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Question Number 175420 Answers: 2 Comments: 0

solve for x 2^x .3^x^2 = 6

$s o l v e f o r x$ $2^{x} . 3^{x^{2}} = 6$

Question Number 175411 Answers: 2 Comments: 3

7+67+667+6667+.....(n terms)=?

$7 + 67 + 667 + 6667 + . . . . . (n t e r m s) = ?$

Question Number 175410 Answers: 1 Comments: 0

A=∫9xcos 2x dx

$A = \int 9 x \cos 2 x d x$

Question Number 175409 Answers: 0 Comments: 3

exercise Consider a polygon with an odd number 𝗻 of vertices. We connect any 3 vertices of this polygon to form a triangle. What is the probability that this triangle contains the center of the circle circumscribing the polygon?

$exercise$ Consider a polygon with an odd number 𝗻 of vertices. We connect any 3 vertices of this polygon to form a triangle. What is the probability that this triangle contains the center of the circle circumscribing the polygon?

Question Number 175407 Answers: 2 Comments: 0

Trouver la somme de S_n Si S_n =1+11+111+...+1111...111

$\underset{―}{Trouver la somme de S_{n}}$ $Si S_{n} = 1 + 11 + 111 + . . . + 1111 . . . 111$

Question Number 175406 Answers: 1 Comments: 0

(d^2 y/dx^2 ) −tan (x)(dy/dx) = 0

$\frac{d^{2} y}{{d x}^{2}} - \tan (x) \frac{d y}{d x} = 0$

Question Number 175402 Answers: 0 Comments: 1

solve for x x_(n+1) =rx_n (1−x_n )

$s o l v e f o r x$ $x_{n + 1} = {r x}_{n} (1 - x_{n})$

Question Number 175399 Answers: 0 Comments: 3

if (x+1)^4 (√4)=x+1 find x

$i f {(x + 1)}^{4} \sqrt{4} = x + 1 f i n d x$

Question Number 175396 Answers: 2 Comments: 0

Question Number 175395 Answers: 0 Comments: 0

f(4) = (1/4) and f(8) = (1/2) ∫_4 ^8 (( [f′(x)]^2 )/([f(x)]^4 ))dx = 1 then f(6)= ?

$f (4) = \frac{1}{4} a n d f (8) = \frac{1}{2}$ $\int_{4}^{8} \frac{{[f^{'} (x)]}^{2}}{{[f (x)]}^{4}} d x = 1 t h e n f (6) = ?$

Question Number 175394 Answers: 0 Comments: 0

Question Number 175393 Answers: 0 Comments: 0

if (x+1)_(√4) ^4 =x+1 find x

$i f {(x + 1)}_{\sqrt{4}}^{4} = x + 1 f i n d x$

Question Number 175387 Answers: 2 Comments: 1

J=∫_0 ^(π/2) ((sin x)/(1+sin x+cos x)) dx

$J = \underset{0}{\int^{π / 2}} \frac{\sin x}{1 + \sin x + \cos x} d x$

Question Number 175386 Answers: 0 Comments: 0

∫_5 ^7 (([ (1/4)x+3 ])/( (√(9x^2 −12x+4)))) dx=? [ ..] =floor function

$\underset{5}{\int^{7}} \frac{[\frac{1}{4} x + 3]}{\sqrt{9 x^{2} - 12 x + 4}} d x = ?$ $[. .] = f l o o r f u n c t i o n$

Question Number 175375 Answers: 1 Comments: 0

Question Number 175369 Answers: 1 Comments: 0

solve for x log_(∣sinx∣ ) (x^2 −8x+23) > (3/(log_2 ∣sinx∣ ))

$solve for x$ $\log_{∣ \sin x ∣} (x^{2} - 8 x + 23) > \frac{3}{\log_{2} ∣ \sin x ∣}$

Question Number 175362 Answers: 1 Comments: 2

lim_(x→(π/4)) ((sin x−cos x)/(tan ((π/8)−(x/2)))) =?

$\lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{\tan (\frac{π}{8} - \frac{x}{2})} = ?$

Question Number 175353 Answers: 1 Comments: 0

ax^2 +bx+c = 0 x = ((−b±(√(b^2 −4ac)))/(2a)) Example: Find the values of x in the equation x^2 +5x+4 = 0 In order to solve for that, let′s first take a look on what are the values of a, b and c 1x^2 + 5x + 4 = 0 a = 1 ; b = 5 ; c = 4 Now, using the quadratic formula x = ((−5±(√(5^2 −4(1)(4))))/(2(1))) = ((−5±(√(25−16)))/2) = ((−5±3)/2)

${ax}^{2} + bx + c = 0$ $x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$ $Example :$ $Find the values of x in the equation$ $x^{2} + 5 x + 4 = 0$ $In order to solve for that, {let}^{'} s first take$ $a look on what are the values of a, b and c$ ${1 x}^{2} + 5 x + 4 = 0$ $a = 1; b = 5; c = 4$ $Now, using the quadratic formula$ $x = \frac{- 5 \pm \sqrt{5^{2} - 4 (1) (4)}}{2 (1)}$ $= \frac{- 5 \pm \sqrt{25 - 16}}{2}$ $= \frac{- 5 \pm 3}{2}$