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Q: Buyer sec 2 1

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1 min 40 sec or 100 sec.

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta

8 qt = 1 gallon 60 seconds = 1 minute 2 qt/min = 2qt/60 sec = 1/30 qt/sec = 1/30 (1/8 gallon)/sec = 1/240 gallons/sec ≈ 0.0042 gallons/sec

1 min 57 secondsHere's how.Convert to Seconds3 Min & 54 Sec = 3 X 60 + 54 = 234 SecDivide234 / 2 = 117 SecRe-convert to minutes and seconds.117 Sec = 1 Min & 57 SecORBorrow 1 minute to add to the Seconds.3 Min & 54 Sec = 2 Min & (60 + 54) Sec = 2 Min & 114 SecDivide by 2 = 1 Min & 57 Sec

we know , 1 min =60 sec 100 sec =1 min 40 sec

1.5 Hz =1.5 1/sec 1.5/sec * 2m= 3m/sec

sec(x) = 2 so cos(x) = 1/2 and so x = pi/3

f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2

Yes, it is. the basic identity is for a double angle relation: cos 2x = 2 cosx cos x -1 since sec x =1/cos x if we multiply both sides by sec x we get cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x

integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C

American Snowmobiler - 2007 Buyer's Guide 1-1 was released on: USA: 2 October 2007

2-7-1 (2-5 sec)

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