Question

Can someone solve this???

Answer 1

Answer:

1. a)x² + y² = 29  ------------------(1)      

y² - x² = 23 ----------------(2)

y²  = 23 + x²

Using the value of y² in the first equation:

x²  + y²  = 29

x²  + (23 + x²)  = 29

2x²  = 6

x² = 3

x = √3

Using the value of x in equation (2)

y²  - x²  = 23

y²  - 3 = 23

y²  = 26

y = √26            and            x = √3

b) ( 1 - 12x) / 3 < 0   =   1-12x < 0  =  x > 1 / 12

(1-x) / 2 ≤ (3-x) 3  =  3 - 3x ≤ 6 - 2x  =   -3 ≤ x

In these 2 equations, the value of x will be the values of x common to both the inequalities

Since the second equation starts at a much lower value than the first equation, and they are both going up, all the values for the first equation will also stand for the second equation

Hence, the value of x for the given system of equations is

x  > 1/12

2.

3x² - 7x  + 2 ≥ 0

3x² -6x -1x  + 2 ≥ 0       (splitting the middle term)

3x(x - 2) -(x - 2) ≥ 0

(3x - 1)(x-2) ≥ 0

(3x-1) ≥ 0         or     (x-2) ≥ 0

x ≥ 1/3       or    x ≥ 2

Written above are the 2 solutions for the given equation

4.

We are given that Sin α = (2√3) / 3  or 2/√3

According to the pythagorean law:

sin²α + cos²α = 1

4 / 3  + cos²α = 1

cos²α = (3-4)/3

cos²α = -1/3

cos α = √(-1/3)      or    i / √3    since √-1 is also known as i

Cosec α = 1/ Sin α

Cosec α = 1 / (2/√3)

Cosec α = √3 / 2

Sec α = 1 / cos α

Sec α = 1 /  √(-1/3)

Sec α = (√3)  / (√-1)

Since (√-1) is also known as i , we can write this as:

Sec α = (√3)  / i

Tan α = Sin α / Cos α

Tan α = (2√3 / 3) / (i / √3)

Tan α = (2√3 * √3) / 3 * i)

Tan α = 2 / i

Cot α = 1/Tan α

Cot α = 1 / (2 / i)

Cot α = i / 2