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3 years ago

Show that the dual of the exclusive-OR is equal to its complement.

1 answer:

7 0

Solution:XOR: X+Y= XY’ + X’YDual of XOR:= (X +Y’)+(X’+Y)= XX’+XY +X’Y’ +YY’= XY + X’Y’ Complement of XOR (XNOR)= (X+Y)’= (XY’ + X’Y)’=(X’+Y)+(X +Y’)= XX’+ XY + X’Y’+YY’= XY + X’Y’
HOPE IT HELPS

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Hey can you help me round the fallowing numbers to the nearest 10

Juliette [100K]

10
20
30 there u go
that the rounding

8 0

3 years ago

Sumi rides at a constant speed of 10 ft/s. Write an equation that expresses the distance d that sumi has traveled as a function

Shtirlitz [24]

Answer:

d = 10s

Step-by-step explanation:

The formula for calculating the speed of Sumi is expressed as;

Speed = distance/time

Given

Speed = 10ft/s

time = s seconds

Required

Distance travelled as a function of time;

from the formula;

Distance = Speed * time

Distance = 10 * s

d = 10s

Hence the equation that expresses the distance d that sumi has traveled as a function of the time s in seconds since she started riding her bike is;

d = 10s

3 0

3 years ago

7/10 + 2/10 , 13/16+ 2/16 , 4/5 + 1/5 , 7/15 + 2/15 , 9/20 + 3/20 , 5/8 + 1/8 . All in simplist form.

blondinia [14]

Hello!

<h3><em><u>Answers</u></em></h3>

1. $\frac{7}{10} + \frac{2}{10} = \frac{9}{10}$ Simplest form: $\frac{9}{10}$

2. $\frac{13}{16} + \frac{2}{16} = \frac{15}{16}$ Simplest form: $\frac{15}{16}$

3. $\frac{4}{5} + \frac{1}{5} = \frac{5}{5}$ Simplest form: $\frac{1}{1}$

4. $\frac{7}{15} + \frac{2}{15} = \frac{9}{15}$ Simplest form: $\frac{3}{5}$

5. $\frac{9}{20} + \frac{3}{20} = \frac{12}{20}$ Simplest form: $\frac{3}{5}$

6. $\frac{5}{8} + \frac{1}{8} = \frac{6}{8}$ Simplest form: $\frac{3}{4}$

<h3><em><u>Explanation:</u></em></h3>

Simply add the numerators of the like fractions together. The denominators remain the same.

4 0

3 years ago

Let f(x) represent a function.

STatiana [176]

Answer:

For $f(x+\frac{5}{4})$ , $f(x)$ is translated $\frac{5}{4}$ units left.

For $f(x)-\frac{5}{4}$ , $f(x)$ is translated $\frac{5}{4}$ units down.

Step-by-step explanation:

The transformation of $f(x)\rightarrow f(x+C)$ means that the graph shift to left by C units if C is positive number and shifts to right by C units if C is a negative number.

The transformation of $f(x)\rightarrow f(x)+C$ means that the graph shifts up by C units if C is positive number and shifts down by C units if C is a negative number.

Here, in the transformation of $f(x)\rightarrow f(x+\frac{5}{4}),C = \frac{5}{4}>0$ , so, the function translates left by $\frac{5}{4}$ units.

Similarly, in the transformation of $f(x)\rightarrow f(x)-\frac{5}{4},C = -\frac{5}{4}$ , so, the function translates down by $\frac{5}{4}$ units.

3 0

3 years ago

Complete the following proof. Prove: In an isosceles triangle, two medians are equal.

Paraphin [41]

Answer:

Midpoint of AB = (0 + 2a / 2 , 0 + 0 / 2) = (2a / 2 , 0 / 2) = (a,0)

x coordinate of point c = a

N = (0 + a / 2 , 0 + b / 2) = (a / 2 , b / 2)

M = ( 2a + a / 2 , 0 + b / 2) = (3a / 2 , b / 2)

MA = √(3a / 2 - 0)² + b / 2 - 0)²

= √(3a / 2 )² + (b / 2) = 9a² / 4 + b² / 4

NB = √(a / 2 - 2a)² + (b / 2 - 0 )²

= √( a / 2 - 4a / 2)² + (b / 2 - 0)²

= √(-3a / 2)² + (b / 2)² = √9a² / 4 + b² / 4

Step-by-step explanation:

I tried my best hope its correct :0

8 0

3 years ago