Make improper fractions.
33/5 - 43/10
Find a common denominator, in this case it is 10.
66/10 - 43/10 = 23/10
This can be simplified to 2 3/10
Hope this helps!
Answer:
parallelogram
Step-by-step explanation:
Answer:
D. B,C,A
Step-by-step explanation:
A= |7|=7
B= -6
C= |-5|=5
so this means that B<C<A = -6<5<7
so; B,C,A = -6,7,5
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
2 n/3 would be the closest answer although it looks as if the 2 is supposed to be multiplied by n/3 which would lead me to write it as 2(n/3)
