1. A debit signifies something owed. So, that means the answer is negative.
-$40
2. A deposit means you are gaining money. So, the answer is positive.
+$225
3. Above sea level signifies to us it is above zero. So, the answer is positive.
+14000 feet
(Keep in mind, whenever they say above, it is always positive.)
4. The temperature increased, meaning that it is positive.
+40°F
5. Withdrawl implies losing money. So, the answer is negative.
-$225
6. Below sea level implies it is below zero. So, the answer is negative.
-14000 feet
Answer:
<h2>76904685 ways</h2>
Step-by-step explanation:
Given data
the number of students n=40
the number of groups r= 8
We are going to use the combination approach to solve the problem
nCr= n!/r!(n-r)!
substituting into the expression for the number of ways we have
40C8= 40!/8!(40-8)!
nCr= 40!/8!(32)!
nCr= 40!/8!(32)!
nCr= 40*39*38*37*36*35*34*33*32!/8!(32)!
nCr= 40*39*38*37*36*35*34*33*/8!
nCr= 40*39*38*37*36*35*34*33*/8*7*6*5*4*3*2
nCr= 3100796899200/40320
nCr=76904685 ways
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
(1/8), (2/8), (3/8), (4/8), (5/8), (6/8), (7/8)
if u want, 2/8=1/4, 4/8=1/2, 6/8=3/4
