2cos[x-pi/8]-sqrt2=0 Pi=3.141592654

2cos[x-pi/8]-sqrt2=0Pi=3.141592654
1. 1 answer
2. asked by Adam
3. 478 views
Assume that no denominator equals 0.sqrt12 - sqrt18 + 3sqrt50 + sqrt75 = (sqrt2^2*3) - (sqrt2*3^2) + (3sqrt2*5^2) + (sqrt3*5^2)
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2. asked by Jon
3. 736 views
Given: f(x) = X^2-xevaluate: f(sqrt2 + 3) I have worked out this and am stuck: (sqrt2+3)^2 - (sqrt2+3) (sqrt2+3)*(sqrt2+3)-
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2. asked by Steve
3. 613 views
Hi--can anybody show me how to do this?Find the length of the curve r(t)= (sqrt2)i + (sqrt2)j + (1-t^2)k from (0,0,1) to (sqrt2,
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2. asked by Madison
3. 566 views
Is this correct:sqrt2/sqrt10=sqrt2/5sqrt2 nope. it would be sqrt2/(sqrt5)(sqrt2) no.d sqrt2/sqrt10= sqrt (2/10)= sqrt (1/5) sqrt
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2. asked by margie
3. 820 views
I have tried solving this problem over and over but I'm still not sure with my answer.The question is sqrt2 / 2 + sqrt2. I think
1. 3 answers
2. asked by Tommy
3. 534 views
Simplify: (2^(1+sqrt2)/2^(1-sqrt2))^sqrt2I need all of the steps. I need this ASAP because this assignment is due tomorrow.
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2. asked by Brandon
3. 499 views
Can someone please check my work on this problem? For some reason, I get the feeling that I did it wrong.sqrt=square root (by
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2. asked by Whitney
3. 747 views
Not sure where to go next. I need to solve this quadratic equation 7x^2-5=0.This is what I have so far, not sure where/what to
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2. asked by dani
3. 1,208 views
cosA= 5/9 find cos1/2Aare you familiar with the half-angle formulas? the one I would use here is cos A = 2cos^2 (1/2)A - 1 5/9 +
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2. asked by Anonymous
3. 847 views