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asambeis [7]
3 years ago
6

1011+111 in binary form

Mathematics
1 answer:
Ann [662]3 years ago
8 0

Answer:

10010

Step-by-step explanation:

1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0

111=1(2)^2+1(2)^1+1(2)^0

So 1011+111 gives us:

1(2)^3+0(2)^2+1(2)^1+1(2)^0

+

1(2)^2+1(2)^1+1(2)^0

-----------------------------------------------------

Combine like terms:

1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0

1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace 2^0 with 1 and 2 with (2)^1:

1(2)^3+1(2)^2+(2)^2+(2)^1(1)

I want a 2^0 number:

1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0

Combine like terms:

1(2)^3+2(2)^2+1(2)^1+0(2)^0

2(2)^2=2^3:

1(2)^3+2^3+1(2)^1+0(2)^0

Combine like terms:

2(2)^3+1(2)^1+0(2)^0

We can rewrite the first term by law of exponents:

2^4+1(2)^1+0(2)^0

1(2)^4+1(2)^1+0(2)^0

So the binary form is:

10010

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

      1     0     1      1

+            1      1      1

------------------------------

     (1)    (1)    (1)

      1     0     1      1

+            1      1      1

------------------------------

     1 0    0     1       0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
kogti [31]

Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

Then, we have the following system of equations:

x = a (2)

(b^{2}-1) = 2\cdot x\cdot y (3)

y = c (4)

By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

b^{2} = 1 + 2\cdot a \cdot c

b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

4 0
3 years ago
More, less or equal to: <br> Absolute 4 or Absolute 1
Sloan [31]

|4| > |1|

Absolute value of 4 is just 4, and the abosulute valvue of 1 is just 1.

So , since 4 is greater than 1 , that would be your answer.

4 > 1


7 0
3 years ago
Please help small brain melted
Sergeu [11.5K]
The correct answer is 4> -3
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2 years ago
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