Answer: The number is 26.
Step-by-step explanation:
We know that:
The nth term of a sequence is 3n²-1
The nth term of a different sequence is 30–n²
We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.
we get:
3n²- 1 = 30–m²
Notice that as n increases, the terms of the first sequence also increase.
And as n increases, the terms of the second sequence decrease.
One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.
if m = 1, then:
3n²- 1 = 30–1²
3n²- 1 = 29
3n² = 30
n² = 30/3 = 10
n² = 10
There is no integer n such that n² = 10
now let's try with m = 2, then:
3n²- 1 = 30–2² = 30 - 4
3n²- 1 = 26
3n² = 26 + 1 = 27
n² = 27/3 = 9
n² = 9
n = √9 = 3
So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.
the number is:
3*(3)² - 1 = 26
30 - 2² = 26
The number that belongs to both sequences is 26.
Answer:
C) Area of Triangle is 102 Square Units
Answer:
Part c: Contained within the explanation
Part b: gcd(1200,560)=80
Part a: q=-6 r=1
Step-by-step explanation:
I will start with c and work my way up:
Part c:
Proof:
We want to shoe that bL=a+c for some integer L given:
bM=a for some integer M and bK=c for some integer K.
If a=bM and c=bK,
then a+c=bM+bK.
a+c=bM+bK
a+c=b(M+K) by factoring using distributive property
Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.
So L=M+K in bL=a+c.
We have shown b|(a+c) given b|a and b|c.
//
Part b:
We are going to use Euclidean's Algorithm.
Start with bigger number and see how much smaller number goes into it:
1200=2(560)+80
560=80(7)
This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.
Part a:
Find q and r such that:
-65=q(11)+r
We want to find q and r such that they satisfy the division algorithm.
r is suppose to be a positive integer less than 11.
So q=-6 gives:
-65=(-6)(11)+r
-65=-66+r
So r=1 since r=-65+66.
So q=-6 while r=1.
X= 3.124 bc ur using Pythagoras Theorem so a squared + b squared = c squared :)
Answer:
68 members
Step-by-step explanation:
80% of 85 =
10% = 8.5
8.5 x 8 = 68.
Hope this helps
