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Question Number 194252 Answers: 2 Comments: 0

Find V = tan 9°−tan 27°−tan 63°+tan 81°

$$\:\:\mathrm{Find}\:\mathrm{V}\:=\:\mathrm{tan}\:\mathrm{9}°−\mathrm{tan}\:\mathrm{27}°−\mathrm{tan}\:\mathrm{63}°+\mathrm{tan}\:\mathrm{81}° \\ $$

Question Number 194250 Answers: 2 Comments: 0

find the value of a for which the limit lim_(x→0) ((sin (ax)−tan^(−1) (x)−x)/(x^3 +x^4 )) is finite and then evaluate the limit

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{ax}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} }\:\mathrm{is}\:\mathrm{finite}\: \\ $$$$\:\mathrm{and}\:\mathrm{then}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{limit}\: \\ $$

Question Number 194248 Answers: 0 Comments: 0

Gyanashram classes weekly test by−Bittu sir CHEMISTRY TEST Electrochemistry 1. 2. 3. 4. 5. 6. ? 7. 8. 9. 10. ? 5 1. ? 2. objective 1. 1f 2. 127 g cu 3 4. ? 5.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Gyanashram}\:{classes} \\ $$$${weekly}\:{test}\:\:\:\:\:\:\:\:\:\:{by}−{Bittu}\:{sir}\:\: \\ $$$$\:\:\:\mathbb{CHEMISTRY}\:\:\mathbb{TEST} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Electrochemistry} \\ $$$$\mathrm{1}.\: \: \: \: \: \: \: \: \\ $$$$\mathrm{2}.\:\: \: \: \: \: \\ $$$$\:\mathrm{3}. \: \: \: \: \\ $$$$\mathrm{4}.\: \: \: \: \\ $$$$\mathrm{5}. \: \: \: \: \: \\ $$$$\mathrm{6}. \: \: \: \: \: \: \: ? \\ $$$$\mathrm{7}. \: \: \: \\ $$$$\mathrm{8}. \: \: \: \: \: \: \: \: \: \\ $$$$\mathrm{9}. \: \: \: \: \: \\ $$$$\mathrm{10}. \: \: \: ? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\: \: \: \\ $$$$\mathrm{1}.\: \: \: \: ?\: \: \: \: \: \: \\ $$$$ \: \: \: \: \: \\ $$$$\mathrm{2}. \: \: \: \: \: \: \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{object}\mathrm{ive} \\ $$$$\mathrm{1}.\:\:\mathrm{1f}\:\: \: \: \: \: \: \\ $$$$\mathrm{2}. \: \: \:\mathrm{127}\:{g}\:{cu}\: \: \: \: \: \\ $$$$\: \: \: \: \\ $$$$\mathrm{3}\: \: \: \: \: \\ $$$$\:\mathrm{4}.\: \: \: \: \: \: ? \\ $$$$\mathrm{5}.\: \: \: \: \: \: \: \\ $$

Question Number 194241 Answers: 1 Comments: 0

Question Number 194240 Answers: 2 Comments: 0

Question Number 194238 Answers: 1 Comments: 0

Question Number 194237 Answers: 0 Comments: 1

how to evaluate 𝚺_(n=0) ^∞ (((−1)^n )/(k^n n!(zn+1)))

$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{k}^{{n}} {n}!\left({zn}+\mathrm{1}\right)} \\ $$

Question Number 194236 Answers: 1 Comments: 0

Question Number 194226 Answers: 3 Comments: 0

If x^2 − 65x = 64(√x) then (√(x − (√x) )) = ?

$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{65}{x}\:=\:\mathrm{64}\sqrt{{x}}\:\mathrm{then}\:\sqrt{{x}\:−\:\sqrt{{x}}\:}\:=\:? \\ $$$$ \\ $$

Question Number 194219 Answers: 1 Comments: 0

Question Number 194218 Answers: 1 Comments: 2

Question Number 194216 Answers: 1 Comments: 0

Question Number 194211 Answers: 1 Comments: 0

(1/x^2 ) +(1/y^2 ) = (1/3) (d^2 y/dx^2 ) =?

$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$

Question Number 194208 Answers: 0 Comments: 0

Question Number 194209 Answers: 0 Comments: 0

((1+kx))^(1/3) +x = 1 has two real roots . ⇒ k=? kx+1= 1−3x+3x^2 −x^( 3) x^( 3) −3x^( 2) + (k+3)x=0 x=0 x^( 2) −3x +k+3=0 1: Δ=0 9 − 4k −12=0 k= ((−3)/4) 2: k=−3 ⇒ x=0 is a root of x^( 2) −3x+k+3=0 ∴ k= ((−3)/4) or k=−3

$$ \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{kx}}\:+{x}\:=\:\mathrm{1}\:\:\:{has} \\ $$$$\:\:\:\:\:\:{two}\:{real}\:{roots}\:.\:\:\Rightarrow\:{k}=? \\ $$$$\:\:\:{kx}+\mathrm{1}=\:\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\:\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{3}{x}^{\:\mathrm{2}} +\:\left({k}+\mathrm{3}\right){x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}\:+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{1}:\:\:\Delta=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{9}\:−\:\mathrm{4}{k}\:−\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{2}:\:\:{k}=−\mathrm{3}\:\Rightarrow\:{x}=\mathrm{0}\:{is}\:{a}\:{root}\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}}\:\:\:\:{or}\:\:\:{k}=−\mathrm{3} \\ $$

Question Number 194207 Answers: 0 Comments: 0

Solve for x x^(x−4) =(√(3 ))

$$\:\:\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{x}−\mathrm{4}} \:\:=\sqrt{\mathrm{3}\:} \\ $$

Question Number 194206 Answers: 1 Comments: 0