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:
7
Step-by-step explanation:
7-7=0 and this cannot happen in the denominator
A^2 + b^ = c^2
c is the longest side of the triangle
a and b are the other two sides
So the given are the hypotenuse leg, and one of the other legs
So
a= ?
b= 12
c= 14
a^2+12^2=14^2
a^2+144=196
-144 -144
a^2=52
So we find the square root of 52
Which is 7.211 which you can round to 7.21 or 7.2
Answer:
-2
Step-by-step explanation:
if showing in graph would be minus 2
Answer:

Step-by-step explanation:
84/4=21
Square root of 4=2
