Answer: 4xy and 3xy are, 2s and 5s are, but -10 and -10b aren't.
Step-by-step explanation: 4xy and 2xy have the same terms. 2s and 5s have the same terms but -10a and -10b have different variables.
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
cot(x)sec⁴(x) cot(x)sec⁴(x)
0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
0 = cos⁴(x)(1 + tan²(x))²
0 = cos⁴(x) or 0 = (1 + tan²(x))²
⁴√0 = ⁴√cos⁴(x) or √0 = (√1 + tan²(x))²
0 = cos(x) or 0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x or √-1 = √tan²(x)
i = tan(x)
(No Solution)
2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
sin²(x) - cos²(x) = sin²(x) - cos²(x)
+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
0 = 0
3. 1 + sec²(x)sin²(x) = sec²(x)
sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
x = 0
4. -tan²(x) + sec²(x) = 1
-1 -1
tan²(x) - sec²(x) = -1
tan²(x) = -1 + sec²
√tan²(x) = √(-1 + sec²(x))
tan(x) = √(-1 + sec²(x))
tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
x = 0
Answer:
AC = 25 units
Step-by-step explanation:
Given ∠ A = ∠ C then the triangle is isosceles and BC = AB , that is
7x - 1 = 5x + 5 ( subtract 5x from both sides )
2x - 1 = 5 ( add 1 to both sides )
2x = 6 ( divide both sides by 2 )
x = 3
Then
AC = 6x + 7 = 6(3) + 7 = 18 + 7 = 25 units
Answer:
y = 5x - 7
Step-by-step explanation:
Using the equation of a line with two points
y_2 - y_1 / x_2 - x_1 = y - y_1 / x - x_1
Giving two points
( -4 , 10)(16 , -2)
x_1 = -4
y_1 = 10
x_2 = 16
y_2 = -2
Using the above formula
y_2 - y_1 / x_2 - x_1 = y - y_1 / x - x_1
-12 - 3 / -1 - 2 = y - 3 / x - 2
-15/-3 = y - 3 /x - 2
Cross multiply
-15(x - 2) = -3(y - 3)
-15x + 30 = -3y + 9
-15x + 30 + 3y - 9 = 0
-15x +30 - 9 + 3y = 0
-15x + 21 + 3y = 0
3y = 15x - 21 ( following y = mx + C
Dividing through by 3 to find y
3y/3 = 15x -21 / 3
y = 15x - 21 / 3
We can still separate
y = 15x / 3 - 21/3
y = 5x - 7
Therefore, the equation of the line is
y = 5x - 7