Answer:
f'(x) = 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Step-by-step explanation:
f' of tan(x) = sec²(x)
f' of csc(x) = -csc(x)cot(x)
General Power Rule: uⁿ = xuⁿ⁻¹ · u'
Step 1: Write equation
2tan³(x) + 5csc⁴(x)
Step 2: Rewrite
2(tan(x))³ + 5(csc(x))⁴
Step 3: Find derivative
d/dx 2(tan(x))³ + 5(csc(x))⁴
- General Power Rule: 2 · 3(tan(x))² · sec²(x) + 5 · 4(csc(x))³ · -csc(x)cot(x)
- Multiply: 6(tan(x))²sec²(x) - 20(csc(x))³csc(x)cot(x)
- Simplify: 6tan²(x)sec²(x) - 20csc⁴(x)cot(x)
- Factor: 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
If I understand the concept of markups correctly I would say 500%
Answer:
Lonnie is not correct
Step-by-step explanation:
Answer:
Per hour decay of the isotope is 0.96%.
Step-by-step explanation:
Amount of radioactive element remaining after t hours is represented by
where a = initial amount
t = duration of decay (in hours)
Amount remaining after 1 hour will be,
y = 0.9904a
So amount of decay in one hour = a - 0.9904a
= 0.0096a gms
Percentage decay every hour = 
= 
= 0.958 %
≈ 0.96 %
Therefore, per hour decay of the radioactive isotope is 0.96%.
Dana can fill 12 planters. Let me know if you want an explanation!
