<span>Int var1 = 0b0001;
int var2 = 0b1111;
int results1 = var1 & var2;
int results2 = var1 | var2;
int results3 = var1 ^ var2;
int printit = results1 + results2 + results3;
what are the values for results1, results2, results3 and printit after executing the code?
notes:
1. faster responses will be obtained if your code is presented line by line (in a file) before posting.
2. please specify language, many languages use the same syntax but could have differences in interpretation
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Assuming Java as the language. C is similar.
</span><span><span>& bitwise AND &
^ </span><span>bitwise exclusive OR
</span><span>| bitwise inclusive OR
So
results1=var1&var2=0b0001&0b1111=0b0001
results2=var1|var2=0b0001&0b1111=0b1111
results3=var1^var2=0b0001&0b1111=0b1110
printit=results1+results2+results3=0b0001+0b1111+0b1110
=0b10000+0b1110
=0b11110
Note: by default, int has 4 signed bytes, ranging from decimal -2147483648 to +2147483647
</span></span>
Answer:
Get all the terms with x on the left. Get all the terms without x on the right. Simplify by adding and subtracting the terms without x and adding and subtracting the terms with x. Finally divide to get the value of x.
Step-by-step explanation:
6x+6 =- 5+3x+17
6x+6 -6 =- 5+3x+17 -6
6x = 5+3x+11
6x = 3x + 16
6x - 3x = 3x + 16 -3x
3x = 16
3x / 3 = 16 / 3
x = 15/3 + 1/3
x = 5 1/3
x = 5.33333...
Answer:
3=1/5x+7
Step-by-step explanation:
when it says "Of a number" it means to add a variable, so i added a x. and when it says "more than" it means to add, so 3=1/5x+7 is the answer
For this case we have that by definition of power properties it is fulfilled that:
We must rewrite the following function:
Using the mentioned property we have:
Solving the operation within the parenthesis we have:
Thus, the correct option is option B
ANswer:
Option B
