Answer:
A = 1.0 L
B = 0.50 atm
C = 0.60 atm
D = 4.0 L
Step-by-step explanation:
According to the Boyle's law, the volume of a gas is inversely proportional to its volume. If we consider an initial state (1) and a final state (2):
P₁ × V₁ = P₂ × V₂
A
P₁ × V₁ = P₂ × V₂
1.5 atm × 2.0 L = 3.0 atm × A
A = 1.0 L
B
P₁ × V₁ = P₂ × V₂
1.5 atm × 2.0 L = B × 6.0 L
B = 0.50 atm
C
P₁ × V₁ = P₂ × V₂
1.5 atm × 2.0 L = C × 5.0 L
C = 0.60 atm
D
P₁ × V₁ = P₂ × V₂
1.5 atm × 2.0 L = 0.75 atm × D
D = 4.0 L
Ok, this definition of f is just a bunch of points (4 to be exact).
so if point (a,b) is part of f, then f(a)=b
f(1) is about point (1,0), so f(1)=0.
g(1) = 1 (due to point (1,1))
g(2/3) = 0
f(2) = 3/4
g(-2) = 3
f(π) = -2
just a lookup once you see what is happening
Answer:
soccer bruv
Step-by-step explanation:
Hello :) ok so to find AC you would have to use sin. It would be sin70= x/4. To find AC you would have to multiply 4 by sin70 which is 3.758 or to the nearest hundredth would be 3.76. I hope I helped, if you need a more in depth explanation lmk
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
