Hello :
if : a <span>≡ 5 ( mod 7) ....(1)
b </span><span>≡ 6 ( mod 7) ...(2)
(1)×(2) : ab </span><span>≡ 30 ( mod 7)
</span>but : 30 = 7×4 + 2 ..... so : 30 ≡ 2 ( mod 7)
then : ab ≡ 2 ( mod 7)
<span>The area of an n-sided regular polygon approaches the area of a circle as n gets very large. ... If an N-gon (polygon with N sides) has perimeter P, then each of the N ... and we can use one of them to derive theequation sin(theta/2) ... 2*pi/N radians), R is the length of the lines to the center (the radius of the ...</span><span>
</span>
$57.18 + 256.79 = 313.97
313.97 - 68.42 = 245.55
245.55 - 50.00 = 195.55
195.55 - 31.06 = 164.49
that’s your answer
In
,
$a$ is the real part
$ib$ is the imaginary part.
Comparing, we get:
Real part: $91$
Imaginary part: $-27i$
Answer: (A) The image of JKL after a 90° counterclockwise about the origin is shown in figure 1. (B) The image of JKL after a reflection across the y-axis is shown in figure 2.
Explanation:
(A)
From the given figure it is noticed that the coordinate points are J(-4,1), K(-4,-2) and L(-3,-1).
If a shape rotate 90 degree counterclockwise about the origin, then,




Therefore, the vertex of imare are J'(-1,-4), K'(2,-4) and L'(1,-3). The graph is shown in figure (1).
(B)
If a figure reflect across the y-axis then,




Therefore, the vertex of imare are J''(4,1), K''(2,-4) and L''(3,-1). The graph is shown in figure (2).