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C2 questions
Help is much appreciated thank you.
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
2. open circuluar cylinder of radius r, and heigh H, surface area including base is 250 cm
show volume is given by V = 125R - PIER^3/2
thanks!!!
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
2. open circuluar cylinder of radius r, and heigh H, surface area including base is 250 cm
show volume is given by V = 125R - PIER^3/2
thanks!!!
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
I assume that here that the 'i' is a typo, rather than indicating the imaginary number 'i'.
If cos(t-20) = -.437 then find the values of t-20 (make sure you get them all using the ASTC quadrants rule), then add 20 degrees to each to find theta. It's important that you apply the quadrants rule before adding 20 degrees.
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
3tanx=2cosx
Can you think of anything that you could multiply by to simplify the equation so that you may then use an identity to get the equation in terms of one type of trigonometric ratio? Remember also that the quadratic equation can be used for things such as a(sinx)^2 + b(sinx) + c = 0 to find sinx and hence x.
2. open circuluar cylinder of radius r, and heigh H, surface area including base is 250 cm
show volume is given by V = 125R - PIER^3/2
show volume is given by V = 125R - PIER^3/2
V=(pi)(r^2)(H)
SA=(pi)(r^2) + 2(pi)(r)(H)=250
Clearly we need to eliminate H from V.
250-(pi)(r^2) = (2)(pi)(r)H
H = 125/(pir) - (1/2)(r)
By substitution: V=(pi)(r^2)(125/pir - (1/2)r) = 125r - (1/2)(pi)r^3
Reply 3
LondonBoy
Help is much appreciated thank you.
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
2. open circuluar cylinder of radius r, and heigh H, surface area including base is 250 cm
show volume is given by V = 125R - PIER^3/2
thanks!!!
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
2. open circuluar cylinder of radius r, and heigh H, surface area including base is 250 cm
show volume is given by V = 125R - PIER^3/2
thanks!!!
3tanx=2cosx
3sinx=2(cosx)^2=2 - 2(sinx)^2
2(sinx)^2 + 3sinx - 2 = 0
You can finish that
Reply 4
8.a solve for 0-360degree, cosi(theta-20)= - 0.437
Interval in range is; -20 --> 340 degrees
cos^-1 -0.437 = 115.9 degrees
cos theta = cos (- theta) = -115.9 degrees
= cos (360 - theta) = 360 - 115.9 = 244.1 degrees
= cos (theta + 360) = -115.9 + 360 = 244.1 degrees
theta - 20 = 115.9
theta = 115.9 + 20 = 135.9 degrees
theta = 244.1 + 20 = 264.1 degrees
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
3(sinx/cosx) = 2cosx
3sin(x) = 2cos^2(x)
3sin(x) = 2(1 - sin^2(x))
2 - 2sin^2(x) = 3sin(x)
2sin^2(x) + 3sin(x) - 2 = 0
(2sin(x)-1)(sin(x)+2) = 0
sin(x) = 1/2 or sin(x) = -2 (not a solution)
sin(x) = 1/2
x = sin^-1(1/2) = 30 degrees
sin(x) = sin(180 - x) => x = 180 - x = 180 - 30 = 150 degrees
Ah, that's my c2 revision done for the day.
Interval in range is; -20 --> 340 degrees
cos^-1 -0.437 = 115.9 degrees
cos theta = cos (- theta) = -115.9 degrees
= cos (360 - theta) = 360 - 115.9 = 244.1 degrees
= cos (theta + 360) = -115.9 + 360 = 244.1 degrees
theta - 20 = 115.9
theta = 115.9 + 20 = 135.9 degrees
theta = 244.1 + 20 = 264.1 degrees
b. find the exact values of x in the interval 0<x<360 for which
3tanx=2cosx
3(sinx/cosx) = 2cosx
3sin(x) = 2cos^2(x)
3sin(x) = 2(1 - sin^2(x))
2 - 2sin^2(x) = 3sin(x)
2sin^2(x) + 3sin(x) - 2 = 0
(2sin(x)-1)(sin(x)+2) = 0
sin(x) = 1/2 or sin(x) = -2 (not a solution)
sin(x) = 1/2
x = sin^-1(1/2) = 30 degrees
sin(x) = sin(180 - x) => x = 180 - x = 180 - 30 = 150 degrees
Ah, that's my c2 revision done for the day.
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