Answer:
264
Step-by-step explanation:
do 20 mulitplyed by 30 then subtracted it by 336
The first equation, make the y as the subject. y = 3 + 5x, then substitute y in the second equation and you will get the value of x. then, substitute the x in the first equation (y = 3 + 5x)
You're looking for the largest number <em>x</em> such that
<em>x</em> ≡ 1 (mod 451)
<em>x</em> ≡ 4 (mod 328)
<em>x</em> ≡ 1 (mod 673)
Recall that
<em>x</em> ≡ <em>a</em> (mod <em>m</em>)
<em>x</em> ≡ <em>b</em> (mod <em>n</em>)
is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly
1 ≡ 4 (mod 41)
is not true.
So there is no such number.
Answer:
A
Step-by-step explanation:
30% of 4
4÷100=0.04
0.04×30=1.2
