Base 10 has the ten digits: {0, 1, 2, 3, 4, 5, 6,7, 8, 9}
Base 11 has the digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A} where A is treated as a single digit number
Base 12 has the digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B}
Base 13 has the digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C}
Base 14 has the digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D}
The digit D is the largest single digit of that last set. So the largest 3-digit base 14 integer is DDD which is the final answer
Note: It is similar to how 999 is the largest 3-digit base 10 integer
Answer: 
<u>Find Common Denominator</u>
2: 2,4,6,8,10
4: 4,8,12,16,20
CD=4
<u>Add</u>
1/4+2/4=3/4
3/4=.75
Answer:
Step-by-step explanation:
circumference = 2πr
2πr = 30π inches

Area = πr² = π*15*15 = 225π square inches
This is false for example, the GCF of 13 and 26 is 13, an odd number.
Answer:
V = 128π/3 vu
Step-by-step explanation:
we have that: f(x)₁ = √(4 - x²); f(x)₂ = -√(4 - x²)
knowing that the volume of a solid is V=πR²h, where R² (f(x)₁-f(x)₂) and h=dx, then
dV=π(√(4 - x²)+√(4 - x²))²dx; =π(2√(4 - x²))²dx ⇒
dV= 4π(4-x²)dx , Integrating in both sides
∫dv=4π∫(4-x²)dx , we take ∫(4-x²)dx and we solve
4∫dx-∫x²dx = 4x-(x³/3) evaluated -2≤x≤2 or too 2 (0≤x≤2) , also
∫dv=8π∫(4-x²)dx evaluated 0≤x≤2
V=8π(4x-(x³/3)) = 8π(4.2-(2³/3)) = 8π(8-(8/3)) =(8π/3)(24-8) ⇒
V = 128π/3 vu