To find which ratio is higher, we can first convert the ratios into fractions (that makes it easier, at least for me) and then simplify the fractions and see which one is greater.
Since ratios are basically division, 15:20 = 
, and
12:16 = 
Now we simplify these fractions. 15 and 20 have a GCF of 5, so taking out the common number gives us the simplified fraction of 
.
12 and 16 have a GCF of 4, so taking out the common number gives us the simplified fraction of . Since = , these ratios are the same.
This is a refreshing question!
We are given that
f(r)=ar+b, and
Sum f(r) =125 for r=1 to 5
Sum f(r) = 475 for r=1 to 10.
and we know, using Gauss's method, that
G(n)=sum (1,2,3.....n) = n(n+1)/2 or
G(n)=n(n+1)/2
Sum f(r) =125 for r=1 to 5
=>
sum=a(sum of 1 to 5) + 5b => G(5)a+5b=125 [G(5)=15]
15a+5b=125 ...................................................(1)
Similarly, Sum f(r) = 475 for r=1 to 10 => G(10)a+5b=475 [G(10)=55]
=>
55a+10b=475.................................................(2)
Solve system of equations (1) and (2)
(2)-2(1)
55-2(15)a=475-2(125) => 25a=225 =>
a=9
Substitute a=9 in 1 => 15(9)+5b=125 => 5b=-10
b=-2
Substitute a and b into f(r),
f(r)=9r-2
check: sum f(r), r=1,5 = (9-2)+(18-2)+(27-2)+(36-2)+(45-2)=135-10=125 [good]
We define the sum of f(r) for r=1 to n as
S(n)=sum f(r) for r=1 to n = 9(sum 1,2,3....n)-2n = 9n(n+1)/2-2n = 9G(n)-2n
S(n)=9n(n+1)/2-2n
checks:
S(5)=9(15)-2(5)=135-10=125 [good]
S(10)=9(55)-2(10)=495-20=475 [good]
Hence
(a)
S(n)=sum f(r) for r=1,n
= 9(sum i=1,n)+n(-2)
= 9(n(n+1)/2 -2n
=(9(n^2+n)/2) -2n
(b) sum f(r) for i=8,18
=sum f(r) for i=1,18 - sum f(r) for i=1,7
=S(18)-S(7)
=(9(18^2-18)/2-2(18))-(9(7^2-7)/2-2(7))
=1503-238
=1265
Answer:
C 24
Step-by-step explanation:
Answer:
23) x= 
24) x= 
Step-by-step explanation:
