Question and Answers Forum

AllQuestion and Answers: Page 1950

Question Number 13152 Answers: 1 Comments: 0

Solve: 5^(log(x)) + logx^5 = 25

$$\mathrm{Solve}:\:\:\mathrm{5}^{\mathrm{log}\left(\mathrm{x}\right)} \:+\:\mathrm{logx}^{\mathrm{5}} \:=\:\mathrm{25} \\ $$

Question Number 13151 Answers: 1 Comments: 0

Question Number 13145 Answers: 1 Comments: 0

((64))^(1/(3)^(1/(√5)) ) = x How to write x into fraction exponent form?

$$\sqrt[{\sqrt[{\sqrt{\mathrm{5}}}]{\mathrm{3}}}]{\mathrm{64}}\:=\:{x} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{write}\:{x}\:\mathrm{into}\:\mathrm{fraction}\:\mathrm{exponent}\:\mathrm{form}? \\ $$

Question Number 13143 Answers: 0 Comments: 1

Find the point (x,y) which lies 8 unit from the origin, along the terminal line of 155°.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{which}\:\mathrm{lies}\:\mathrm{8}\:\mathrm{unit}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin},\:\:\mathrm{along}\:\mathrm{the}\:\mathrm{terminal}\:\mathrm{line} \\ $$$$\mathrm{of}\:\mathrm{155}°.\: \\ $$

Question Number 13142 Answers: 0 Comments: 1

Find the height PQ of a tower of an observant at a point O, 135 m from the foot of the tower. Determine the angle of elevation of the tower.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{height}\:\mathrm{PQ}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{of}\:\mathrm{an}\:\mathrm{observant}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{O},\:\mathrm{135}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}.\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}. \\ $$

Question Number 13141 Answers: 1 Comments: 3

Question Number 13140 Answers: 2 Comments: 1

Question Number 13223 Answers: 0 Comments: 1

If a, b, c are sides of triangle show that (1 + ((b−c)/a))^a (1 + ((c−a)/b))^b (1 + ((a−b)/c))^c < 1

$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\:+\:\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}\:+\:\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}\:+\:\frac{{a}−{b}}{{c}}\right)^{{c}} \:<\:\mathrm{1} \\ $$

Question Number 13128 Answers: 0 Comments: 0

If [x] stands for the greatest integer function, the value of ∫_( 4) ^( 10) (([x^2 ])/([x^2 −28x+196]+[x^2 ])) dx is

$$\mathrm{If}\:\:\left[{x}\right]\:\mathrm{stands}\:\mathrm{for}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer} \\ $$$$\mathrm{function},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\: \\ $$$$\underset{\:\mathrm{4}} {\overset{\:\:\:\:\mathrm{10}} {\int}}\:\frac{\left[{x}^{\mathrm{2}} \right]}{\left[{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{196}\right]+\left[{x}^{\mathrm{2}} \right]}\:{dx}\:\mathrm{is} \\ $$

Question Number 13127 Answers: 0 Comments: 0

Question Number 13121 Answers: 1 Comments: 0

Find the value of : (2/(15)) + (2/(35)) + (2/(63)) + (2/(99)) + ... + (2/(9999))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::\:\:\frac{\mathrm{2}}{\mathrm{15}}\:+\:\frac{\mathrm{2}}{\mathrm{35}}\:+\:\frac{\mathrm{2}}{\mathrm{63}}\:+\:\frac{\mathrm{2}}{\mathrm{99}}\:+\:...\:+\:\frac{\mathrm{2}}{\mathrm{9999}} \\ $$

Question Number 13117 Answers: 1 Comments: 0

Question Number 13107 Answers: 1 Comments: 0

A man moves 20m north then 12m east and finally 15m south. His displacement from the starting point is ?

$$\mathrm{A}\:\mathrm{man}\:\mathrm{moves}\:\mathrm{20m}\:\mathrm{north}\:\mathrm{then}\:\mathrm{12m}\:\mathrm{east}\:\mathrm{and}\:\mathrm{finally}\:\mathrm{15m}\:\mathrm{south}.\:\:\mathrm{His}\:\mathrm{displacement} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:? \\ $$

Question Number 13103 Answers: 2 Comments: 0

∫_( −1) ^2 (( ∣ x ∣ )/x) dx =

$$\:\underset{\:−\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{\:\mid\:{x}\:\mid\:}{{x}}\:{dx}\:=\: \\ $$

Question Number 13102 Answers: 0 Comments: 4

Find the sum of the nth term : 1^6 + 2^6 + 3^6 + 4^6 + ... + n^6

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\::\:\:\mathrm{1}^{\mathrm{6}} \:+\:\mathrm{2}^{\mathrm{6}} \:+\:\mathrm{3}^{\mathrm{6}} \:+\:\mathrm{4}^{\mathrm{6}} \:+\:...\:+\:\mathrm{n}^{\mathrm{6}} \\ $$

Question Number 13099 Answers: 2 Comments: 0

If f(x + 5) = g(2x −1) Find 2f^(−1) (x) (A) g^(−1) (x) + 11 (D) g^(−1) (x/2) + 6 (B) g^(−1) (x) + 9 (E) g^(−1) (2x) + 6 (C) g^(−1) (x) + 6

$$\mathrm{If}\:{f}\left({x}\:+\:\mathrm{5}\right)\:=\:{g}\left(\mathrm{2}{x}\:−\mathrm{1}\right) \\ $$$$\mathrm{Find}\:\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{11}\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{g}^{−\mathrm{1}} \left({x}/\mathrm{2}\right)\:+\:\mathrm{6} \\ $$$$\left(\mathrm{B}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:{g}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\:+\:\mathrm{6} \\ $$$$\left(\mathrm{C}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{6} \\ $$

Question Number 13098 Answers: 1 Comments: 0

Question Number 13097 Answers: 1 Comments: 0

S=Σ_(x_2 =1) ^x_1 Σ_(x_3 =1) ^x_2 ∙∙∙Σ_(x_n =1) ^x_(n−1) Σ_(t=1) ^x_n t Can you evaluate S?

$${S}=\underset{{x}_{\mathrm{2}} =\mathrm{1}} {\overset{{x}_{\mathrm{1}} } {\sum}}\underset{{x}_{\mathrm{3}} =\mathrm{1}} {\overset{{x}_{\mathrm{2}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t} \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}\:{S}? \\ $$

Question Number 13091 Answers: 1 Comments: 0

A motor car moves with a velocity of 20m/s on a rough horizontal road and covers a displacement of 50m. Find the coefficient of dynamic friction between the tyre and the ground (g = 10m/s^2 ).

$$\mathrm{A}\:\mathrm{motor}\:\mathrm{car}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{20m}/\mathrm{s}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\:\mathrm{horizontal}\:\mathrm{road}\:\mathrm{and} \\ $$$$\mathrm{covers}\:\mathrm{a}\:\mathrm{displacement}\:\mathrm{of}\:\mathrm{50m}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{dynamic}\:\mathrm{friction}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{tyre}\:\mathrm{and}\:\mathrm{the}\:\mathrm{ground}\:\:\left(\mathrm{g}\:=\:\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right). \\ $$

Question Number 13081 Answers: 2 Comments: 0

{ ((x+y+z=[1]_5 )),((xy=[2]_5 )),((yz=[1]_5 )) :} Solve system on Z_5

$$\begin{cases}{{x}+{y}+{z}=\left[\mathrm{1}\right]_{\mathrm{5}} }\\{{xy}=\left[\mathrm{2}\right]_{\mathrm{5}} }\\{{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} }\end{cases} \\ $$$${Solve}\:{system}\:{on}\:\mathbb{Z}_{\mathrm{5}} \\ $$

Question Number 13075 Answers: 0 Comments: 5

S(x) is the sum of 49 terms of AP The first term is (1/2)x^3 and the difference is (7 − x) If S(x) maximum, the value of 10^(th) term is ...

$${S}\left({x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{49}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{AP} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{term}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{is}\:\left(\mathrm{7}\:−\:{x}\right) \\ $$$$\mathrm{If}\:{S}\left({x}\right)\:\mathrm{maximum},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{10}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{is}\:... \\ $$

Question Number 13059 Answers: 2 Comments: 0

please help for ∫_(−3π ) ^( 3π) sin^(2009) x dx

$$\mathrm{please}\:\mathrm{help}\:\mathrm{for} \\ $$$$\int_{−\mathrm{3}\pi\:} ^{\:\:\:\mathrm{3}\pi} \mathrm{sin}^{\mathrm{2009}} \mathrm{x}\:\mathrm{dx} \\ $$

Question Number 13054 Answers: 1 Comments: 0

An object is placed between a converging lens and a plane mirror. Explain how two real images of the object may be produced by the system. If the focal length of the lens is 15cm and the object is 20cm from both the lens and the mirror.Calculate the distance of the two images from the lens.. pls help me with this....

Question Number 13049 Answers: 1 Comments: 0

Two charges q_1 = 10 μC and q_2 = 5 μC are placed on the axis at A (10, 0) cm and (20, 0) cm respectively. Determine a position between the two charges where the electric field intensity is 0.

$$\mathrm{Two}\:\mathrm{charges}\:\mathrm{q}_{\mathrm{1}} \:=\:\mathrm{10}\:\mu\mathrm{C}\:\mathrm{and}\:\:\mathrm{q}_{\mathrm{2}} \:=\:\mathrm{5}\:\mu\mathrm{C}\:\:\mathrm{are}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{A}\:\left(\mathrm{10},\:\mathrm{0}\right)\:\mathrm{cm} \\ $$$$\mathrm{and}\:\left(\mathrm{20},\:\mathrm{0}\right)\:\mathrm{cm}\:\:\mathrm{respectively}.\:\mathrm{Determine}\:\mathrm{a}\:\mathrm{position}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{charges} \\ $$$$\mathrm{where}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{field}\:\mathrm{intensity}\:\mathrm{is}\:\mathrm{0}.\: \\ $$

Question Number 13068 Answers: 2 Comments: 0

If y = (x)^(1/3) Find (dy/dx) from the first principle

$$\mathrm{If}\:\:\:\mathrm{y}\:=\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:\:\:\:\:\mathrm{Find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{principle} \\ $$

Question Number 13067 Answers: 0 Comments: 4

Pg 1945 Pg 1946 Pg 1947 Pg 1948 Pg 1949 Pg 1950 Pg 1951 Pg 1952 Pg 1953 Pg 1954

Contact: info@tinkutara.com