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Reiny2

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Recall that sin^2 x = 1/2 - (1/2)cos (2x) so... ∫(sin(x^2))dx from 0 to 1/2 = ∫(1/2 - (1/2)cos (2x) dx from 0 to 1/2 = x/2 - (1/4)sin (2x) from 0 to 1/2 = ......
slope = (6-(-2))/(-4-(-8) = 2 using the point (-4,6) y-6 = 2(x+4) y-6 = 2x + 8 y = 2x + 14 to check, sub both given points into this equation. Both points work!
#1 take the derivative, sub in x = 1 Now you have the slope of the tangent Sub x=1 into the original equation to get y So now you have the slope and a point. Use the method you learned probably in grade 9 to find the equation of the tangent. #2 I would use
Use the cosine law to find angle Q, the smallest angle 10^2 = 17^2 + 21^2 - 2(17)(21)cosQ cosQ = 630/714 = .... Q = ..... sinQ = h/17 h = 17(sinQ)
(I am Reiny , but not using my regular IP address) This is a poorly worded question, so you are not stupid. 1st equation: 20m + 40w = 240 divided each term by 20 m + 2w = 12 2nd equation : 5m + 15w = 110 divide each term by 5 m + 3w = 22 now subtract them,

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