Math log equations/exponential form
Beats me. A log is a log, not an exponential form. You have 1/2 log2 + 2logw + 3logx - 5logy - 7logz similarly for (b) do you have an example of another problem which is solved, indicating what you wish to achieve?
sin2x = cos(x+30) when x=120, sin 240 = cos 150 There are other values, but you have to solve a 4th-degree equation to get them. 2sinx cosx = cos30 cosx - sin30 sinx square both sides, collect terms, convert sin^2 = 1-cos^2, and square both sides again. Gets messy
Algebra II Part 2
An excellent grapher can be found at http://rechneronline.de/function-graphs/ Just enter -3*cos(x) as your function
algebra II Part 2
for h=1, 1 = 7 cos(pi/3 t) cos(pi/3 t) = 1/7 pi/3 t = 1.427 t = 1.363 and so on for the others
visit http://rechneronline.de/function-graphs/ enter your function. Change the range to be -20 to 20 and you will see the graph.
recall that a^3-b^3 = (a-b)(a^2+ab+b^2) You have (6x)^3 - y^3 = (6x-y)((6x)^2 + (6x)y + y^2) = (6x-y)(36x^2 + 6xy + y^2)
sin 237 = sin(180+57) = -sin 57 = -m cos 33 = cos(90-57) = sin 57 = m cos 3 = cos(60-57) = cos60cos57+sin60sin57 = m/2 + m√3/2 = m/2(1+√3) cos 66 = cos(60+6) = cos60cos6 - sin60sin6 Use your double angle formulas to get sin(6) and cos(6) from cos(3) and sin(3)
dilate by 2: (x,y) -> (2x,2y) up 5: (x,y) -> (x,y+5) rotate: (x,y) -> (-y,x) combining them, (x,y) -> (2x,2y) -> (2x,2y+5) -> (-(2y+5),2x) so, (1,4) -> (-13,2) and so on for the other points
Algebra II Part 2
I think she means θ when using 0 I'll use x to avoid copy/paste of special characters. one way: recall the definition of co-functions: function of complementary angle. So, cos(x) = sin(pi/2 - x) cot(x) = tan(pi/2 -x) so, cot(x-pi/2) = -cot(pi/2 - x) = -tan(x) or, use ...
2sin^2 x - 5sinx - 3 = 0 (2sinx+1)(sinx-3) = 0 so, sinx=3 or sinx = -1/2 now, sinx is never 3, so that's out sin π/6 = 1/2, so that's your reference angle sin x < 0 in QII,QIV, so we want x = π + π/6 x = 2π - π/6 in general, then recalling t...
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