The diagnonals of a parallelogram bisect each other ,
so
=》 BD = 3x × 2
=》BD = 6x
and there's given that
=》
=》
=》
=》
so, if x = 4 , then the given quadrilateral is a parallelogram.
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:
(X+2)(v-5)
Step-by-step explanation:
first, group and factor out the greatest common factor and combine to get (x+2)(v-5)
Answer:
The answer of this question is D
Step-by-step explanation:
Time 2
