I only
II only
III only
I and III only
Correct answer is B
Using the formula, \((n - 2) \times 180°\) to get the sum of the interior angles. Then we can have
\((n - 2) \times 180° = 108n\) ... (1)
\((n - 2) \times 180° = 116n\) ... (2)
\((n - 2) \times 180° = 120n\) ... (3)
Solving the above given equations, where n is not a positive integer then that angle is not the interior for a regular polygon.
(1): \(180n - 360 = 108n \implies 72n = 360\)
\(n = 5\) (regular pentagon)
(2): \(180n - 360 = 116n \implies 64n = 360\)
\(n = 5.625\)
(3): \(180n - 360 = 120n \implies 60n = 360\)
\(n = 6\) (regular hexagon)
Hence, 116° is not an angle of a regular polygon.
If \(\frac{3^{(1-n)}}{9^{-2n}}=\frac{1}{9}\) find n
\(-\frac{3}{2}\)
\(\frac{1}{3}\)
-1
-3
Correct answer is C
\(\frac{3^{(1-n)}}{9^{-2n}}=\frac{1}{9}\\
3^{1-n}\times 3^{-2(-2n)} = 3^{-2}\\
1-n-2(-2n)= -2\\
1-n+4n=-2\\
n=-1\)
0.01014
0.01021
0.01015
0.01016
Correct answer is A
option A CANNOT BE BECAUSE THE LAST NUMBER BEFORE 1 CAN ONLY BE ROUNDED DOWN TO ZERO.
Evaluate \((111_{two})^2 - (101_{two})^2\)
10two
100two
1100two
11000two
Correct answer is D
\((111_{2})^2 - (101_{2})^2\)
Difference of two squares
\((111 - 101)(111 + 101)\)
= \((10)(1100)\)
= \(11000_{2}\)
85o
60o
55o
45o
Correct answer is C
< QPS = < PRS = 65° (angles in the same segment)
< PSR + 40° + 65° = 180°
< PSR + 105° = 180°
< PSR = 75°
< PSR = < PSQ + < QSR
75° = < PSQ + 20° \(\implies\) < PSQ = 75° - 20° = 55°
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