Question and Answers Forum

AllQuestion and Answers: Page 1801

Question Number 24696 Answers: 0 Comments: 7

App Update App has been updated with the following changes • Long image upload issue with submit button • Added an option to rotate image while uploading • Added ability to set a password for reusing id across devices • line thickness for square root • Some minor cosmetic changes (font size, colors, menus texts)

$$\boldsymbol{\mathrm{App}}\:\boldsymbol{\mathrm{Update}} \\ $$$$\mathrm{App}\:\mathrm{has}\:\mathrm{been}\:\mathrm{updated}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{changes} \\ $$$$\bullet\:\mathrm{Long}\:\mathrm{image}\:\mathrm{upload}\:\mathrm{issue}\:\mathrm{with} \\ $$$$\:\:\:\:\mathrm{submit}\:\mathrm{button} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{an}\:\mathrm{option}\:\mathrm{to}\:\mathrm{rotate}\:\mathrm{image} \\ $$$$\:\:\:\:\mathrm{while}\:\mathrm{uploading} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{ability}\:\mathrm{to}\:\mathrm{set}\:\mathrm{a}\:\mathrm{password}\:\mathrm{for} \\ $$$$\:\:\:\:\mathrm{reusing}\:\mathrm{id}\:\mathrm{across}\:\mathrm{devices} \\ $$$$\bullet\:\mathrm{line}\:\mathrm{thickness}\:\mathrm{for}\:\mathrm{square}\:\mathrm{root} \\ $$$$\bullet\:\mathrm{Some}\:\mathrm{minor}\:\mathrm{cosmetic}\:\mathrm{changes} \\ $$$$\:\:\:\left(\mathrm{font}\:\mathrm{size},\:\mathrm{colors},\:\mathrm{menus}\:\mathrm{texts}\right) \\ $$

Question Number 24684 Answers: 0 Comments: 0

Let ABCD be a square and M, N points on sides AB, BC respectably, such that ∠MDN = 45°. If R is the midpoint of MN show that RP = RQ where P, Q are the points of intersection of AC with the lines MD, ND.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square}\:\mathrm{and}\:{M},\:{N}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{sides}\:{AB},\:{BC}\:\mathrm{respectably},\:\mathrm{such}\:\mathrm{that} \\ $$$$\angle{MDN}\:=\:\mathrm{45}°.\:\mathrm{If}\:{R}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$${MN}\:\mathrm{show}\:\mathrm{that}\:{RP}\:=\:{RQ}\:\mathrm{where}\:{P},\:{Q} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{AC}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{lines}\:{MD},\:{ND}. \\ $$

Question Number 24680 Answers: 1 Comments: 2

Question Number 24677 Answers: 2 Comments: 0

Let T_k is the k^(th) term and S_k is the sum of the first k term in arithmetic progression If T_3 + T_6 + T_9 + T_(12) + T_(15) + T_(18) = 45 Find S_(20)

$$\mathrm{Let}\:{T}_{{k}} \:\mathrm{is}\:\mathrm{the}\:{k}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{and}\:\mathrm{S}_{{k}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:{k}\:\mathrm{term}\:\mathrm{in}\:\mathrm{arithmetic}\:\mathrm{progression} \\ $$$$\mathrm{If}\:{T}_{\mathrm{3}} \:+\:{T}_{\mathrm{6}} \:+\:{T}_{\mathrm{9}} \:+\:{T}_{\mathrm{12}} \:+\:{T}_{\mathrm{15}} \:+\:{T}_{\mathrm{18}} \:=\:\mathrm{45} \\ $$$$\mathrm{Find}\:{S}_{\mathrm{20}} \\ $$

Question Number 24670 Answers: 1 Comments: 0

Question Number 24667 Answers: 0 Comments: 0

Question Number 24663 Answers: 1 Comments: 0

solve ∣x−3∣=∣3x+2∣−1

$$\boldsymbol{{solve}}\:\mid{x}−\mathrm{3}\mid=\mid\mathrm{3}{x}+\mathrm{2}\mid−\mathrm{1} \\ $$

Question Number 24661 Answers: 1 Comments: 2

Question Number 24662 Answers: 1 Comments: 0

solve ∣x^2 −4x−5∣=7

$$\boldsymbol{{solve}}\:\mid{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\mid=\mathrm{7} \\ $$$$ \\ $$

Question Number 24651 Answers: 2 Comments: 0

∫sin x+cos y dx

$$\int\mathrm{sin}\:{x}+\mathrm{cos}\:{y}\:{dx} \\ $$

Question Number 24647 Answers: 0 Comments: 0

Question Number 24643 Answers: 1 Comments: 0

Question Number 24635 Answers: 0 Comments: 0

Calculate the electric potential at a point P at a distance of 3m of either charges of +20 μC and − 15μC. which are 25cm apart. Also calculate potential energy of a +3.5μC placed at point P.

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{potential}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{P}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{3m}\:\mathrm{of}\:\mathrm{either} \\ $$$$\mathrm{charges}\:\mathrm{of}\:\:+\mathrm{20}\:\mu\mathrm{C}\:\mathrm{and}\:\:−\:\mathrm{15}\mu\mathrm{C}.\:\:\mathrm{which}\:\mathrm{are}\:\mathrm{25cm}\:\mathrm{apart}. \\ $$$$\mathrm{Also}\:\mathrm{calculate}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{a}\:\:+\mathrm{3}.\mathrm{5}\mu\mathrm{C}\:\mathrm{placed}\:\mathrm{at}\:\mathrm{point}\:\mathrm{P}. \\ $$

Question Number 24644 Answers: 2 Comments: 0

The density of a non-uniform rod of length 1 m is given by ρ(x) = a(1 + bx^2 ) where a and b are constants and 0 ≤ x ≤ 1. The centre of mass of the rod will be at (1) ((3(2 + b))/(4(3 + b))) (2) ((4(2 + b))/(3(3 + b))) (3) ((3(3 + b))/(4(2 + b))) (4) ((4(3 + b))/(3(2 + b)))

$$\mathrm{The}\:\mathrm{density}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{uniform}\:\mathrm{rod}\:\mathrm{of} \\ $$$$\mathrm{length}\:\mathrm{1}\:\mathrm{m}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\rho\left({x}\right)\:=\:{a}\left(\mathrm{1}\:+\:{bx}^{\mathrm{2}} \right) \\ $$$$\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{1}.\:\mathrm{The}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rod} \\ $$$$\mathrm{will}\:\mathrm{be}\:\mathrm{at} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{3}\left(\mathrm{2}\:+\:{b}\right)}{\mathrm{4}\left(\mathrm{3}\:+\:{b}\right)} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{4}\left(\mathrm{2}\:+\:{b}\right)}{\mathrm{3}\left(\mathrm{3}\:+\:{b}\right)} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{3}\left(\mathrm{3}\:+\:{b}\right)}{\mathrm{4}\left(\mathrm{2}\:+\:{b}\right)} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{4}\left(\mathrm{3}\:+\:{b}\right)}{\mathrm{3}\left(\mathrm{2}\:+\:{b}\right)} \\ $$

Question Number 24631 Answers: 1 Comments: 0