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Question Number 194240 Answers: 2 Comments: 0
Question Number 194238 Answers: 1 Comments: 0
Question Number 194237 Answers: 0 Comments: 1
$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(β\mathrm{1}\right)^{{n}} }{{k}^{{n}} {n}!\left({zn}+\mathrm{1}\right)} \\ $$
Question Number 194236 Answers: 1 Comments: 0
Question Number 194226 Answers: 3 Comments: 0
$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:β\:\mathrm{65}{x}\:=\:\mathrm{64}\sqrt{{x}}\:\mathrm{then}\:\sqrt{{x}\:β\:\sqrt{{x}}\:}\:=\:? \\ $$
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
$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$
Question Number 194208 Answers: 0 Comments: 0
Question Number 194209 Answers: 0 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{kx}}\:+{x}\:=\:\mathrm{1}\:\:\:{has} \\ $$$$\:\:\:\:\:\:{two}\:{real}\:{roots}\:.\:\:\Rightarrow\:{k}=? \\ $$$$\:\:\:{kx}+\mathrm{1}=\:\mathrm{1}β\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} β{x}^{\:\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:β\mathrm{3}{x}^{\:\mathrm{2}} +\:\left({k}+\mathrm{3}\right){x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{2}} β\mathrm{3}{x}\:+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{1}:\:\:\Delta=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{9}\:β\:\mathrm{4}{k}\:β\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}=\:\frac{β\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{2}:\:\:{k}=β\mathrm{3}\:\Rightarrow\:{x}=\mathrm{0}\:{is}\:{a}\:{root}\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} β\mathrm{3}{x}+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:{k}=\:\frac{β\mathrm{3}}{\mathrm{4}}\:\:\:\:{or}\:\:\:{k}=β\mathrm{3} \\ $$
Question Number 194207 Answers: 0 Comments: 0
$$\:\:\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{x}β\mathrm{4}} \:\:=\sqrt{\mathrm{3}\:} \\ $$
Question Number 194206 Answers: 1 Comments: 0
Question Number 194204 Answers: 1 Comments: 0
Question Number 194201 Answers: 1 Comments: 3
Question Number 194197 Answers: 1 Comments: 0
Question Number 194194 Answers: 0 Comments: 0
Question Number 194193 Answers: 0 Comments: 0
Question Number 194191 Answers: 1 Comments: 0
Question Number 194190 Answers: 0 Comments: 1
$${Explanation}\:{Why}: \\ $$$${While}\:{f}\left({ax}+{b}\right)+{f}\left({cx}+{d}\right)={ex}+{g} \\ $$$${then}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C}\:ΒΏ \\ $$
Question Number 194185 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:β} } \:\left\{\:\frac{\:{x}^{\:\mathrm{2}} \:+\mathrm{2}{cos}\left({x}\right)\:+\:\lfloorβ\frac{{tan}\left({x}\right)}{{x}}\:\rfloor}{{ax}^{\:\mathrm{4}} }\:\right\}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:{a}:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:β\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}:\:\:\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}:\:\:β\mathrm{12}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Question Number 194183 Answers: 1 Comments: 0
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)} \\ $$
Question Number 194176 Answers: 3 Comments: 0
$${Hello}\:{everyone} \\ $$$${I}\:{try}\:{to}\:{solve}\:\mathrm{4}^{{x}+\mathrm{1}} +\mathrm{2}^{\mathrm{2}β{x}} =\mathrm{65} \\ $$$${Thx}\:{in}\:{advance} \\ $$
Question Number 194170 Answers: 0 Comments: 0
Question Number 194169 Answers: 0 Comments: 0
Question Number 194174 Answers: 1 Comments: 0
$${Know}\:{f}\left({x}^{β\mathrm{1}} \right)={f}^{β\mathrm{1}} \left({x}\right)\:\left({f}^{β\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{{f}\left({x}\right)}\right) \\ $$$${Find}\:{f}\left({x}\right)ΒΏ \\ $$
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