52 is half of 104.
Is ur answer.
Answer:
x=±1. are the factors of the quadratic equation.
Step-by-step explanation:
Given quadratic expression, f(x)=-12x - 2x + 60x² +14x-60
Rearranging and adding the terms in the expression and equating to zero.
f(x)= 60x² -60=0
60(x² - 1) =0
The zero product property states that if the product of a⋅b=0 then either a or b equal zero or both of them must be equal to zero. This basic property helps us solve the quadratic equations like (x+2)(x-5)=0 where x =-2,5.
from the zero product property we can infer that 60≠0⇒x² - 1=0
⇒(x+1)×(x-1) = 0
⇒x=±1.
Therefore, x=±1. are the factors of the quadratic equation.
Parallel = same slope
Y = -4x + b
(0,8) y intercept; plug in
Solution: y = -4x + 8
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
The equation of a horizontal line is
y
=
c
for some constant
c
Since we are told that the line passes through
(
x
,
y
)
=
(
−
4
,
6
)
Then for at least one point on the line
y
=
6
But for a horizontal line this value is a constant for all points on the line.
So the equation for all points on the line is
y
=
6
