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a doctor has 12000 patients 4560 of these patients are male what percentage are female

First work out how many are female

then it will be  

number of females / total number of patients *100   %

4x^4 y^3 m^2 over 8x y r^0

do you mean   (4x^4 y^3 m^2) over (8x y r^0)

\(\quad \frac{4x^4 y^3 m^2 }{ 8x y r^0}\\ =\frac{x^4 y^3 m^2 }{ 2x^1 y^1 *1}\\ =\frac{x^3 y^2 m^2 }{ 2}\\\)

isnt that x?

k can be any number. 

You cannot find any distances with this information.  You need at least one distance to work the others out :)

I can tell you that when you get to the first checkpoint that you have covered 2/20   or 10% of the race but I cannot tell you how far this is ://

3 sides  180*1

4 sides   180*2

5 sides    180*3

30 sides  180*28

-1 to 1 (sin^2 (pi*x))*(sin^2(pi*y))

\( (sin^2 (\pi x))*(sin^2(\pi y)) \)

This makes no sense what so ever!  What is your question ?????

There are 105 lettered tiles in a board game. You choose the tiles shown(r,a,m,e,I,l,b) who many of the 105 tiles would you except to be vowels

Well  2 out of the 7 were vowels.  So it may be vaguely reasonable to consider that 2/7 of all the tiles are vowels.

2/7*105 = 30 vowels.

In reality that sample is probably not statistically significant.  But that is the answer your teacher wants.  :)

960^2=720^2+c^2−2(720)(c)*cos(20) solve for c with steps

This is a triangle ( cos - rule )

We have \(a = 960,\ b = 720,\ A = 20^{\circ}\)

1. sin - rule B?

\(\begin{array}{rcll} \frac{ \sin{(B)} } {720} &=& \frac{ \sin{ ( 20^{\circ} )} } {960} \\ \sin{(B)} &=& \frac{720}{960} \cdot \sin{(20^{\circ} ) } \\ \sin{(B)} &=& 0.75 \cdot 0.34202014333\\ \sin{(B)} &=& 0.25651510749\\ B &=& \arcsin{ (0.25651510749) } \\ B &=& 14.8633809491^{\circ} \\ \end{array}\)

2.  C?

\(\begin{array}{rcll} C &=& 180^{\circ} - (A+B) \\ C &=& 180^{\circ} - (20^{\circ}+14.8633809491^{\circ}) \\ C &=& 180^{\circ} - 34.8633809491^{\circ} \\ C &=& 145.136619051^{\circ} \\ \end{array}\)

3. sin - rule c?

\(\begin{array}{rcll} \frac{c}{ \sin{(C)} } &=& \frac{960}{ \sin{ ( 20^{\circ} )} } \\ c &=& 960 \cdot \frac{\sin{(C)}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{\sin{(145.136619051^{\circ} )}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{0.57162157869}{0.34202014333} \\ c &=& 960 \cdot 1.67130968701 \\ c &=& 1604.45729953 \\ \end{array}\)

If F1 and F2 are at the same angle then they must have the same magnitude to prevent any horizontal movement.

The vertical components of force require that 2F1*sin(48.5) = 9.8*2.8

or:  F1 = 9.8*2.8/(2sin(48.5))

Can you take it from here?

The domain is x<0 and x>3

The region between 0 and 3 produces values of 3x^2 - 9x for which the logarithm isn't defined (for Real numbers).

a/b+b/a

Put each term in a/b+b/a over the common denominator a b: a/b+b/a  =  a^2/(a b)+b^2/(a b):
a^2/(a b)+b^2/(a b)

a^2/(a b)+b^2/(a b) = (a^2+b^2)/(a b):
Answer: |  (a^2+b^2)/(a b)

Solve for c:
921600 = 518400+c^2-1440 c cos(20°)

921600 = 518400+c^2-1440 c cos(20°) is equivalent to 518400+c^2-1440 c cos(20°) = 921600:
518400+c^2-1440 c cos(20°) = 921600

Subtract 518400 from both sides:
c^2-1440 c cos(20°) = 403200

Add 518400 cos(20°)^2 to both sides:
c^2-1440 c cos(20°)+518400 cos(20°)^2 = 518400 cos(20°)^2+403200

Write the left hand side as a square:
(c-720 cos(20°))^2 = 518400 cos(20°)^2+403200

Take the square root of both sides:
c-720 cos(20°) = sqrt(518400 cos(20°)^2+403200) or c-720 cos(20°) = -sqrt(518400 cos(20°)^2+403200)

Add 720 cos(20°) to both sides:
c = sqrt(518400 cos(20°)^2+403200)+720 cos(20°) or c-720 cos(20°) = -sqrt(518400 cos(20°)^2+403200)

Add 720 cos(20°) to both sides:
Answer: |  c = sqrt(518400 cos(20°)^2+403200)+720 cos(20°)       or      c = 720 cos(20°)-sqrt(518400 cos(20°)^2+403200)

This has already been answered at: http://web2.0calc.com/questions/an-automobile-is-depreciating-at-10-per-year-every-year-a-22-000-car-depreciating-at-this-rate-can-be-modeled-by-the-equation-v-t-22-000-0-90-t-what#r1 

The term (5/2)^x is common to both terms as (5/2)^(x+3) can be written in the form (5/2)^x*(5/2)^3

So (5/2)^x + (5/2)^(x+3)   →   (5/2)^x*(1 + (5/2)^3)   →   (5/2)^x*(1 + 125/8)  →   (5/2)^x*(133/8)

So with A = 133/8 we have  (5/2)^x + (5/2)^(x+3) = A*(5/2)^x

.

There is a ceiling function (ceil(x)) if that's what you mean:

The horizontal distance as a function of time is given by x = v0*cos(60°)*t

The vertical distance as a function of time is given by y = v0*sin(60°)*t - g*t²/2    where g is gravitational acceleration.

The above assumes air resistance is negligible.

25^1312000

let  

\(y=25^{1312000}\\ log\;y=log\;25^{1312000}\\ log\;y=1312000log\;25\\ \)

1312000*log(25) = 1834097.2913777134624

log y= 1834097.2913777134624

log y= 0.2913777134624+1834097

\(y = 10^{(0.2913777134624+1834097)}\\ y = 10^{0.2913777134624}*10^{1834097}\)

10^0.2913777134624 = 1.9560399181356844

\(y\approx 1.9560399181356844\times 10^{1834097}\)

That is in radians.

-cos(2 radians)

im x → 1, ((sin(x+1))-(sin2x))/(x-1)

\(\displaystyle \lim_{ x → 1} \; \frac{sin(x+1)-sin2x}{x-1}\\ \mbox{This is the indeterminate form of 0/0 so use l'hopital's rule}\\ =\displaystyle \lim_{ x → 1} \; \frac{cos(x+1)-2cos2x}{1}\\ =\displaystyle \lim_{ x → 1} \; cos(x+1)-2cos2x\\ = cos(2)-2cos2\\ =-cos(2)\)