Answer:
Step-by-step explanation:
Let x = third side
Using the Triangle Inequality theorem which states that the sum of two sides of a triangle must be longer than the third side and the difference of the two sides is the lower limit of the third side, the answer to your question is that the third side must be between 3 and 13, or written using inequalities, 3 < third side (or x) < 13 is the range.
Answer: She has 1 5/6 exercises for homework.
Step-by-step explanation:
1/2 + 2/3 = 1 1/6
2 * 3 = 6 3 * 2 = 6
1 * 3 = 3 2 * 2 = 4
4 + 3/ 3 + 3
7/6 = 1 1/6
So, she finished 1 1/6 of her problems.
3 - 1 1/6 = 1 5/6
3 - 1 = 2
2 - 1/6 = 1 5/6
9a + (9t+27)=x
To write it another way you could flip the numbers around:
(9t+27)+9a=x
Or flip the variable to the other side.
x=9a + (9t+27)
Any way you put the numbers as long as x is on the either side, you will get the same answer.
Hope this helps!
Answer:
True
Step-by-step explanation:
The size of the vertex cover is atleast twice the size of the maximum edge. Vertex must contain atleast one vertex from the matched edge. Vertex are never the strong dual solutions of each other.
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
