Answer:
Step-by-step explanation:
Given : universal set in this diagram is the set of integers from 1 to 15.
Solution :
The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15
The intersection of odd integer and Factors of 15 are 1,5
The intersection of odd integer,multiples of 3 is 9
The remaining multiples of 3 are 6,12
The remaining odd integers are 7,11,13
Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .
Refer the attached figure for the answer.
Answer:
11
4/5
Step-by-step explanation:
there u go thats quick maths
Answer: 28 dots
Step-by-step explanation: In this pattern,
we can see the the first figure is just 1 dot.
The second figure has a new row on the bottom with 2 dots.
The third figure has a new row on the bottom with 3 dots
and the fourth figure has a new row on the bottom with 4 dots.
So continuing with this pattern,
the next figure will have a new row with 5 dots,
the next will have a new row with 6 dots,
and the next will have a new row with 7 dots.
So in the 7th picture, we will have 7 + 6 + 5 + 4 + 3 + 2 + 1 dots.
This simplifies to 28 dots.
Take a look below.
Answer:
126 km/h
Step-by-step explanation:
Sorry if I'm wrong:/
Answer:
E) 1 and 2
Step-by-step explanation:
We are given that there are two integers (s and t) and they are factors of another integer (n). For example if s = 3 and t = 2, we can have n = 6.
Thus:
n^(st) = 6^(2*3) = 6^6 = (2^6)(3^6)
For the first condition: s^t = 3^2 is a factor of (2^6)(3^6)
For the second condition: (st)^2 = (3*2)^2 = 6^2 is a factor of 6^6
For the third condition: s+t = 3+2 = 5 is not a factor of 6^6 or (2^6)(3^6)
Therefore, only 1 and 2 are factors of n^(st)
