrm{C}_{\mathrm{2}} } \\ $$

Question Number 13498    Answers: 1   Comments: 1

Question Number 13548    Answers: 1   Comments: 11

Question Number 13491    Answers: 1   Comments: 0

Test 1. Solve equation (k^2 −1)x^2 +(k−1)x+(k+1)=0 k∈R (30) 2. Prove ((sin(x))/(1+cos(x)))=((1−cos(x))/(sin(x))) (35) 3.P(x)=−2x^3 −2x^2 −x+2409 Find P(−11) (35) Evaluate other answers and give marks I want to see how math teachers evaluate in other countries Sorry foy my english

$${Test} \\ $$$$\mathrm{1}.\:{Solve}\:{equation} \\ $$$$\left({k}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} +\left({k}−\mathrm{1}\right){x}+\left({k}+\mathrm{1}\right)=\mathrm{0}\:\:{k}\in\mathbb{R} \\ $$$$\left(\mathrm{30}\right) \\ $$$$\mathrm{2}.\:{Prove} \\ $$$$\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}=\frac{\mathrm{1}−{cos}\left({x}\right)}{{sin}\left({x}\right)} \\ $$$$\left(\mathrm{35}\right) \\ $$$$\mathrm{3}.{P}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2409} \\ $$$${Find}\:{P}\left(−\mathrm{11}\right) \\ $$$$\left(\mathrm{35}\right) \\ $$$$ \\ $$$${Evaluate}\:{other}\:{answers}\:{and}\:{give}\:{marks} \\ $$$${I}\:{want}\:{to}\:{see}\:{how}\:{math}\:{teachers}\:{evaluate}\:{in}\:{other}\:{countries} \\ $$$${Sorry}\:{foy}\:{my}\:{english} \\ $$

Question Number 13490    Answers: 1   Comments: 0

The area of a rectangle is 255 m^2 . If its length is decreased by 1 m, it becomes a square. The perimeter of the square is ____ m.

$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{255}\:\mathrm{m}^{\mathrm{2}} .\:\mathrm{If} \\ $$$$\mathrm{its}\:\mathrm{length}\:\mathrm{is}\:\mathrm{decreased}\:\mathrm{by}\:\mathrm{1}\:\mathrm{m},\:\mathrm{it}\: \\ $$$$\mathrm{becomes}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\_\_\_\_\:\mathrm{m}. \\ $$

Question Number 13478    Answers: 1   Comments: 0

The initial velocity of a particle is u (at t = 0) and acceleration f is given by f = at where t is time and ′a′ is a constant. Which of the following relations is valid? (1) v = u + at^2 (2) v = u + (1/2) at^2 (3) v = u + at (4) v = u

$$\mathrm{The}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is}\:{u} \\ $$$$\left(\mathrm{at}\:{t}\:=\:\mathrm{0}\right)\:\mathrm{and}\:\mathrm{acceleration}\:{f}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${f}\:=\:{at}\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{and}\:'{a}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relations}\:\mathrm{is} \\ $$$$\mathrm{valid}? \\ $$$$\left(\mathrm{1}\right)\:{v}\:=\:{u}\:+\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{v}\:=\:{u}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{v}\:=\:{u}\:+\:{at} \\ $$$$\left(\mathrm{4}\right)\:{v}\:=\:{u} \\ $$

Question Number 13475    Answers: 1   Comments: 0

An object starts from rest with constant acceleration 4 m/s^2 , then find the distance travelled by object in 5^(th) half second.

$$\mathrm{An}\:\mathrm{object}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{acceleration}\:\mathrm{4}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{travelled}\:\mathrm{by}\:\mathrm{object}\:\mathrm{in}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{half}\:\mathrm{second}. \\ $$

Question Number 13510    Answers: 1   Comments: 1

A body of mass 2 kg has an initial velocity of 3 ms^(−1) along OE and it is subjected to a force of 4 newton in OF direction perpendicular to OE. The distance of the body from O after 4 second will be (a) 12 m (b) 28 m (c) 20 m (d) 48 m

$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{has}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{3}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{along}\:{OE}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{subjected} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{4}\:\mathrm{newton}\:\mathrm{in}\:{OF}\:\mathrm{direction} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:{OE}.\:\mathrm{The}\:\mathrm{distance}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{body}\:\mathrm{from}\:{O}\:\mathrm{after}\:\mathrm{4}\:\mathrm{second}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{12}\:\mathrm{m} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{28}\:\mathrm{m} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{20}\:\mathrm{m} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{48}\:\mathrm{m} \\ $$

Question Number 13449    Answers: 2   Comments: 0

Four particles A, B, C and D are situated at the corners of a square ABCD of side a at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC, C along CD and D along DA. At what time will these particles meet each other?

$$\mathrm{Four}\:\mathrm{particles}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{are}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{corners}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:{ABCD}\:\mathrm{of}\:\mathrm{side} \\ $$$${a}\:\mathrm{at}\:{t}\:=\:\mathrm{0}.\:\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particles}\:\mathrm{moves} \\ $$$$\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{v}.\:{A}\:\mathrm{always}\:\mathrm{has}\:\mathrm{its} \\ $$$$\mathrm{velocity}\:\mathrm{along}\:{AB},\:{B}\:\mathrm{along}\:{BC},\:{C}\:\mathrm{along} \\ $$$${CD}\:\mathrm{and}\:{D}\:\mathrm{along}\:{DA}.\:\mathrm{At}\:\mathrm{what}\:\mathrm{time}\:\mathrm{will} \\ $$$$\mathrm{these}\:\mathrm{particles}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}? \\ $$

Question Number 13438    Answers: 1   Comments: 0

7x^5 −4x^4 +9x^3 +12x^2 +5x−9=0 How many roots of this equation are Negative?

$$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{9}=\mathrm{0} \\ $$$${How}\:{many}\:{roots}\:{of}\:{this}\:{equation} \\ $$$${are}\:{Negative}? \\ $$$$ \\ $$

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