
Let a, b, and c be positive real numbers. Prove that sqrt(a^2  ab + b^2) + sqrt(a^2  ac + c^2) is greater or equal to sqrt(b^2 + bc + c^2). Under what conditions does equality occur? That is, for what values of a, b, and c are the two sides equal? This

Use Property 2 to simplify each of the following radical expressions. sqrt (10)/ sqrt(49) My answer: sqrt (10) / (7) THis next one i need help: Use the properties for radicals to simplify each of the following expressions. Assume that all variables

Prove: [1/sqrt(2)] [sqrt(a) + sqrt(b)] <= sqrt(a + b) <= sqrt(a) + sqrt(b) for all nonnegative real numbers a and b.

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.

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=


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

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

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] =

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?

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<x<sqrt 3 The answer is given

the problem reads: evaluate sqrt (4) if possible. my answer: sqrt (4)<0 therefore this is not a real number looks correct you cannot take the square root of a negative number in the real number set. Have you learned the imaginary number i = sqrt(1)

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

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))

sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So you can see the steps — sqrt 6 * sqrt 8 = sqrt 48 sqrt 7 * sqrt 5 = sqrt 35 I hope this helps a little more. Thanks for asking.

I'm trying to get a handle on what square roots are (I think that's what they are called). Here is an example. 4(almost like a division sign but it is more like a check mark)3. I don't understand what it is, its a square root? and How do I deal with these


I need to simply this equation, but I got stuck. h/(4sqrt(16+h)) = y First, I multiplied (4+sqrt(16+h)/(4+sqrt(16+h) to both sides, and I ended up with h(4+sqrt(16+h)/h. Is this correct? (I tried to graph both equations to see if I would get the same

l = lim as x approaches 0 of x/(the square root of (1+x)  the square root of (1x) decide whether: l=1 or l=0 or l=1 Let me make sure I understand the question. Do we have lim x>0 x/[sqrt(1+x)  sqrt(1x)] ? If so then multiply the expression by

Let {An} be the sequence defined recursively by A1=sqr(2) and A(n+1) = sqr(2+An) for n is bigger and equal to 1. Show that An < 2. What is An? and how do I find it? Thank you for your time. An is the nth number in the sequence, and is defined by the

f(x)= 4x^2 and g(x)= sqrt (x) find the implied domain of fg(x) fg(x)= f(sqrt(x)) fg(x)= 4(sqrt(x))^2 fg(x)=4(sqrt x)(sqrt x) fg(x)=4x domain= (x:x=all real numbers) Am I correct?

am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2 2y sqrt 135*** 27

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.

Evaluate the indefinite integral: 8xx^2. I got this but I the homework system says its wrong:sqrt((x8)x)/(2*sqrt(x8)*sqrt(x))*(((sqrt(x8)*(x4)*sqrt(x))32*log(sqrt(x8)+sqrt(x))

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

Simplify: 2 sqrt (3) + 6 sqrt(2)  4 sqrt(3) + sqrt (2) a) 8 sqrt(2)  3 sqrt(3) b) 6 sqrt(2)  8 sqrt(3) c) 5 sqrt(6) d) 7 sqrt(2)  2 sqrt(3) the answer i picked was d

given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b  sqrt xa >= sqrt x+a  sqrt xb


given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b  sqrt xa >= sqrt x+a  sqrt xb

15sqrt8x^15/5sqrt2x^4 remember that sqrt(a)/sqrt(b) = sqrt(a/b) simplify the inside. also notice that sqrt(x^11) = sqrt(x^10) * sqrt(x) = x^5 * sqrt(x)

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

Use the GramSchmidt process to transform the basis [1 1 1] , [0 1 1] , [2 4 3] for the Euclidean space R3 into an orthonormal basis for R3. (Enter each vector in the form [x1, x2, ...]. Enter your answers as a commaseparated list.) so i went through the

Simplify: sqrt (21) (sqrt(7) + sqrt(3)) a) 7 sqrt (3) + 3 sqrt(7) b) 7 sqrt(3) + sqrt (3) c) sqrt (210) d) sqrt (147) + sqrt (63)''. a

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

how would you simplify this equation: y = (x+3)/[(4sqrt(16+h))] please help me! you have three variables. I am not certain "simplify" is an appropriate term here. ohhhh it was my mistake. I meant: y = h/[(4sqrt(16+h))] y = h/[(4sqrt(16+h))] rationalize

Given f(x)=x^4(2x^215). On what interval(s) is the graph of f concave upwards? A. (0, sqrt(3)) B. (sqrt(3), 0) C. (sqrt(3), 0) and (0, sqrt(3)) D. (sqrt(3), sqrt(3)) E. (Negative infinity, sqrt(3)) and (sqrt(3), infinity) I got E

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

Multiplying sq rts sqrt18a^7b times sqrt27a^8b^6 Jake 1214 18 = 9 * 2 then sqrt 18 = 3sqrt2 sqrt a^7 = a^3 * sqrt a 27 = 9 * 3 then sqrt 27 = 3 sqrt3 sqrt a^8 = a^4 sqrt b^6 = b^3 Now just multiply the liketerms together. 3sqrt6 and sqrt2a^4?


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,

Find the roots of the function f(x)= x^2+2x+2 Determine f(x) a) (x+1i)(x+1+i) b)(x+1i sqrt of 2)(x+1+i sqrt of 2) c)x1+i)(x1i) d)x1+i sqrt of 2) (x1i sqrt of 2) Please Help! For all of these, use the rule [f(x) + ai]*[(f(x)  ai} = [f(x)]^2 + a^2

prove that d/dx 4x .√(x + √x) = 6x+5 (x)1/2/√(x + √x) solution is d/dx 4x.[x+(x)1/2]1/2 = d/dx 4x.[x+(x)^1/2]^1/2 d/dx 4x.[x+(x)^1/2]^1/2 Product Rule = 4x[1/2(x+(x)^1/2)^1/2 * (1+1/2x^1/2) +[x+(x)^1/2]^1/2*4 d/dx(4 x

Simplify each expression. All variables represent positive numbers. Sqrt of 75p^3q^2 divide by sqrt of p^5q^4 My answer: 5q sqrt of 3 / p^2q^2 Is this right?

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)

The geometric mean of two postitive numbers a and b is sqrt(ab). Show that for f(x) = 1/x on any interval [a,b] of positive numbers, the value of c in the conclusion of the mean value theorem is c = sqrt(ab) I have no idea how to do this! If the mean of a

A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10? B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10? * calculus  Damon,

Which of these expressions is in simplified form? A. a ^3 sqrt 4  b ^3 sqrt 2 / 2 B. sqrt 1/2x + sqrt 1/2z C. x^2  3x sqrt y / sqrt 3 D. sqrt 125x  x^2

Rationalize the denominator, assume that all expressions under radicals represent postive numbers. sqrt:a  sqrt:b/sqrt:a + sqrt:b

I asked a question and it was answered by Steve. He had the numbers wrong. The problem should be sqrt of 18 over (sqrt of 8)3. not sqrt of 18 3. I got 2+sqrt of 8. Is it correct?


I need help writing the series 4 + 1/5 + .3 + 1/(3 + sqrt 2) + 1/(9+ sqrt 3) + 1/(27 + sqrt 4) + 1/(81 + sqrt 5 .... I have played with using irrational numbers, natural log and a vast variety of exponential arrangements. Any help to get me going in the

Can you check the following? A1. (2x + 3)^2 = 25 sqrt(2x + 3)^2 = +sqrt(25) 2x + 3 = + 5 2x + 3 = 5, 2x + 3 = 5 2x = 2, 2x = 8 x = 1, x = 4 {4, 1} A2. x^2 + 14x + 45 = 0 x^2 + 14x + __ = 45 + __ x^2 + 14x + 49 = 45 + 49 (x +7)^2 = 4 sqrt(x+7)^2 =

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"

1 last square root : Rationalize the denominator 5/sqrt[3]+sqrt[5]= 5*sqrt[3]sqrt[5]/ sqrt[3]sqrt5[5]*sqrt[3]sqrt[5]= 5sqrt[3]5sqrt[5]/sqrt[9]sqrt[15]+sqrt[15] sqrt[25]= 5sqrt[3]5sqrt[5]/16 = 5sqrt[3]5sqrt[5]/ 2

simplify: sqrt ((7)/(100)) my work: 7 = 1sqrt (7) 100= 10 sqrt (1) (1)/(10) sqrt (7) my answer: (sqrt (7))/(10) Your answer is right but your statements: 7 = 1sqrt (7) 100= 10 sqrt (1) make no sense. how can sqrt(7)=7 ? you are saying 2.6457..=7 and

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?

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

simplify by combining like terms: x sqrt(18)  3 sqrt(8x)sqrd If <sqrt(8x)sqrd> means [sqrt(8x)]^2, then that also equals 8x. So x sqrt(18)  3 sqrt(8x)sqrd = x[sqrt(9*2)3*8] = 3x(sqrt 2 3

so we are doing integrals and I have this question on my assignment and I can't seem to get it, because we have the trig substituion rules, but the number isn't even so its not a perfect square and I just cant get it, so any help would be greatly

Prove algebraically that w+z is less than or equal to w+z for any complex numbers w and z, where  is the magnitude. After letting w = a+bi and z = c + di and doing some plugging in etc. I got that acbd <= b^2 + sqrt (a^2+b^2)(c^2+d^2) + d^2 But


Which of the following definite integrals gives the length of y = e^(e^x) between x=0 and x=1? All the answers are preceded by the integral sign from 0 to 1. (a) sqrt[1 + e^(2*(x+e^x))] dx (b) sqrt[1 + e^(4x)] dx (c) sqrt[1 + e^(x+e^x)] dx (d) sqrt[1 +

Simplify 11 sqrt 112. 44 sqrt 7******** 176 sqrt 7 27 sqrt 7 4 sqrt 7 Is this right? and I need to show my work, idk the steps

I have a question if someone could help, please explain it. Given b = 1 and h = 1, what is the equation of the graph if the parent function is y = sqrt (x) Answers: a.y = sqrt(x1) b.y = sqrt (x+1) c.y =  sqrt (x1) d.y = sqrt(x1) I think it is "b"

show lim x>3 (sqrt(x)) = (sqrt(c)) hint: 0< sqrt(x)sqrt(c)= (xc)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*xc

show lim x>3 (sqrt(x)) = (sqrt(c)) hint: 0< sqrt(x)sqrt(c)= (xc)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*xc

Which is the exact value of the expression sqrt 32 sqrt 50 + sqrt 128 2 sqrt 7 7 sqrt 2 22 sqrt 2 2 sqrt 55

find the domain of the real valued function; f(x) = sqrt(5  (sqrtx)) my solution: 5  (sqrt x) >=0 (sqrt x) >= 5 (sqrt x) <= 5 x <= 25 (infinity, 25] is this correct?

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

Could someone show me how to get the work for: 3/4 (sqrt)16/25 ? The answer says 3/5i. Could anybody tell me how to get there and work with complex fractions and simplify them? Thanks! 3/4 SQRT(16/25) 3/4 SQRT( 16/25*1) 3/4 *4/5 *SQRT( 1) but the sqrt

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


Posted by Megan on Tuesday, October 12, 2010 at 3:44pm. I have a question if someone could help, please explain it. Given b = 1 and h = 1, what is the equation of the graph if the parent function is y = sqrt (x) Answers: a.y = sqrt(x1) b.y = sqrt (x+1)

For what values of x is the graph of y = 8e^−x^2 concave down? (Enter your answer using interval notation.) I started by finding the second derivative and factoring and ended up getting 16e^x^2 (2x^21) and I know up till this part that I'm doing

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.

Find of g(f(x)) when f(x)=sqrt(x+3) and g(x)=(x^2+2)/x. a. g(f(x))=(x^2+2)(sqrt(x+3))/x b. g(f(x))=(x+5)/(sqrt(x+3)) c. g(f(x))=(x^2+6x+11)/(sqrt(x+3)) d. g(f(x))=(sqrt(((x^2+2)/x)+3) Every time I work this problem, I get some crazy answer. Can someone

Write the following with rational, positive exponents. a. sqrt of 8 = 8^1/2 b. seventh root of 5^2 = 5^2/7 c. sqrt sqrt 6 = ??? d. fourth root of sqrt of 5^3 = ??? Can someone please check my work and help with "c and d"? Thanks.

2^x+20=15+3^x A)1 B)16 C)5 D)25 I chose c but i got it wrong i am having a problem understanding it. Let me rewrite the question in a more standard format. 2* sqrt(x)+20=15 + 3*sqrt(x) where * means multiply, and sqrt means the square root. Your use of ^

SIMPLIFY: ROOT 0F 7 * ROOT OF 14 SIMPLIFY: ROOT OF 1/5 EVALUATE, IF POSSIBLE: ROOT OF 9/16 (I GOT 3/4 BUT I WAS NOT SURE IF THAT WAS CORRECT OR NOT) SIMPLIFY. ASSUME X REPRESENTS A POSITIVE NUMBER: ROOT OF 45X^4 ROOT 0F 7 * ROOT OF 14 = sqrt 7 * sqrt 7 *

simplify: sqrt (28) a) 4 sqrt (7) b) 7 sqrt (2) c) 2 sqrt (7) d) sqrt (7) c im just making show that they are correct sorry for posting up so many

How would I do: sqrt32a^8b + sqrt50a^16b Are the exponents 8b and 16b or 8 and 16? I'll assume they are 8a and 16a sqrt 32 = sqrt (16*2) = 4 sqrt 2 sqrt 50 = sqrt (25*2) = 5 sqrt 2 sqrt32 a^(3b) + sqrt50a^(2b) = sqrt2*a^(8b)[4 + 5(a^8b)] There are many

d/dx x(1+y)^(1/2) + y(1+x)^(1/2)=0. Can some1 show me the calculation because i didn't get like the answer. [ans=1/(1+x)^2 but my calculation=(sqrt(y+1) (2 sqrt(x+1) sqrt(y+1)+y)/(sqrt(x+1) (2 sqrt(x+1) sqrt(y+1)+x)]


d/dx x(1+y)^(1/2) + y(1+x)^(1/2)=0. Can some1 show me the calculation because i didn't get like the answer. [ans=1/(1+x)^2 but my calculation=(sqrt(y+1) (2 sqrt(x+1) sqrt(y+1)+y)/(sqrt(x+1) (2 sqrt(x+1) sqrt(y+1)+x)]

Integral of: __1__ (sqrt(x)+1)^2 dx The answer is: 2ln abs(1+sqrt(x)) + 2(1+sqrt(X))^1 +c I have no clue why that is! Please help. I used substitution and made u= sqrt(x)+1 but i don't know what happened along the way! Your first step was a good one.

Solve by whatever Method 1. X^2 +8X = 16 X^2 +8X +16 = 16 +16 (X +4)^2 =0 X+4 =+/ SQRT 0 X=4+/ 0 X=4, X=4 2. 3x^22x5=0 X=b+/ sqrt (b)^24ac 2a a=3 ,b=4,c=5 X=(2)+/ Sqrt (2)^24(3)(5) 2(3) X=2+/sqrt64 = 2+/8 6 X= 2+8=10/6=1 2/3,

Verify that the function satisfies the three hypotheses of Rolle's Therorem on the given interval. Then find all numbers c that satisfy the conclusiton of Rolle's Theorem. f(x)= x*sqrt(x+6) [6,0] f is continuous and differential f(6) =6*sqrt(6+6) =0

Determine dy/dx if: (i) Y= sqrt x^sqrt x (ii) Y= xarcsin(x) + sqrt(1x) Please show all steps. Thanks 1) http://www.analyzemath.com/calculus/Differentiation/first_derivative.html 2) an antiderivative of arcsin(x) is xarcsin(x)sqrt(1x^2); are you certain

Simplify the number using the imaginary unit i. (sqrt)of 28 •2(sqrt)7 •2 (sqrt)7 •i (sqrt)28 •2i (sqrt)7

Find the product and enter the real and imaginary parts below. Note that you do not need to type in the since it is already given to you. Simplify your square roots as much as possible. sqrt27((sqrt5)(sqrt11))

Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x: âˆ«[4,1] sqrt(1+(dy/dx)^2) dy/dx =

Find the product and enter the real and imaginary parts below. Note that you do not need to type in the since it is already given to you. Simplify your square roots as much as possible. sqrt27((sqrt5)(sqrt11))

Graph the curve and find its exact length. x = e^t + e^t, y = 5  2t, from 0 to 3 Length = Integral from 0 to 3 of: Sqrt[(dx/dt)^2 + (dy/dt)^2] dx/dt = e^t  e^t, correct? dy/dt = t^2  5t, correct? So: Integral from 0 to 3 of Sqrt[(e^t  e^t)^2 +


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

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

Integrate: 1/(xsqrt(x+2) dx I came up with: (2/3)(2*ln((sqrt(x+2))2)+ln((sqrt(x+2))1)) but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how? I found: Ln[xsqrt(x+2)] +

For f(x) = 2sinx + (sinx)^3 + tanx find f'(pi/3). Ok, so what I tried was... f'(x) = 2cosx + cosx(3(sinx)^2) + (sinx/cosx) pi/3 = (1/2, sqrt(3)/2) therefore, 2(1/2) + 1/2(3(sqrt(3)/2)(sqrt(3)/2) + (sqrt(3)/2 (2/1)) 1 + .5(3 (3/4)) + sqrt(3) 1 + .5(9/4) +

Question: ∫(x^2)/sqrt(x^2+1) u=x^2+1 , x^2= u1, du=2xdx ∫(u1)/sqrt(u) , expand ∫u/sqrt(u)  1/sqrt(u) Integrate: 2/3(u^(3/2))  2u^(1/2) + c My answer: [ 2/3(x^2+1)^(3/2)  2(x^2+1) + c ] When I took the derivative of this to check my answer, it

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 186 N/m. The motion of the object on spring 1 has 3 times the amplitude as the motion of the object on spring

Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from sqrt(1x^2) to sqrt(1x^2) The inner one goes from sqrt(1x^2z^) to sqrt(1x^2z^) for 1/sqrt(x^2+y^2+z^2) dydzdx I don't

Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from sqrt(1x^2) to sqrt(1x^2) The inner one goes from sqrt(1x^2z^) to sqrt(1x^2z^) for 1/sqrt(x^2+y^2+z^2) dydzdx I don't

Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from sqrt(1x^2) to sqrt(1x^2) The inner one goes from sqrt(1x^2z^) to sqrt(1x^2z^) for 1/sqrt(x^2+y^2+z^2) dydzdx I don't

Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from sqrt(1x^2) to sqrt(1x^2) The inner one goes from sqrt(1x^2z^) to sqrt(1x^2z^) for 1/sqrt(x^2+y^2+z^2) dydzdx I don't


A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10? B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10? C) They give a

A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10? B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10? C) They give a

A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10? B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10? C) They give a

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

If y = sqrt(x1)+sqrt(x+1),prove that sqrt(x^21)dy/dx=1y/2