Answer:
21500 m
Step-by-step explanation:
We will divide this point into four parts and we will pass it by the meter:
Total: (20 km * 1000 m / 1 km) + (75 hm * 100 m / 1 hm) + (75 dam * 10 m / 1 dam) + 250 m = 20000 m + 7500 m + 750 m + 250 m = 28500 m
First stage: (5 km * 1000 m / 1 km) + (5 hm * 100 m / 1 hm) = 5000 m + 500 m = 5500 m
Second stage: (1 km * 1000 m / 1 km) + (50 dam * 10 m / 1 dam) = 1000 m + 500 m = 1500 m
The value of the third stage would be the subtraction between the total and the first and second stage:
Third stage: 28500 m - 5500 m - 1500 m = 21.500 m
He traveled 21500 meters in the third stage
12, 106, 42
hope this helps!
Answer:
Result after multiplication of polynomial is: 
Degree of polynomial = 3
Step-by-step explanation:
The given polynomials are:

In order to multiply the given polynomials we have to work step by step. First of all the polynomials in the bracket will be multiplied and then their result will be multiplied with x.
So, multiplying the polynomials in round brackets first

Now multiplying with x

Degree of a polynomial is the highest exponent of variable in the polynomial.
In the acquired result, the highest exponent of x is 3 so the degree is 3.
Hence,
Result after multiplication of polynomial is: 
Degree of polynomial = 3
Answer:
<u>Length</u><u> </u><u>is</u><u> </u><u>6</u><u>√</u><u>2</u><u> </u><u>units</u><u> </u><u>or</u><u> </u><u>8</u><u>.</u><u>4</u><u>9</u><u> </u><u>units</u>
Step-by-step explanation:
Length formular:

x1 is 2
x2 is -4
y1 is 3
y2 is 9

<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
-------------------------------------------------------------------------------
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>