If we take the Pythagorean identity identity sin^2 x + cos^2 x = 1 then
<span>(cos^2 x + sin^2 x) / (cot^2 x - csc^2 x)
The numerator becomes 1 since addition order matters not.
1 / </span>(cot^2 x - csc^2 x)
If we factor the denominator out a negative
1 / -(<span>csc^2 x - cot^2 x)
Consider </span><span>sin^2 x + cos^2 x = 1. Divide both sides by sin^2 x to get
1 + cot^2 x = csc^2 x
Subtract both sides by cot^2 x to get 1 = csc^2 x - cot^2 x.
Replace the denominator
1 / -(1) = -1
For cos</span>^2 θ / sin^2 θ + csc θ sin θ, we use cscθ = 1/sinθ and cosθ/sinθ = cotθ so
= cos^2 θ / sin^2 θ + 1
= cot^2 θ + 1
We use 1 + cot^2 <span>θ = csc^2 </span>θ to simplify this to
= csc^2 θ
Answers: -1
csc^2 θ
Answer:
-2 ≤x ≤10
Step-by-step explanation:
the domain refers to the x
the x is in between -2 and 10
Answer: Approximately 6.3876 years
When rounding to the nearest whole number, this rounds up to 7 years.
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Work Shown:
We'll use the compound interest formula
A = P*(1+r/n)^(n*t)
where,
- A = amount of money after t years
- P = initial deposit amount or principal
- r = interest rate in decimal form
- n = compounding frequency
- t = number of years
In this case, we know that,
- A = 2P, since we want the initial amount to double. P can be any positive real number you want and it doesn't affect the answer.
- r = 0.11
- n = 4, since we're compounding 4 times a year
- t = unknown, what we want to solve for
So,
A = P*(1+r/n)^(n*t)
2P = P*(1+r/n)^(n*t)
2 = (1+r/n)^(n*t)
2 = (1+0.11/4)^(4*t)
2 = 1.0275^(4t)
Ln(2) = Ln(1.0275^(4t))
Ln(2) = 4t*Ln(1.0275)
4t*Ln(1.0275) = Ln(2)
t = Ln(2)/(4*Ln(1.0275))
t = 6.38758965414661
It takes roughly 6.3876 years for the deposit to double. If you need this to the nearest whole number, then round up to 7. We don't round to 6 because then we would come up short of the goal of doubling the deposit.
The eighth term is 0.0625
If the student got 123 marks and failed by 39, your first step would be to add the two together, getting 162. This is the amount of marks needed to pass the test.
You would then divide 162 by .36 (.36 is equivalent to 36%). This would result in a maximum of 450 marks.
The maximum marks on this test are 450 marks.
