Answer:
Step-by-step explanation:
Quadratic formula: 
when the equation is 
The given equation is 
. Let's first arrange this so its format looks like 
:
Subtract 1 from both sides of the equation
Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:
I hope this helps!
Answer:
19.2873015 (Done by a calculator) <---- could be wrong
Step-by-step explanation:
And fun fact:
Negative numbers don't have real square roots since a square is either positive or 0. The square roots of numbers that are not perfect square are members of irrational numbers. This means that they can't be written as the quotient of two integers.
Hope it helps!
We can write a system of equations:
1x + 10y = 182
x + y = 56
Where 'x' is the number of $1 bills, and 'y' is the number of $10 bills.
To find this we can solve using substitution.
Re-arrange the 2nd equation:
x + y = 56
Subtract 'y' to both sides:
x = -y + 56
Now we can plug in '-y + 56' for 'x' in the first equation.
1x + 10y = 182
1(-y + 56) + 10y = 182
-y + 56 + 10y = 182
Subtract 56 to both sides:
-y + 10y = 126
Combine like terms:
9y = 126
Divide 9 to both sides:
y = 14
Now we can plug this into any of the two equations to find the 'x' value.
x + y = 56
x + 14 = 56
Subtract 14 to both sides:
x = 42
So our final answer is (42, 14).
This means that the motel clerk had 42 $1 bills, and 14 $10 bills.
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
That would be the average of total time paddled upstream & downstream . . . 12 miles / 7 hours = approx 1.71 miles per hr
