Answer: a) 83, b) 28, c) 14, d) 28.
Step-by-step explanation:
Since we have given that
n(B) = 69
n(Br)=90
n(C)=59
n(B∩Br)=28
n(B∩C)=20
n(Br∩C)=24
n(B∩Br∩C)=10
a) How many of the 269 college students do not like any of these three vegetables?
n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C)
n(B∪Br∪C)=
So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83
b) How many like broccoli only?
n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C))
n(only Br)=
c) How many like broccoli AND cauliflower but not Brussels sprouts?
n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C)
n(Br∩C-B)=
d) How many like neither Brussels sprouts nor cauliflower?
n(B'∪C')=n(only Br)= 28
Hence, a) 83, b) 28, c) 14, d) 28.
Suppose
is a solution to the ODE. Then
and
, and substituting these into the ODE gives

Then the particular solution to the ODE is

Answer:
Step-by-step explanation:
Answer:
See solutions below
Step-by-step explanation:
a) Given the expressions
x³y and x⁴y³
Find their factors
x³y = (x * x * x * y)
x⁴y³ = (x * x * x * y) * x * y * y
GCF = x*x*x*y
GCF = x³y
The expressions in parenthesis is the GCF. Hence the GCF of x³y and x⁴y³ is x³y
b) For 48m⁶n⁵p⁴ and 20m⁴n⁵p⁶
48m⁶n⁵p⁴ =12 * (4 * m⁴ * n⁵ * p⁴) * m²
20m⁴n⁵p⁶ = 5 * (4 * m⁴ * n⁵ * p⁴) * p²
GCF is the expression common to both terms
GCF = 4 * m⁴ * n⁵ * p⁴
GCF = 4m⁴n⁵p⁴