Physics(please check)

X=(Vo cos Q) t

X=(3.107 cos 0)(0.443)

I just wanted to make sure that I put this equation into my calculator correctly. I have a TI-30 so first I put in 0 then cos to get 1. Then I multiplied by 3.107 to get 3.107 and then multiplied by 0.443 to get 1.376 .

Did I do this correctly? Thank you!

asked by Hannah
  1. Please diregard. Thank you!

    posted by Hannah

Respond to this Question

First Name

Your Response

Similar Questions

  1. TRIG!

    Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
  2. math

    Determine exact value of cos(cos^-1(19 pi)). is this the cos (a+b)= cos a cos b- sina sin b? or is it something different. When plugging it in the calculator, do we enter it with cos and then the (cos^-1(19 pi)).
  3. pre-cal

    Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x
  4. math

    Find the exact value of cos 300 degrees. thanks guys cos 300 = 1/2 = 0.500 how do you know? I am supposed to show my work. You ought to know the rule on 30-60-90 triangles. If the hyp is 2, the shorter side is 1, and the longer
  5. 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)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)
  6. 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)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)
  7. maths

    Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) -cos(pi/6)][(cos(x) -cos(pi/2)][(cos(x) -cos(5pi/6)]
  8. maths

    Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) -cos(pi/6)][(cos(x) -cos(pi/2)][(cos(x) -cos(5pi/6)]
  9. 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
  10. 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 -

More Similar Questions