So let's take a peek at both's ages, keep in mind, every year, is 1year added to Irene and 1year added to Fred
so... if we look at their ages

notice, Fred is always 40years older than Irene
thus, whatever age Irene is, let's say "i", then Fred is " i + 40 "
now, when is Fred 5 times Irene's age or 5*i or 5i? well,
f = fred's age i = irene's age
f = i + 40
now if f = 5i
5i = i + 40 <--- solve for "i" to see how old Irene was then
Answer:
a) 12
b) 4
Step-by-step explanation:
:)
The answer is 1/3 of 12 its just easy math for me
B1 to H1, B3 to H2, B2 to H3.
Answer:
solution:-We know that for any two finite sets A and B, n(A∪B)=n(A)+n(B)−n(A∩B).
Here, it is given that n(A)=20,n(B)=30 and n(A∪B)=40, therefore,
n(A∪B)=n(A)+n(B)−n(A∩B)
⇒40=20+30−n(A∩B)
⇒40=50−n(A∩B)
⇒n(A∩B)=50−40
⇒n(A∩B)=10
Hence, n(A∩B)=10
Step-by-step explanation:
hope it helps you friend ☺️