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  1. Question #33eb8 - Socratic

    First, subtract #color (red) ($145)# from each side of the equation to isolate the #w# term while keeping the equation balanced:

  2. The sum of the first nth term of a geometric series is 145 ... - Socratic

    The sum of the first nth term of a geometric series is 145 and the sum of the reciprocal is 145/33. The first term is 1. What is n and the common ratio? Precalculus

  3. An isosceles triangle has sides A, B, and C, such that sides

    12.04 (approx) Let the length of side A or B = x and base is 16. So the height is sqrt [x^2 - (16/2)^2] Now the area of triangle = 1/2 * base * height = 1/2 * 16 * sqrt [x^2 - (16/2)^2]. …

  4. Question #933a2 - Socratic

    Solve: 1. sin x = 0.5623 2. cos x = 0.8963 3. tan x = -2.89 1. Calculator gives --> sin x = 0.5623 --> x = 34.22 and x = 180 - 34.22 = 145.78 deg. Extended answers: x = 34.22 + k360 deg x = …

  5. Solve # { (sqrtx (1-4x-y)=2), (sqrty (1+4x+y)=6):}# - Socratic

    No real solutions. Calling alpha = 4x+y we have { (sqrtx (1-alpha)=2), (sqrty (1+alpha)=6):} then 2 alpha = (1+alpha)^2- (1-alpha)^2=36/y-4/x or 4x+y = 9/y-1/x we ...

  6. Can you help me with finding perimeter of this sector fragment?

    Aug 26, 2017 · Just Q8. I'll ask more questions if I need more help.

  7. If the measures of a triangle are 90°, 55°, and 3x - Socratic

    => x = 10 The measures (angles) of a triangle sum to 180^o. =>90 + 55 + (3x+5) = 180 =>145 + (3x+5)=180 =>3x + 5 = 35 =>3x = 30 => x = 10

  8. What is the formula for the sequence #4,2,3,4,3,2,...# - Socratic

    a_n = 1/120 (3 n^5-40 n^4+145 n^3+40 n^2-868 n+1200) If you want a formula for a sequence with these as the first six terms then you can proceed as follows: Write down the original …

  9. Sans the lone satellite Luna, our planet Earth had cleared ... - Socratic

    Another (2015 TB 145 ) came a little beyond Moon's maximum apogee distance of 405400 km. In view of all these findings, it is reasonable to admit that the Earth is yet to clear some NEOs like …

  10. Circle A has a center at # (12 ,9 )# and an area of #16 pi#. Circle B ...

    Yes. First we need to find the radii of the circles. We can do this using the formula for area: "Area=pir^2 Circle A pir^2=16pi r^2=16=>r=sqrt(16)=4 Circle B pir^2=67pi r^2=67=>r=sqrt(67) …