Answer:
can you please describe this more and what it is then i will answer it in the comments
Step-by-step explanation:
Answer:
No answer for you >:) <3
Step-by-step explanation:
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)]
Answer:
5.36
Step-by-step explanation:
Given that:
<BAD = <CAE, therefore, BD = EC
Let's take x to be the length of BD = EC
BD + DE + EC = BC
BC = 20,
BD = EC = x
DE ≈ 9.28
Thus,
x + 9.28 + x = 20
x + x + 9.28 = 20
2x + 9.28 = 20
Subtract 9.28 from both sides
2x + 9.28 - 9.28 = 20 - 9.28
2x = 10.72
Divided both sides by 2 to solve for x
BD ≈ 5.36
*Answer* :
5.75
Explanation:
Let the unknown side = X
Cos() = adj/hyp
Cos(44)= X/8
X= 5.75