((2X14Cos45)/(49+X^214XCos45))+0.3536 = 0 Solve to find the value of X Change cos 45 to sqrt 2 and rewrite the equation in the form a x^2 + bx + c = 0 Start out by rewriting it as 2x  9.8995
133,634 results
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 correct f'(x=1) = 1/2 The

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 r=sqrt 5^2+4^2 r=sqrt 41

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.) Simplify (tan^2 theta +

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). Is my method right? Or

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 Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int

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. u=3[sqrt(5)/5]i2[sqrt(5)/5]j d.

math
Solve for the indicated variable A=1/2 bh Solve for b 2A = bh, so b = 2A/h ok thanks so on this example C=2pie r Solve for R Would the answer be 2C/pie=r ???? ok thanks so on this example C=2pie r Solve for R Would the answer be 2C/pie=r ???? Divide both

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

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 completed solution? a.

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 3/2 which is D 2)Find the

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 sin^2(theta) = 1  (sqrt 25/9)

algebra
simplify: square root of 5 + square root of 20  square root of 27 + square root of 147. simplify: 6/4square root of 2 sqrt 5 + sqrt 20  sqrt 27 + sqrt 147 = sqrt 5 + sqrt (4*5)  sqrt (3*9) + sqrt (49*3) = 3 sqrt 5  3 sqrt 3 + 7 sqrt 3 = 3 sqrt 5 + 4

Algebra
Evaluate sqrt7x (sqrt x7 sqrt7) Show your work. sqrt(7)*sqrt(x)sqrt(7)*7*sqrt(7) sqrt(7*x)7*sqrt(7*7) sqrt(7x)7*sqrt(7^2) x*sqrt 7x49*x ^^^ would this be my final answer?

Vectors/Tension
A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 2 meters. What is the magnitude of the tension on the ends of the clothesline? Alright, so i have

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) (dy/dx)(dx/du)=

math
solve 2x^2+3x+8=0 and express the solutions in a+bi form. Let's use the quadratic formula to solve for x and express those solutions in a+bi form. x = [b + or  sqrt(b^2  4ac)]/2a Note: sqrt = square root. a = 2, b = 3, and c = 8 from your problem.

Math(answer check)
1. Find the modulus and argument of the following complex numbers, and write them in trigonometric form: a. 5 – 8i Answer = Sqrt{89} ( cos (1.01219701) + i sin ( 1.01219701)) b. –1 – i answer = Sqrt2 (cos (5pi/4) + i sin (5pi/4)) c. –5 + 12i

Calculus  Second Order Differential Equations
Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, c2=4

calc check: average value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(ba))*inegral of a to b for: f(x) dx f ave = (1/(20))*integral

Inequality
When I solve the inquality 2x^2  6 < 0, I get x < + or  sqrt(3) So how do I write the solution? Is it (+sqrt(3),sqrt(3)) or (infinity, sqrt(3))? Why? Thanks. So would this work? abs x < ( sqrt 3 ) or  sqrt 3

calc: avg value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(ba))*inegral of a to b for: f(x) dx f ave = (1/(20))*integral

Algebra 2
Operations with Complex Numbers Simplify. 1. sqrt(144) 2. sqrt(64x^4) 3. sqrt(13)*sqrt(26) 4. (2i)(6i)(4i) 5. i13 6. i38 7. (5 – 2i) + (4 + 4i) 8. (3 – 4i) – (1 – 4i) 9. (3 + 4i)(3 – 4i) 10. (6 – 2i)(1 + i) 11. (4i)/(3+i) 12. (10+i)/(4i)

Trig Practice
Find the exact value of sinA where a=9 and b=10 and angle C is a right angle. a. sin A= 9/sqrt 181, cos A= sqrt 181/10 b. sin A= sqrt 181/9, cos A= 10/sqrt 181 c. sin A= 9/sqrt 181, cos A= 10/sqrt 181 d. sin A=sqrt181/10, cos A= 9/sqrt 181

math!!
I have no idea how to do this!!! Let ΔABC be a triangle such that b=sqrt(3), c=sqrt(10), and cos(B)=3/sqrt(10). Find all possible values for side length a. Please show work

Precal (Please Check)
1) Evaluate (if possible) the sine, cosine, and tangent of the angles without a calculator. a) 10pi/3 Sin = sqrt 3/2 Cos = 1/2 Tan = sqrt 3 Are these correct ? I do not understand when to make them negative. b) 17pi/3 Sin = sqrt 3/2 Cos = 1/2 Tan = 

Math:)
A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12 radians? a.

trigonometry (please double check this)
Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2ƒÆ = (sqrt 3)/2 2. sin^2ƒÆ = cos^2ƒÆ + 1/2 3. sin 2x

Math
If a and B are two angles in Quadrant 2 such that tan a=1/2 and tan B= 2/3, find cos(a+b)? tan(a) = 1/2 oppsite side = 1: adjacentside = 2 hypotenuse = sqrt(1+4) = sqrt(5) sin(a) = 1/ã5 cos(a) = 2/ã5 tan(b) = 2/3 opposite side = 2 and

Math(Roots)
sqrt(24) *I don't really get this stuff.Can somebody please help me? The square root of 24 is 4.898979485566356 I know that..lol,but it says not to use decimals.Here is an example they gave me. Ex.sqrt(18)=sqrt(2*3*3)=3sqrt(2) sqrt[24] = sqrt[4*2*3] =

Solving for x
((2X14Cos45)/(49+X^214XCos45))+0.3536 = 0 Solve to find the value of X Change cos 45 to sqrt 2 and rewrite the equation in the form a x^2 + bx + c = 0 Start out by rewriting it as 2x  9.8995 = (0.3536)(49 + x^2  9.8995 x) Conimue until you get the

URGENT  Trigonometry  Identities and Proofs
Okay, today, I find myself utterly dumbfounded by these three questions  Write a proof for  2/(sqrt(3)cos(x) + sin(x))= sec((pi/6)x) Solve the following equation  2sin(2x)  2sin(x) + 2(sqrt(3)cos(x))  sqrt(3) = 0 Find all solutions (exact) to the

Precalculs
I have no idea how to do these type of problems. Problem Solve each equation on the interval 0 less than or equal to theta less than 2 pi 42. SQRT(3) sin theta + cos theta = 1  There is an example prior to the

Calc.
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) (dy/dx)(dx/du)=

Precalculus check answers help!
1. Which expression is equivalent to cos 2theta for all values of theta ? cos^2 theta – sin^2 theta ~ cos^2 theta – 1 1 – 2 sin^2 theta 2 sin theta cos theta 2. Use a half–angle identity to find the exact value of sin 105°. 1/2(sqrt)(2 + Sqrt3)

CALC
1. The problem statement, all variables and given/known data cos(x) = 2 I was trying to solve that problem ignoring what people say about it not being defined and got a number strangely. I showed step by step so you can tell me what I did wrong as I don't

Math
So I am supposed to solve this without using a calculator: Sqrt[20]/10  Sqrt[10]/Sqrt[32]  Sqrt[0.3125] + Sqrt[3 + 1/5] You can put this into WolframAlpha as is to make it prettier. Answer given is 1/2 * SQRT(5) I really don't know where to start here. I

Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1  cos^2(x)) = 4cos^4(x)  4cos^2(x)

Trig
Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v  u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos v + cos v sin u. cos (v  u) = cos u

PreCalculus
The question is: 5e^(4x+2) +3= 1/2log(1+x^2) Solve with a calculator for the smallest possible solution, with steps. So this hasn't really been covered in my schoolwork so I'm trying to work off what I have learned. I started with the change of base

math Trigonometry
If cos degree equals to 0.8641 What is Sin degree? I have no idea how to find this. Please help me. I got help from two people, but I'm not getting the answer and how they got the numbers either. Someone says: cos^2+sin^2=1 sinDegree=sqrt(1cos^2degree)

Calculus
Please look at my work below: Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2

Calculus  Second Order Differential Equations
Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i,

Math/Calculus
Solve the initialvalue problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(4 +/ sqrt(164(1)(6))/2 r=2 +/ sqrt(2)*i , alpha = 2, beta = 2(sqrt(2))

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 4cos(x)cos(2x)cos(4x)=cos(ax)+cos(bx)+cos(cx)+cos(dx) for all

Vectors
The angle between the vectors a=(6,2,3) and b=(1,p,2) is cos^1(1/14). Determine the value of p. a=sqrt(6^2+(2)^2+3^2)=7 b=sqrt(1^2+p^2+(2)^2)=5+p^2 theta=cos(cos^1(1/14))=1/14 a x b = (43p,15,6p+2) a x b = abcos(theta) a x b = sqrt(5/2)

math
A trigonmetric polynomial of order n is t(x) = c0 + c1 * cos x + c2 * cos 2x + ... + cn * cos nx + d1 * sin x + d2 * sin 2x + ... + dn * sin nx The output vector space of such a function has the vector basis: { 1, cos x, cos 2x, ..., cos nx, sin x, sin 2x,

calc
find the area between the xaxis and the graph of the given function over the given interval: y = sqrt(9x^2) over [3,3] you need to do integration from 3 to 3. First you find the antiderivative when you find the antiderivative you plug in 3 to the

Calculus AP
hi again im really need help TextBook: James Stewart:Essential Calculus, page 311. Here the problem #27: First make a substitution and then use integration by parts to evaluate the integral. Integral from sqrt(pi/2) TO sqrt(pi)of θ^3 cos(θ^2)dθ. i did

Calculus
In these complex exponential problems, solve for x: 1)e^(i*pi) + 2e^(i*pi/4)=? 2)3+3=3i*sqrt(3)=xe^(i*pi/3) MY attempt: I'm not really sure of what they are asking. For the 1st one I used the e^ix=cos(x)+i*sin(x) and got 1+sqrt(2) +sqrt(2)i 2) I solved

Trigenometry
VERY URGENT! PLEASE HELP SOLVE! tan(x)+1=sqrt(3)+sqrt(3)cot(x)as well as sin(2x)= 2 cos^(x) 4cos2(x) = 8sin(x)cos(X)

inverse
find f^1 (x). (this is asking me to find the inverse) f(x) = (x2)^2, x If f(x)=(x2)^2 x

maths
Choose the option that gives the solution of the initialvalue problem dy dx=(1 + 2 cos2x^2)/y (y > 0), y= 1 when x = 0. Options A y = 1+ 2sin^2 x B y = (1 + 2 sin x)^2 C y = (4x + cos(2x))^2 D y = 4x + cos^2 x E y =sqrt(4x + cos(2x) F y =sqrt(1 + 4x +

Math Help please!!
Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine them. sqrt 3* sqrt 15=

Algebra
Solve for s: h=(square root of 3)times s/2 and solve for h V= (pi)r squared h / 3 Solve for s: h=(square root of 3)times s/2 Multiply both sides by 2. 2h = (sqrt 3)*s*2/2 which cancels the 2 on the right. 2h = (sqrt 3)*s Now divide the right side by

some algebra help (radicals)
I hope I am writing this down right.. I am trying to do some practice questions to learn 10^5 (sqrt)2y  4^5 (sqrt)2y I am trying to figure out how to solve this They gave us some answers to choose from, but I am clueless on how to solve this 6y ^5 (sqrt)2

Calculus
Find the derivative of cos(sqrt(e^(x^3)cos(x)) I got sin(sqrt(e^(x^3)cos(x))*((sqrt(e^(x^3)cos(x)) Do I just leave the e^(x^3)cosx alone since it's "e" or do I still have the find the derivative for it?

Math
Hi, I wanted to know how wanted to know if i was doing this step right for my homework. the question is Given rt=

calculus
pleaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasw help can you pleaaaaase help me find the area between y=cos(4x) and y=1cos(4x) 0

Calculus
Differentiate. f(x)=(cos x)/x I got to here: (x sin x  cos x)/x^2 Find the equation of the tangent line to f(x)=sec x  2 cos x at (pi/3, 1). Is the slope 3 times sqrt of 3? Equation is y=3 times sqrt of 3x  pi times sqrt of 3 +1??

Algebra 2: Radicals URGENT!!
Could some kind, saintly soul help me solve this problem? Simplify: 8w sqrt(48w^5)  x^2 sqrt(3xw^2) . . =8w(√16)(√3)(√w^4)(√w)  x^2(√3)(√x)(√w^2) =32w^3(√3w)  wx^2(√3x) not much of a "simplification" really 8w sqrt(16*3w^5)  x^2 w

calculus
Find the length of the curve given by the equation y= intergral from pi to x of sqrt(cos(t)) dt for x between pi and pi. I think I know to do this at least part of it. I am using the fundamental theorem of calculus and the arc length formula, but I

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.) Simplify (tan^2 theta +

math calc
find the linear approximation to square root(a+x) for x near o. a is a constant (pos.) L(x)= sqrt(a)+1/2(a+0)^1/2 (x0) is this correct? use you approximation to estimate sqrt (4+0.1) How would I do this part ? Also confused with this? find the linear

Trigonometry
If cos(a)=1/2 and sin(b)=2/3, find sin(a+b), if 1) Both angles are acute; Answer: (sqrt(15)+2)/6 ii) a is an acute angle and pi/2 < b < pi; Answer: (2sqrt(15))/6 2. Find the exact value of the six trigonometric functions of 13pi/12. Partial answer:

Trig
Compute inverse functions to four significant digits. cos^2x=35cosx rewrite it as.. cos^2x + 5cosx 3=0 now you have a quadratic, solve for cos x using the equation cosx=(5 + sqrt (25 +12)/2

TRIG
what is the exact value of sin^2 pi/62 sin pi/6 cos pi/6 +cos^2 (pi/6)? a) 2 sqrt 3 b) 2 sqrt 3/ 2 c) 2+ sqrt 3/2 d) 2+ sqrt 3 I am not sure how to work this out and would appreciate you how to show me the process. Thank you in advanced!!!!!!!

maths
Find all possible solutions in the range 0°≤ x ≤ 180° of the following equations: (a) cos x = sqrt(2) / 2 (b) tan x = sqrt(3) / 3 (c) sin x = sin60° (d) cos x = 1/2 (e) sin x = 1/2

MATHS
Find all possible solutions in the range 0°≤ x ≤ 180° of the following equations: (a) cos x = sqrt(2) / 2 (b) tan x = sqrt(3) / 3 (c) sin x = sin60° (d) cos x = 1/2 (e) sin x = 1/2

Math
cos(2 tan^1 1/3) cos(sin^1 1/5 +cos^1 1/3) For this one I got the answer to be sqrt 24/15 sqrt 8/15, is that right? I have no clue how to start the first one

Calc II
For the following question, we need to find the length of the polar curve: r= 2/(1cosx) from pi/2

Math(Please check!!!)
1) Evaluate (if possible) the sine, cosine, and tangent of the angles without a calculator. a) 10pi/3 Sin = sqrt 3/2 Cos = 1/2 Tan = sqrt 3 Are these correct ? I do not understand when to make them negative. b) 17pi/3 Sin = sqrt 3/2 Cos = 1/2 Tan = 

Calc
1. The problem statement, all variables and given/known data cos(x) = 2 I was trying to solve that problem ignoring what people say about it not being defined and got a number strangely. I showed step by step so you can tell me what I did wrong as I don't

Trigonometry
I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi2arctan(b) b=sin(pi2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b)) sin(ab)=sinacosbcosasinb

Math
How do you find a square root of a number that's not a perfect square? I'm very confused. The book doesn't explain it too well. You can approximate it or simplify it in terms of (products of) square roots of smaller numbers. E.g. consider sqrt[117] The

Surds
Solve in the exact form. (sqrt of 4x+1)+(sqrt of x+1)=2 Someone showed me to do this next: Square both sides..so.. 4x+1+2((sqrt of 4x+1)•(sqrt of x+1))=4 I do not understand where the 2 come from ..and why do we need to multiply the sqrt of 4x+1 and sqrt

Math sin/cos
On a piece of paper draw and label a right triangle using the given sides, solve for the unknown side and write the trigonometric functions for angles A and B, if a=5 and c=7. I already found side b which equals 2 sqrts of 6. Now I need to find the sin/cos

Calculus  MathMate Please help
ok, i tried to do what you told me but i cant solve it for c because they cancel each others out! the integral for the first one i got is [sin(c)cos(x)cos(c)sin(x)+sin(x)+c] and the integral for the 2nd one i got is [sin(c)cos(x)+cos(c)sin(x)sin(x)+c] I

math,algebra,help
Directions are simplify by combining like terms. x radiacal 18 3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which reduces to x sqrt (9*3) or 3x

MATH
cos(x)=2 and sin(x)=5 Thanks for your previous answers mathmate. No. It's not a typo. I was wondering if someone could explain to me when in real life would I need to solve such problems, or similar, besides answering problems from a textbook. When ever I

Pre Calc.
Use the sum or difference identity to find the exact value of sin255 degrees. My answer: (sqrt(2) sqrt(6)) / (4) Find the value of tan (alphabeta), if cos alpha= 3/5, sin beta= 5/13, 90

Homework Help Calculus
Find the linear approximation L(x)of the function f(x)=cos(pi/(6)x) at the point x=1 and use it to estimate the value of cos(13pi/72). Here's what I did so far: L(x)=sqrt(3)/21/12pi(x1)+0((x1)^2) How do I find cos(13pi/72)

Calc
Can anyone help me find the derivative of these functions? y = x^2x y = x^cos(x) y = ln(7x^2 + 5y^2) sqrt(3x+y)=2+x^2y^2 ysec(x) = 3xtan(y) y = arctan[(sqrt((1 + x)/(1  x)))]

Trigonometry
Find all the angles θ in the interval [2pi, 2pi] for which cos(3θ) = 1/sqrt(2) I'm not sure why I would have to look for angles in 3 revolutions of the circle (3 positive, 3 negative). Also how would I convert cos(3θ) = 1/sqrt(2) to a radian value? Do

calculus
find max, min and saddle points of the give function f(x,y)=sin(x)+sin(y)+sin(x+y) 0

Calculus
Hi, been tearing my hair out over this question. I think I'm on the right track, but just not sure how to finish it. Anyway, here it is: A particle is moving along the ellipse 4x^2 + 16y^2 = 64. At any time 'T' its x and ycoordinates are given by x =

find dy/ds
y = s*square root of(1s^2) + cos inverse(s) Just give me some hints and I will do it. Thanks. You have y = s*square root of(1s^2) + cos inverse(s) which I would write as y = s*sqrt(1s2) + cos1(s) and you want dy/ds For the first term use the product

mathematics
Just a few questions im having trouble with 1. sqrt(63x)2=0 A. 2/3 B. 2 C. 10/3 D. 4 2. Using a reference triangle, find the exact value of the expression: sin(cos^1(1x)) A. sqrt(x(2+x) B.1/sqrt(x(2+x)) C.1/sqrt(2+x) D.sqrt(x(2x)) 3. The ladder of a

PreCalc
Find the point (x, y) on the unit circle that corresponds to the real number 5pi/6. Use it to evaluate cos n. Is cos n = (sqrt(3)/2)?

Precalculus
Solve Cos^2(x)+cos(x)=cos(2x). Give exact answers within the interval [0,2π) Ive got the equation down to cos^2(x)+cos(x)+1=0 or and it can be simplified too sin^2(x)+cos(x)=0 If you could tell me where to go from either of these two, it would be great

Calculus
Use a trig identity to combine two functions into one so you can solve for x. (The solution should be valid for any value of t). 3cos(t) + 3*sqrt(3)*sin(t)=6cos(tx) I know that 6 cos(tx) can be 6(cos(t)cosx(x)+sin(t)sin(x)) I don't know where to go from

math
determine the area of triangle abc when side ab= square root of x, side bc= square root of x, and side ca= the square root of the square root of x. So how do I solve it. How do I add the square root of x + the square root of x + the square root of the

trig
Assume that we know sin(18 degrees)=(sqrt(5)1)/4, cos(18 degrees)=(sqrt(sqrt(5)+5))/8, find sin(42 degrees) exactly, without using a calculator. Also find cos(42 degrees).

trig question
Compute inverse functions to four significant digits. cos^2x=35cosx cos^2x + 5cosx 3=0 now you have a quadratic, solve for cos x using the equation cosx=(5 + sqrt (25 +12)/2 the back of the book says .9987+2kpi and .09987+2kpi when i do it i don't get

AlgebraPlease check
1. solve sqrt (11x) = sqrt(x7) choices are 6,7,8 or9 I think according to my calcs the answer is (9) The range of f(x) = sqrt(3x6)  2 answers would be y= (2) y> (2) y

rationalizing
sqrt(18)  sqrt(17) How does this expression equal 1/(sqrt(18) + sqrt(17)) ? How would you change it like that? And why is this good to use when not using a calculator?

trig
2sin(x)cos(x)+cos(x)=0 I'm looking for exact value solutions in [0, 3π] So I need to find general solutions to solve the equation. But do I eliminate cos(x), like this... 2sin(x)cos(x)+cos(x)=0 2sin(x)cos(x)= cos(x) 2sin(x) = 1 sin(x) = 1/2 at 4pi/3

math
Suppose z is a complex number such that z^5 = \sqrt{3} + i and z = \sqrt[5]{2} ( \cos n^{\circ} + i \sin n^{\circ}) , where 270 < n < 360 . Find the exact value of n .

MATH
Solve the equation algebraically for x where 0

MATH
For each set of numbers, draw your own number line on a piece of paper, taking care to plot each pair of irrational numbers. Then write a statement comparing the position of the two given numbers on a number line. Also write an inequality comparing the two

math,help,algebra I
I need help can someone help me get unstuck and let me know if i am correct.thank you. solve by completing the square. 4x^2+2x3=0 this is what i am doing: i used the quadratic equation. so where i am is in this step: x = (2 (+) sqrt (44))/(8) from this