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what is 852185.81 in money terms

$8,521,185.81 dollars and cents.

1 1/8 POUNDS=1.125 pounds

what is the answer to 1+i divided by 1-2i

\(\frac{1+i }{1-2i}\\ =\frac{1+i }{1-2i}\times\frac{1+2i}{1+2i}\\ =\frac{(1+i)(1+2i) }{(1-2i)(1+2i)}\\ =\frac{1(1+2i) +i(1+2i)}{1-4(i^2)}\\ =\frac{1+2i +i-2}{1+4}\\ =\frac{-1+3i }{5}\\ =\frac{-1 }{5}+\frac{3 }{5}i \)

Simplify the following:
(6-1/x)/(1/x)

Multiply the numerator of (6-1/x)/(1/x) by the reciprocal of the denominator. (6-1/x)/(1/x) = (6-1/x) x:
x (6-1/x)

Put each term in 6-1/x over the common denominator x: 6-1/x  =  (6 x)/x-1/x:
(6 x)/x-1/x x

(6 x)/x-1/x = (6 x-1)/x:
(6 x-1)/x x

((6 x-1) x)/x = x/x×(6 x-1) = 6 x-1:
Answer: |  6 x-1

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42. 

Don't question it.

If the width is 18 then the diameter is 18 (if i've understood you right). The radius is therefore 9. The area of a circle (as the base of a cylinder is a circle) is calculated with the formula A = r * r * pi = 9*9*3.1415926535 = 254.47 length units. 

If you want to calculate the volume too you use the formula V = B * H = Base * Height = 254.47 * 10 = 2544.7 Area units. 

What does it takento become a site admin

C=2*pi*r

sub your values in and solve ro r

then double it to get your diameter.

Yes if you divide both sides of the circumference equation by 2pi you will get

r= c/(2pi)       You need tge brackets there :)

cos2theta = -4/5 and 90* < theta < 135*, what is the value of costheta

 \(cos2\theta = \frac{-4}{5}\\ cos^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-sin^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-2sin^2\theta = \frac{-4}{5}\\ 1+\frac{4}{5} = 2sin^2\theta\\ \frac{9}{5} = 2sin^2\theta\\ \frac{9}{10} = sin^2\theta\\ sin\theta=\pm\frac{3}{\sqrt{10}} \\ \mbox{But theta is in the second quadrant}\\ sin\theta=\frac{3}{\sqrt{10}} \\ \theta = 180-sin^{-1}\frac{3}{\sqrt{10}} \\ \theta \approx 108^026'58" \)

Sometimes it isn't possible to use a measuring tool. It will also likely be more accurate asssuming the other measurements are accurate.

solve for x: cos(x)=0.342/0.798

\(\begin{array}{rcll} \cos(x) &=& \frac{0.342}{0.798} \\ \cos(x) &=& 0.42857142857 \qquad & | \qquad \pm\arccos{()}\\ x &=& \pm\arccos{( 0.42857142857 )}\\ x &=& \pm64.6230664748^{\circ}\\ \mathbf{ x_1 } & \mathbf{=} & \mathbf{64.6230664748^{\circ} }\\\\ x_2 &=& -64.6230664748^{\circ} \\ x_2 &=& -64.6230664748^{\circ}+ 360^{\circ} \\ \mathbf{ x_2 } & \mathbf{=} & \mathbf{295.376933525^{\circ}} \end{array} \)

its 10 duuuuuude

this is not a arcus function "arc cosh" is wrong, this is a area - function area cosh(x) or arcosh(x) is right!

just re-arrange to get

cosh(0.438X) = 143.3/15.25

then

(0.438X) = arc cosh( 143.3/15.25),and just do the arithmetic.

The cosh function is never zero,by the way,as it is e^x + e^(-x).The x axis is an asymptote of the function.

2m-4n when m = 1.8 and n = -0.7

2[1.8] - 4 [-0.7]  =

3.6 + 2.8 =

6.4

143.3-15.25 cosh(0.0438x)=0

Subtract   143.3 from both sides

-15.25cosh(0.0438x)  =  - 143.3     

Divide both sides by -15.25

cosh(0.0438x)  = 143.3/15.25

Take the arccosh to isolate x

arccosh(143.3/15.25) = 0.0438x

Divide both sides by 0.0438

arccosh(143.3/15.25)/0.0438  = x  = about 66.9102

As I'm not overly familiar with these functions, WolframAlpha  also gives the negative of this as the other [real solution] x intercept......here's a pic :

(82/100)x = 328

82x =32800

x = 32800/82

x = 400

82% of 400 is 328

i have no idea

I assume this is supposed ro be :

12 - x³  = 0       if so....    add x³ to both sides

12 = x³     take the cube root of both sides

∛12 =  x   = about 2.29

3(x^2) - 5 = 0

3(x²)  = 5

x² = 5/3

x = √(5/3)

x = ± 1.29099444874

asinus :- )

  !

15%  =  .15

Guest......here's an informal proof of why 0!   = 1

Note :

[n] !  = [n + 1 ] !  / [ n + 1]   ......so......

2!  =   3! / 3     =  2

1!  =   2! / 2    =  1

0!  = 1!  /  1     = 1

31.6916676717452859083 sq miles