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AllQuestion and Answers: Page 1898

Question Number 9387    Answers: 0   Comments: 2

Question Number 9349    Answers: 0   Comments: 0

We know that R=Q∪Q^− ⊂ C. Is there another set V such that C⊂V ?

$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathbb{R}=\mathbb{Q}\cup\overset{−} {\mathbb{Q}}\:\subset\:\mathbb{C}.\:\mathrm{Is}\:\mathrm{there}\: \\ $$$$\mathrm{another}\:\mathrm{set}\:\mathbb{V}\:\mathrm{such}\:\mathrm{that}\:\mathbb{C}\subset\mathbb{V}\:? \\ $$$$ \\ $$

Question Number 9348    Answers: 0   Comments: 0

pH = pK_a + lg(C_b /C_a )

$$\mathrm{pH}\:=\:\mathrm{pK}_{\mathrm{a}} \:+\:\mathrm{lg}\frac{\mathrm{C}_{\mathrm{b}} }{\mathrm{C}_{\mathrm{a}} } \\ $$

Question Number 9347    Answers: 0   Comments: 0

pH = (1/2) (14 − pK_b + pK_a )

$$\mathrm{pH}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{14}\:−\:\mathrm{pK}_{\mathrm{b}} \:+\:\mathrm{pK}_{\mathrm{a}} \:\right) \\ $$

Question Number 9346    Answers: 0   Comments: 0

pH = 14 − (1/2) ( pK_b − lgC_b )

$$\mathrm{pH}\:=\:\mathrm{14}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\mathrm{pK}_{\mathrm{b}} \:\:−\:\:\mathrm{lgC}_{\mathrm{b}} \:\right) \\ $$

Question Number 9342    Answers: 1   Comments: 0

Solve for n n^2 ≥n^3 −1

$$\mathrm{Solve}\:\mathrm{for}\:{n} \\ $$$${n}^{\mathrm{2}} \geqslant{n}^{\mathrm{3}} −\mathrm{1} \\ $$

Question Number 9334    Answers: 1   Comments: 0

express the following in form of log a(f(x)) (i) 4log_a x−log_a (x^2 +x^3 ) (ii) log_a x + log_a (x+3)

$$\mathrm{express}\:\mathrm{the}\:\mathrm{following}\:\mathrm{in}\:\mathrm{form}\:\mathrm{of}\:\mathrm{log}\:\mathrm{a}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{4log}_{\mathrm{a}} \mathrm{x}−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} \right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{log}_{\mathrm{a}} \mathrm{x}\:+\:\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}+\mathrm{3}\right) \\ $$

Question Number 9333    Answers: 1   Comments: 0

given that log 2=0.3.show that log 5=0.7

$$\mathrm{given}\:\mathrm{that}\:\mathrm{log}\:\mathrm{2}=\mathrm{0}.\mathrm{3}.\mathrm{show}\:\mathrm{that}\:\mathrm{log}\:\mathrm{5}=\mathrm{0}.\mathrm{7} \\ $$

Question Number 9332    Answers: 1   Comments: 0

Question Number 9330    Answers: 0   Comments: 0

Question Number 9325    Answers: 1   Comments: 0

Question Number 9324    Answers: 1   Comments: 0

if x=((a(1−r^n ))/(1−r)) make r the subject of the formula

$${if}\:\:{x}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\:{make}\:{r}\:{the}\: \\ $$$${subject}\:{of}\:{the}\:{formula} \\ $$

Question Number 9323    Answers: 0   Comments: 1

if x=((a(1−r^n ))/(1−r)) make r the subject of the formula

$${if}\:\:{x}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\:{make}\:{r}\:{the}\: \\ $$$${subject}\:{of}\:{the}\:{formula} \\ $$

Question Number 9313    Answers: 0   Comments: 1

∫_0 ^∞ {{

$$\int_{\mathrm{0}} ^{\infty} \left\{\left\{\right.\right. \\ $$

Question Number 9314    Answers: 1   Comments: 0

solve for M: c=Mln (z)+q

$${solve}\:{for}\:{M}: \\ $$$$ \\ $$$${c}={M}\mathrm{ln}\:\left({z}\right)+{q} \\ $$

Question Number 9306    Answers: 1   Comments: 0

Differentiate from the first principle: y = tan2x

$$\mathrm{Differentiate}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{principle}: \\ $$$$\mathrm{y}\:=\:\mathrm{tan2x} \\ $$

Question Number 9298    Answers: 1   Comments: 0

Question Number 9312    Answers: 0   Comments: 6

Solve simultaneously xy + x + y = 23 ....... (i) xz + x + z = 41 ........ (ii) yz + y + z = 27 ........ (iii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{xy}\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{23}\:\:\:\:.......\:\left(\mathrm{i}\right) \\ $$$$\mathrm{xz}\:+\:\mathrm{x}\:+\:\mathrm{z}\:=\:\mathrm{41}\:\:\:\:........\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{yz}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{27}\:\:\:\:\:........\:\left(\mathrm{iii}\right) \\ $$

Question Number 9287    Answers: 2   Comments: 0

find the determinant of the matrix below determinant ((0,4,0,0,0),(0,0,0,2,0),(0,0,3,0,0),(0,0,0,0,1),(5,0,0,0,0))

$${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$

Question Number 9286    Answers: 1   Comments: 0

find the determinant of the matrix below determinant ((0,0,0,0,1),(0,0,0,2,0),(0,0,3,0,0),(0,4,0,0,0),(5,0,0,0,0))

$${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$

Question Number 9285    Answers: 1   Comments: 0

find the determinant of the matrix below [(1,4,(−3),1),(2,0,6,3),(4,(−1),2,5),(1,0,(−2),4) ]

$${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{0}}&{\mathrm{6}}&{\mathrm{3}}\\{\mathrm{4}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{4}}\end{bmatrix} \\ $$

Question Number 9279    Answers: 3   Comments: 0

evalute the value of Σ_(m=2 ) ^5 m^4

$$\mathrm{evalute}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\overset{\mathrm{5}} {\sum}_{\mathrm{m}=\mathrm{2}\:} \mathrm{m}^{\mathrm{4}} \\ $$

Question Number 9278    Answers: 0   Comments: 1

represent in sigma notation −1+4−9+16..................

$$\mathrm{represent}\:\mathrm{in}\:\mathrm{sigma}\:\mathrm{notation}\: \\ $$$$−\mathrm{1}+\mathrm{4}−\mathrm{9}+\mathrm{16}.................. \\ $$

Question Number 9275    Answers: 1   Comments: 1

About the Euler-Mascheroni Constant: γ = ∫_0 ^1 (1/(1 − x)) + (1/(ln x)) dx We can see that 1− x ≠ 0 ⇔ 1 ≠ x ; x = 0 → ln x ∄ . If x ≠ 0 and x ≠ 1, in the Cartesian Plane, this function has singularity x=0 and x=1. So, I could write f(x) = (1/(1 − x)) + (1/(ln x)) ∫_0 ^1 f(x) dx = lim_(A→0^+ ) lim_(B→1^− ) ∫_A ^B f(x) dx ? PS: Sorry by my worse English

$$\mathrm{About}\:\mathrm{the}\:\mathrm{Euler}-\mathrm{Mascheroni}\:\mathrm{Constant}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\:{dx} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that}\: \\ $$$$\:\mathrm{1}−\:{x}\:\neq\:\mathrm{0}\:\Leftrightarrow\:\mathrm{1}\:\neq\:{x}\:; \\ $$$${x}\:=\:\mathrm{0}\:\rightarrow\:\mathrm{ln}\:{x}\:\nexists\:. \\ $$$$\mathrm{If}\:{x}\:\neq\:\mathrm{0}\:\mathrm{and}\:{x}\:\neq\:\mathrm{1},\:\mathrm{in}\:\mathrm{the}\:\mathrm{Cartesian}\:\mathrm{Plane}, \\ $$$$\mathrm{this}\:\mathrm{function}\:\mathrm{has}\:\mathrm{singularity}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}. \\ $$$$\mathrm{So},\:\mathrm{I}\:\mathrm{could}\:\mathrm{write} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{ln}\:{x}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}\:=\:\underset{{A}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\underset{{B}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\int_{{A}} ^{{B}} {f}\left({x}\right)\:{dx}\:\:? \\ $$$${PS}:\:{Sorry}\:{by}\:{my}\:{worse}\:{English} \\ $$

Question Number 9274    Answers: 0   Comments: 0

e^t = t

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{t}} \:=\:{t} \\ $$

Question Number 9271    Answers: 1   Comments: 0

Find the determinant of the matrix below. determinant (((3 1 5 3)),((4 3 8 5)),((6 2 1 7)),((8 5 8 1)))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{below}. \\ $$$$\begin{vmatrix}{\mathrm{3}\:\:\mathrm{1}\:\:\mathrm{5}\:\:\mathrm{3}}\\{\mathrm{4}\:\:\mathrm{3}\:\:\mathrm{8}\:\:\mathrm{5}}\\{\mathrm{6}\:\:\mathrm{2}\:\:\mathrm{1}\:\:\mathrm{7}}\\{\mathrm{8}\:\:\mathrm{5}\:\:\mathrm{8}\:\:\mathrm{1}}\end{vmatrix} \\ $$

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