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AlgebraQuestion and Answers: Page 264

Question Number 86825 Answers: 1 Comments: 1

a^3 +(1/a^3 )=18 a^4 +(1/a^4 )=?

$a^{3} + \frac{1}{a^{3}} = 18$ $a^{4} + \frac{1}{a^{4}} = ?$

Question Number 86779 Answers: 1 Comments: 0

ssolve 1)x−[x]≥0 2)x−[x]≤0 3)x+[x]≥0 4)x+[x]≤0

$s s o l v e$ $1) x - [x] ⩾ 0$ $2) x - [x] ⩽ 0$ $3) x + [x] ⩾ 0$ $4) x + [x] ⩽ 0$

Question Number 86741 Answers: 1 Comments: 4

{ ((x+10y+50z=500)),((x+y+z=100)) :} find x,y,z

${\begin{cases} x + 10 y + 50 z = 500 \\ x + y + z = 100 \end{cases}$ $f i n d x, y, z$

Question Number 86737 Answers: 2 Comments: 4

prove that 1/cos2x+cosx+1=((sin((5x)/2))/(2sin(x/2)))+(1/2) 2/((cos(x)+isin(x)−1)/(cos(x)+isin(x)+1))=−i tan(x) 3/((cos(5x)+isin(5x)+1)/(cos(5x)−isin(x)+1))=cos(5x)+isin(5x)

$p r o v e t h a t$ $1 / c o s 2 x + c o s x + 1 = \frac{s i n \frac{5 x}{2}}{2 s i n \frac{x}{2}} + \frac{1}{2}$ $2 / \frac{c o s (x) + i s i n (x) - 1}{c o s (x) + i s i n (x) + 1} = - i t a n (x)$ $3 / \frac{c o s (5 x) + i s i n (5 x) + 1}{c o s (5 x) - i s i n (x) + 1} = c o s (5 x) + i s i n (5 x)$

Question Number 86723 Answers: 0 Comments: 1

Question Number 86610 Answers: 0 Comments: 4

Question Number 86602 Answers: 0 Comments: 0

Question Number 86586 Answers: 1 Comments: 4

Question Number 86541 Answers: 0 Comments: 1

Question Number 86426 Answers: 2 Comments: 0

solve in R x^3 −5=[x]

$s o l v e i n R$ $x^{3} - 5 = [x]$

Question Number 86396 Answers: 0 Comments: 0

Question Number 86298 Answers: 0 Comments: 5

let x^x^x^⋰ =2 x^2 =2 x=±(√2) then let x^x^x^⋰ =4 x^4 =4 x=±(4)^(1/4) =±(√2) so we had prove 2=4 right?

$l e t x^{x^{x^{\iddots}}} = 2$ $x^{2} = 2$ $x = \pm \sqrt{2}$ $t h e n l e t x^{x^{x^{\iddots}}} = 4$ $x^{4} = 4$ $x = \pm \sqrt[4]{4} = \pm \sqrt{2}$ $s o w e h a d p r o v e 2 = 4 r i g h t ?$

Question Number 86230 Answers: 1 Comments: 0

is (−1)^(m/n) =((((−1 )^m ))^(1/n) ) or =(((−1))^(1/n) )^m or both of them are fault and why ?

$i s {(- 1)}^{\frac{m}{n}} = (\sqrt[n]{{(- 1)}^{m}}) o r = {(\sqrt[n]{- 1})}^{m}$ $o r b o t h o f t h e m a r e f a u l t a n d w h y ?$

Question Number 86206 Answers: 0 Comments: 3

1.line:y=−x+4 ,meets : xy=1 at:A,B. ⇒ S_(OA^△ B) =? (O=origin of cordinates) 2.find :center area of region bonded by corve: (√(x/a))+(√(y/b))=1,and x,y axes. (a≠b)∈R^+

$1 . line : y = - x + 4, meets : xy = 1 at : A, B .$ $\Rightarrow S_{O \overset{△}{A B}} = ? (O = origin of cordinates)$ $2 . find : center area of region bonded by$ $corve : \sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1, and x, y axes .$ $(a \neq b) \in R^{+}$