Recent content by neutron star
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Derivative ^x, is this correct?
Homework Statement y=\sqrt{x} * 19^x Homework Equations The Attempt at a Solution y=x^{1/2}*19^x y'=1/2x^{-1/2}*19- neutron star
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- Derivative
- Replies: 3
- Forum: Calculus and Beyond Homework Help
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Derivative of a function, Is my answer correct?
Ugh, it said incorrect...- neutron star
- Post #4
- Forum: Calculus and Beyond Homework Help
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What is the derivative of z=(te^6t + e^7t)^5?
Homework Statement Find the derivative of the given function z=(te^6t + e^7t)^5 Homework Equations The Attempt at a Solution 5(te^6t + e^7t)^4- neutron star
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- Derivative
- Replies: 5
- Forum: Calculus and Beyond Homework Help
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Derivative of a function, Is my answer correct?
Homework Statement y=e^9Θsin(8Θ) Homework Equations The Attempt at a Solution dy/dΘ=e^9Θ(sin(8Θ)+8cos(8Θ))- neutron star
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- Derivative Function
- Replies: 5
- Forum: Calculus and Beyond Homework Help
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Derivative, Is my answer correct?
Homework Statement y=sin^6[SIZE="2"](4Θ) Homework Equations The Attempt at a Solution dy/d\theta=24sin^5(4\theta)cos(4\theta)- neutron star
- Thread
- Derivative
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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How in the blazes is this the derivative of f(x)?
Homework Statement f(x)=(2x+1)^11 (5x-1)^9Homework Equations The Attempt at a Solution f'(x)=[(2x+1)^10 (5x-1)^8] * (200x+23) Where does the +23 come from, I don't get it. I get the rest.- neutron star
- Thread
- Derivative
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Forces on an inclined plane, kinetic friction How is this wrong?
I don't think it's possible to solve that without having a. You have: T-38sin(a)-(0.26)(N)=3.8a + 55-T=5.5a It would be 38sin(a)-(0.26)(N)+55=9.3a You can't put in the normal force because you need acceleration to solve m1gcos(a) and you can't get rid of that 38sin(a). I'm guessing...- neutron star
- Post #3
- Forum: Introductory Physics Homework Help
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Solving a Strange Derivative Problem
Hey, thanks that was really helpful! I didn't think of dividing by t^7. That made it easy! Thank you! I'll remember this if I get another similar problem!- neutron star
- Post #4
- Forum: Calculus and Beyond Homework Help
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Forces on an inclined plane, kinetic friction How is this wrong?
Homework Statement In the figure below, m1 = 3.8 kg, m2 = 5.5 kg, and the coefficient of kinetic friction between the inclined plane and the 3.8-kg block is μk = 0.26. Find the magnitude of the acceleration of the masses and the tension in the cord...- neutron star
- Thread
- Forces Friction Inclined Inclined plane Kinetic Kinetic friction Plane
- Replies: 3
- Forum: Introductory Physics Homework Help
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Graphing a Piecewise Function: Is it Continuous and Differential?
Homework Statement This is the last of 10 calculus problems I have to do this week (besides the one I have half finished) 7/8 so far :). This problem asks if the function is continuous and differential. I don't know how to graph this though. Can somebody help explain, I only need to know how...- neutron star
- Thread
- Function Graphing
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Solving a Strange Derivative Problem
Homework Statement f(t)=\frac{t^5 + t^6 - 1}{t^7} Homework Equations The Attempt at a Solution This is different than the other problems I've been doing. My first guess would be that I would do this: f(t)=\frac{5t^4 + 6t^5}{7t^6} Is that the final answer or is there another...- neutron star
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- Derivative Strange
- Replies: 6
- Forum: Calculus and Beyond Homework Help
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Is This Derivative Calculation Correct?
Homework Statement y= 4^x + 9/x^3 Homework Equations The Attempt at a Solution y'= ln(4)4^x-27x^-^4- neutron star
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- Derivative
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Calc Tension of 2 Angles: 590-N Ball Suspended from Cords
Wait, ok so T_{2}=T_{1}cos(50) so T_{1}sin(90)+(T_{1}cos(50))sin(50)-F_{g}=0 right? T_{1}=0.670F_{g} T_{1}=590/.670=880.597N This doesn't seem right, what did I do wrong?- neutron star
- Post #5
- Forum: Introductory Physics Homework Help
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Quick thing How did they get this?
https://www.physicsforums.com/showthread.php?t=259689 In that thread how did the person figure out the contact force between mA and mB? I know they got the acceleration by doing F/mA+mB. But I don't get how they got the contact force, because all the blocks are equal mass. Something like...- neutron star
- Thread
- Replies: 1
- Forum: Introductory Physics Homework Help
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Newton's second law and direction of force
0=22.1m/s^2+2a(55.3) -110.6a=488.41m/s a=-4.416m/s 0=22.1m/s^2+(-4.416m/s)t 22.1m/s^2/4.416m/s=5.004s t=5.004s Is this right? How do I get mass from this? F=ma right, so m=F/a or 12.6=m(-4.416) or m=12.6/-4.416. But it doesn't seem to work...- neutron star
- Post #3
- Forum: Introductory Physics Homework Help