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xxTIMURxx [149]
3 years ago
6

I need help with this please help me

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer: 85 = 48+37

Step-by-step explanation:

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What is the solution to this system of linear equations? x y = 4 x − y = 6.
nexus9112 [7]

Answer:

Step-by-step explanation:x y = 4 x - y = 6 = get you some b1tc1es

5 0
3 years ago
What is the next term in the sequence 4,-12,36,-108 use inductive reasoning
aksik [14]
The is answer is 324. The previous number is multiplied by -3.
3 0
4 years ago
Can u help me with this question
Troyanec [42]

Answer:

14

Step-by-step explanation:

The question is asking for the value of A when n = 2

So we just need to plug in 2 for n

A(2) = 10 + (<em>2</em>-1)(4) = 10 + (1)(4) = 10 +4 = 14

6 0
3 years ago
For each value of 1 ≤ n ≤ 100, the highest common factor of 8n + 3 and 5n + 4 is written down. What is the sum of these values
DerKrebs [107]

Let gcd(8n + 3, 5n + 4) = d

⟹d|8n+3∧d|5n+4

⟹d|8(5n+4)−5(8n+3)

⟹d|17

Therefore highest common factor of 8n + 3 and 5n + 4 is either 1 or 17 for all n

learn more of gcd here brainly.com/question/25550841

#SPJ9

6 0
2 years ago
Find the HCF of 2^120-1 and 2^100-1
zlopas [31]
There may be more brilliant solution than the following, but here are my thoughts.

We make use of Euclid's algorithm to help us out.
Consider finding the hcf of A=2^(n+x)-1 and B=2^(n)-1.

If we repeated subtract B from A until the difference C is less than B (smaller number), the hcf between A and B is the same as the hcf between B and C.

For example, we would subtract 2^x times B from A, or
C=A-2^xB=2^(n+x)-2^x(2^n-1)=2^(n+x)-2^(n+x)+2^n-1=2^n-1
By the Euclidean algorithm, 
hcf(A,B)=hcf(B,C)=hcf(2^n-1,2^x-1)
If n is a multiple of x, then by repetition, we will end up with
hcf(A,B)=hcf(2^x-1,2^x-1)=2^x-1

For the given example, n=100, x=20, so
HCF(2^120-1, 2^100-1)=2^(120-100)-1=2^20-1=1048575
(since n=6x, a multiple of x).

4 0
3 years ago
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