Solving for x
 👍
 👎
 👁
Respond to this Question
Similar Questions

calculus
Find complete length of curve r=a sin^3(theta/3). I have gone thus (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int

Trig
Find the exact values of the six trigonometric functions 0 if the terminal side of 0 in standard position contains the points(5,4). (0 is not the number zero I don't know what its called) I have to find r first. r=sqrt x^2+y^2

Calculus
Find the rate of change of f(x,y,z) = xyz in the direction normal to the surface yx^2 + xy^2 + yz^2 = 75 at (1,5,3). I keep getting this wrong and I don't know what I did right. The answer's 147/sqrt(101) but I got 357/sqrt(1361).

Precalculus check answers help!
1.) Find an expression equivalent to sec theta sin theta cot theta csc theta. tan theta csc theta sec theta ~ sin theta 2.) Find an expression equivalent to cos theta/sin theta . tan theta cot theta ~ sec theta csc theta 3.)

math;)
Find the unit vector in the direction of u=(3,2). Write your answer as a linear combination of the standard unit vectors i and j. a. u=3[sqrt(13)/13]i+2[sqrt(13)/13]j b. u=3[sqrt(5)/5]i+2[sqrt(5)/5]j c.

Trig(3 all with work)
1)Find the exact value of cos 105 by using a halfangle formula. A)sqrt 2  sqrt 3 /2 B)sqrt 2  sqrt 3 /2 C)sqrt 2 + sqrt 3 /2 D)sqrt 2 + sqrt 3 /2 cos 105 cos 105 = cos 210/2 sqrt 1 + 210/2 sqrt 1 + sqrt 3/2 /2 sqrt 2 + sqrt

math;)
The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the

finding c: calc
let f(x)=square root x. if the rate of change of at x=c is twice its rate of change at x=1, then c=? what???? i don't get get it. to begin with, i took the derivative if that helps which is 1/(2*sqrtx) . f'(x) = 1/(2*sqrtx) is

Trig
Find the exact value of sin2(theta) if cos(theta) = sqrt 5/3 and 180 < theta < 270. A)1/9 B)4 sqrt 5/9 C)1/9 D)4 sqrt 5/9 B? sin^2(theta) + cos^2(theta) = 1 sin^2(theta) = 1  cos^2(theta) sin^2(theta) = 1  (sqrt 5/3)^2

Precalculus
solve 4 sin^2x + 4 sqrt 2 cos x6=0 for all real values of x. Going by the example in the book I got to: 4 cos^2x + 4sqrt2 cosx 10 = 0 but do not know how to proceed. Any help would be great. Thanks

Trig/Precalc
So I have two questions that have been puzzling me for quite some time and would really appreciate any help with either of them! (a) There are four positive intergers a, b, c, and d such that

calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)
You can view more similar questions or ask a new question.