Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

135 and 315degrees

Step-by-step explanation:

Given the expression

4tan²θ+tanθ=−4tanθ−1

Equate to zero

4tan²θ+tanθ+4tanθ+1 = 0

4tan²θ+5tanθ+1 = 0

Let x = tanθ

4x²+5x+1 = 0

Factorize;

4x²+4x+x+1 = 0

4x(x+1)+1(x+1) = 0

4x+1 = 0 and x+1 = 0

4x = -1 and x = -1

x = -1/4 and -1

Since x = tanθ

-1 =tanθ

θ = arctan(-1)

θ = -45degrees

Since tan is negative in the 2nd and fourth quadrant;

In the second quadrant

θ = 180 - 45 = 135degrees

In the fourth quadrant;

θ = 360 - 45 = 315degrees

Hence the required angles are 135 and 315degrees

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.