(a) This is the same as computing 4⁵⁵ (mod 11). We have
4² ≡ 16 ≡ 5 (mod 11)
4³ ≡ 4 • 5 ≡ 20 ≡ 9 (mod 11)
4⁴ ≡ 4 • 9 ≡ 36 ≡ 3 (mod 11)
4⁵ ≡ 4 • 3 ≡ 12 ≡ 1 (mod 11)
Then from here,
4⁵⁵ ≡ (4⁵)¹⁰ • 4⁵ ≡ 1¹⁰ • 1⁵ ≡ 1 (mod 11)
(b) Each term in the sum
4ⁿ + 4ⁿ⁺¹ + 4ⁿ⁺² + 4ⁿ⁺³ + 4ⁿ⁺⁴
has a common factor of 4ⁿ, so this sum is the same as
4ⁿ (1 + 4 + 16 + 64 + 256) = 4ⁿ • 341 = 4ⁿ • 11 • 31
So the sum is indeed divisible by 11 for all integers <em>n</em>.
(c) Since 4ⁿ = (2ⁿ)², we know the sum is also divisible by 11 when <em>a</em> = 2.
1 is 16 units and 2 is idk
The y-intercept appears to be at y=-1. The line appears to also go through grid point (3, 1), which is 2 units up and 3 units over. The slope is 2/3.
In slope-intercept form the equation of the line is
y = (2/3)x - 1
Answer:
A
Step-by-step explanation:
Draw an angle of 15° like this (∠) on a piece of paper, label this angle J, extend the base of the triangle and label the hypotenous 11cm (JK). Do you get the solution now? If no, here is an explanation.
This problem can be explained with sin law,
[Assume KL = 4cm]
(sin15)/4 = (sinL)/11
L = 45.4° or 135° (3 sig. fig.)
Since we got 2 possible case for the angle L, there are two possible set of triangles. Hope this will help you understand more about the problem, comment below if you still have any questions.
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