Mathematics

Given that tan t = -1 and that the terminal side of t lies in Quadrant IV, find sin t

Thanks for your help!-sin t = - (√2)/2 ts reference angle is 45º...in the fourth quadrant, that is 315º @Σ-sin(t)=-1/sqrt(2).............Ans because in IV- Quadrant sin(t)and tan(t) are negative and

What are the domain, range, and equation of the asymptote for f(x) = 2^x

If you seen this question before, I made a mistake on the other one!-Domain: ℝ Range: y > 0 Asymptote: y = 0

Trigonometry: Solve the equation for all angles

Not too sure what to do here.Please help me out.Thank you. Solve the equation for all angles, x, in the interval [0degrees, 360degrees). Equation: -cos2theta = 3 - 3sintheta-There is no need to use t

How to solve this quadratic math problem

How to obtain the quadratic equation whose roots are the squares of the roots of the equation 2x^2 + 26x + 80=0?thank you!-2x^2 + 26x +80 = 0 x^2 + 13x + 40 = 0 (x+5)(x+8) = 0 x = -5; -8 so you want

Math problem about secant lines

The graph g(x)=x^3-2 passes through point T(1,-1). Determine an algebraic expression that represents the slope of any secant line TS that passes through T and S(x, x^3-2). I tried this using the slop

How do I find g(x) = ∫(-inf,inf){f(x,y)}dy, given the following f(x,y)

f(x,y) = exp( - (x^2 + y^2 - 2Rxy) / 2(1-R^2)) (-1 Ive already worked out ∫(-inf,inf) ∫(-inf,inf){f(x,y)} dxdy = 2pi for R=0, and shown that ∫(-inf,inf){exp(-ax^2)} dx = sqrt(pi/a) for R=0 and a>0.

2^(2x+1)-2^(x-y+3)-2^(x+y+1)+2^3=0 find the x value

Here is the choices x,y are different from 0. A)1 B)2 C)3 D)4 E)5-Update: This has been fixed (so x = y = 1). ----------------------------- Note that we can rewrite this as 2^((x-y)+(x+y)+1) - 2^(x

To the parabola (x+2y)^2+2x-y-3=0 parallel to the line 4x+3y=2.

find the tangent lines as directed. THANK YOU-eqn of line is 4x + 3y = 2 => 3y =-4x + 2 => y = -(4/3)x + (2/3) so slope of tangent line = - 4/3 (since tangent and given line are parallel) eqn

Please help with this math question: Vectors, 10 points to best

The position vectors A, B, C are given by: (i + j + k), (2i + 2j - 3k), (i - j + 2k) respectively. AB = i + j - 4k,and AC= -2j + k: Let L be the staight line passing through A and C. The perpendicula

Find and indicate, any ordered pair that satisfies the equation 2y=14-3x

Could someone please break this down step by step on how to solve this? How do you find an ordered pair from this, im so confused. Thank you-1) choose any value for x (x = 2 for ex) 2) solve for y :

How do I find ∫dy /(y^2 - y)

Use partial fraction decomposition.First write 1 / (y^2 - y) = A / (y-1) + B / y Then solve for A and B to get A = 1, B = -1.The question then becomes how to solve ∫ [ 1 / (y-1) - 1 / y ] dy, which y

Asymptotes question (easy)

graph y=4^x y=10^x y=(1/2)^x On http://my.hrw.com/math06_07/nsmedia/tool… what is the asymptote equation?-horizontal asymptote for all 3 is y = 0-for all of them the horizontal asymptote is y=infin

How do you solve this system using any algebraic method

4x+2y-z=4 2x-3y+2z=4 x+y-z=-1-4x + 2y - z = 4 . . . . . [1] 2x - 3y + 2z = 4 . . . . [2] x + y - z = 1 . . . . . . . [3] From [1] -----> z = 4x + 2y - 4 . . . [4] From [3] -----> z = x + y + 1. . . .

What are the lengths of the Isosceles Triangle

The lengths of two sides of an isosceles triangle are 15 and 22 respectively. What are the possible values of the perimeter? (The answers are 52 and 59. I was able to get the 52, but not the 59).-sinc

I need some help with an algebra Q

x+1/9=4/3 I got... 4/3-1/9 = 36/27-3/27 = 33/27 x = 11/9 = 1 2/9.... now im suppose to check it but i dont know how...-Your working is correct Put 11/9 into first line....[replace x] and see if