Answer:
0_10 =0_2
Step-by-step explanation:
Convert the following to base 2:
0_10
Hint: | Starting with zero, raise 2 to increasingly larger integer powers until the result exceeds 0.
Determine the powers of 2 that will be used as the places of the digits in the base-2 representation of 0:
Power | \!\(\*SuperscriptBox[\(Base\), \(Power\)]\) | Place value
0 | 2^0 | 1
Hint: | The powers of 2 (in ascending order) are associated with the places from right to left.
Label each place of the base-2 representation of 0 with the appropriate power of 2:
Place | | | 2^0 |
| | | ↓ |
0_10 | = | ( | __ | )_(_2)
Hint: | Divide 0 by 2 and find the remainder. The remainder is the first digit.
Determine the value of 0 in base 2:
0/2=0 with remainder 0
Place | | | 2^0 |
| | | ↓ |
0_10 | = | ( | 0 | )_(_2)
Hint: | Express 0_10 in base 2.
The number 0_10 is equivalent to 0_2 in base 2.
Answer: 0_10 =0_2
Answer:
7a ( 1,1)
7b ∅ or no solution
Step-by-step explanation:
The solution to the system is where the two functions intersect
For 7a. The parabola and the line intersect at (1,1)
For 7b The parabola and the line do not intersect so there is no solution
Answer: There are 10 sides
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x = measure of each interior angle
y = measure of each exterior angle
We know that x = 4y as "each interior angle is four times the measure of each exterior angle"
The interior and exterior adjacent angles are supplementary
x+y = 180
4y+y = 180
5y = 180
5y/5 = 180/5
y = 36
If y = 36, then
n = 360/y
n = 360/36
n = 10
Answer:
B. The rate of change of item II is greater than the rate of change of them I
Using BEDMAS, it would be brackets first, so anything Inside a set of brackets would be solved first, and would follow the same rule. Once the brackets have been solved, next would be the exponents. Since your brackets have been solved you can expand with the exponents. After exponents, would be anything that has to be divided or multiplied. These two operations are interchangeable based on what is more convenient or beneficial to do first. And lastly, addition and subtraction, which once again are interchangeable operations
