V (cone) = (1÷3) × (r×r×h) л cm3
3 cone = 1 cylinder (if "h" and "r" is same)
There is two cone...
1st one: (1÷3) × 3×3×5 л cm3 = 15л cm3
2nd one: (1÷3) × 3×3×8 л cm3 = 24л cm3
15 + 24 = 39л cm3
The second bubble is the correct answer i believe...
Answer: Fraction they both eat =
.
Fraction of bag remains = 
Step-by-step explanation:
Given: Part of cookies eaten by Scott = 
Part of cookies eaten by sister = 
Fraction they both eat = 

Fraction of the bag remain = 
[Taking fraction of whole bag =1]
Hence, Fraction they both eat =
.
Fraction of bag remains = 
Answer:
Options (2) and (3)
Step-by-step explanation:
Let, 

-8 + 8i√3 = a² + b²i² + 2abi
-8 + 8i√3 = a² - b² + 2abi
By comparing both the sides of the equation,
a² - b² = -8 -------(1)
2ab = 8√3
ab = 4√3 ----------(2)
a = 
By substituting the value of a in equation (1),


48 - b⁴ = -8b²
b⁴ - 8b² - 48 = 0
b⁴ - 12b² + 4b² - 48 = 0
b²(b² - 12) + 4(b² - 12) = 0
(b² + 4)(b² - 12) = 0
b² + 4 = 0 ⇒ b = ±√-4
b = ± 2i
b² - 12 = 0 ⇒ b = ±2√3
Since, a = 
For b = ±2i,
a =
= 
= 
But a is real therefore, a ≠ ±2i√3.
For b = ±2√3
a = 
a = ±2
Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)
Options (2) and (3) are the correct options.