The expressions that are equivalent when m = 1 and m = 4 is;
Option B: 3m + 4 and m + 4 + 2m
We are given m = 1 and m =4;
A) 5m - 3 and 2m + 5 + m
B) 3m + 4 and m + 4 + 2m
C) 2m + 7 and 3m - 3 + m
D) 5m + 3 and 4m + 2 + 2m
For option B; 3m + 4 and m + 4 + 2m
Let's put m = 1
3(1) + 4 = 7
Also, 1 + 4 + 2(1) = 7
Similarly, let us put 4 for m to get;
3(4) + 4 = 16
Also, 4 + 4 + 2(4) = 16
In both cases, the expressions are equivalent and as such option B is the right one.
Read more about algebra simplifications at; brainly.com/question/4344214
3 less than twice a number x.
Twice a number would be the x multiplied by 2, and 3 less would be 3 subtracted by 2*x.
2x-3 <- The numeral form.
I hope this helps!
~kaikers
Problem: find 0 ≤ x ≤ 28 such that x^85 ≡ 6 modulo 35.
By Fermat-Euler theorem:
If a and n are coprime, i.e. (a,n)=1, then
a^phi(n) ≡ 1 mod n
where phi(n)=totient function, the number of positive integers less than n that is coprime with n.
for n=35, phi(35)=24 calculated as follows:
There are 10 positive integers from 1 to 34 which are NOT coprime with 35, namely {5,7,10,14,15,20,21,25,28,30}.Therefore phi(35)=34-10=24
From Fermat-Euler theorem,
x^(phi(35) = x^24 ≡ 1 modulo 35 since (24,35)=1, i.e. 24 and 35 are coprime.
=>
x^12 ≡ ± 1 modulo 35. ...........(1)
and
x^85 ≡ x^(85-3*24) ≡ x^(85-72) ≡ x^(13) ≡ 6 mod 35 ............(2)
Substituting (1) in (2)
x^(12)*x ≡ 6 mod 35
=>
(+1)*x = 6 mod 35 or (-1)*x ≡ 6 mod 35
x ≡ 6 mod 35 x ≡ -6 mod 35 (rejected)
=> x=6
So
6^85 ≡ 6 mod 35
If any clarifications are needed or if you find any errors, please post.
|x| = 6
{ x= 6
{ x= -6
X= { 6,-6}
Answer:
96%
Step-by-step explanation:
assume the whole number is x
the first thing is decreasing by 20 %
x - 20%x = 80% x
then now the 80%x is the whole number
now It's the turn of increasing
80%x + 20% * (80% x) = 80%x + 16%x = 96% x
I hope this is hlepful to you !!!!
Feel free to ask me in the comments
