Use point-slope form, y + 6 = (-1/4)(x + 1) -------- Attn: You don't need to change it to slope-intercept form.
Use point-slope form, y + 6 = (-1/4)(x + 1) -------- Attn: You don't need to change it to slope-intercept form.
Let x be the amount contributed by her parents. Using proportion, x/203 = 3/7 Solving for x, x = $87
-216x^3+1 = 1 - (6x)^3 = (1-6x)(1+6x+36x^2) = (-6x+1)(1+6x+36x^2) So, the answer is A.
Use mental substitution, ∫e^(x+e^x)dx = ∫e^x * e^(e^x) dx = ∫e^(e^x) de^x = e^(e^x) <==Answer
Let v = the rate of the holicopter. So 4v = the rate of the jet. Balance by the total time, 1080/(4v) + 180/v = 5 Multiply both sides by 4v, 1080+720 = 20v v = 90 mph 4v = 360 mph Answer: The rate of the jet is 360 mph.
To answer 10 out 12 questions, you can take 2 away out of 12. So the number of ways is 12C2 = 12*11/2 = 66 ways. If you have to answer 1-3, then you need to take 2 away out the remaining 9 questions. So the number of ways is 9C2 = 9*8/2 = 36 ways
Take away the 40 butterflies collected by the teacher, and divide the result by 12. That's the answer. (100-40)/12 = 5 butterflies collected by each child.
y = ax2+bx+c Plug in (0, 8), 8 = c Plug in (2, 20), (-2,-4), 20 = 4a+2b+c ......(1) -4 = 4a-2b+c ......(2) (1)-(2): 24 = 4b b = 6 From (2), a = (1/4)(2b-c-4) = (1/4)(2*6-8-4) = 0 Answer: y = 6x+8 (This is a straight line.)
Let x be the original amount. Balance by spending, x - (4/5)^2 x = 72, focusing on the remaining part each time. Solve for x, x = $200 Answer: The original amount was $200. ------- Hi Murtaza; You are correct. I misread the problem. This problem can be done as above...
sin(sin^-1x-cos^-1x) = sin(sin^-1x)cos(cos^-1x) - cos(sin^-1x)sin(cos^-1x) = x^2 - (1-x^2) = 2x^2 - 1 ------ Attn: cos(sin^-1x) = sqrt(1-x^2), since the domain of sin^-1x is in the first and fourth quadrants, where cos x is positive. sin(cos^-1x) = sqrt(1-x^2),...
Let x = fada (tanx + cotx)/(secx cscx) = tanx/(secx cscx) + cotx/(secx cscx) = sin^2x + cos^2x, since tanx/(secx cscx) = sin^2x, cotx/(secx cscx) = cos^2x = 1
csc [cos^-1(-square root of 3 divided by 2)] = csc(5pi/6) = 1/sin(5pi/6) = 2
tan^4x+2tan^2x+1 = (tan^2x+1)^2 = sec^4x, since tan^2x+1 = sec^2x
By completing the square, (tanx+3)^2 = 5 tanx = -3+/-sqrt(5) Can you get x now?
(sin9x+sin5x)/(cos9x-cos5x) = (sin(7x+2x) + sin(7x-2x))/(cos(7x+2x)-cos(7x-2x)) = 2sin7x cos2x / (2sin7x sin 2x) = cot2x
sin x = 4/5 cos x = -3/5, by 3-4-5 Pythagorean triple and Q. II cos x/2 = sqrt[(1/2)(1+cos x)] = sqrt[(1/2)(1-3/5)] = sqrt(5)/5
Completing the square, (tanx-3)^2 = 9-4 = 5 tan x = 3+sqrt(5), x = 1.38 rad, 4.52 rad or tan x = 3-sqrt(5), x = .65 rad, 3.79 rad. Answer: x = .65 rad, 1.38 rad, 3.79 rad, 4.52 rad.
Method 1. sin(x+pi)+sin(x-pi) = 2sin x cos pi, after expanding and collecting like terms = -2sin x Method 2. sin(x+pi)+sin(x-pi) = -sin x - sin x, using reference angle = -2sin x
Factor, (tan x + sqrt3)(tan x - 1) = 0 tan x = -sqrt3, => x = 2pi/3, 5pi/3 or tan x = 1, => x = pi/4, 5pi/4 Answer: x = pi/4, 2pi/3, 5pi/4, 5pi/3
sin^2x = 1/4 sin x = 1/2, x = pi/6, 5pi/6 or sin x = -1/2, x = 7pi/6, 11pi/6 Answer: x = pi/6, 5pi, 7pi.6, 11pi/6.