<span>binomial </span>is an algebraic expression containing 2 terms. For example, (x + y) is a binomial.
We sometimes need to expand binomials as follows:
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
<span>(a + b)4</span> <span>= a4 + 4a3b</span><span> + 6a2b2 + 4ab3 + b4</span>
<span>(a + b)5</span> <span>= a5 + 5a4b</span> <span>+ 10a3b2</span><span> + 10a2b3 + 5ab4 + b5</span>
Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.
Pascal's Triangle
We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
You can use this pattern to form the coefficients, rather than multiply everything out as we did above.
The Binomial Theorem
We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
<span>Properties of the Binomial Expansion <span>(a + b)n</span></span><span><span>There are <span>\displaystyle{n}+{1}<span>n+1</span></span> terms.</span><span>The first term is <span>an</span> and the final term is <span>bn</span>.</span></span><span>Progressing from the first term to the last, the exponent of a decreases by <span>\displaystyle{1}1</span> from term to term while the exponent of b increases by <span>\displaystyle{1}1</span>. In addition, the sum of the exponents of a and b in each term is n.</span><span>If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.</span>
Answer:
Equilateral Triangle, Isosceles Triangle, Scalene Triangle
Step-by-step explanation:
Equilateral: 3 lines of symmetry
Isosceles: 1 line of symmetry
Scalene: no lines of symmetry
Answer:
6 units
Step-by-step explanation:
At the start, the tank contains
(0.25 lb/gal) * (100 gal) = 25 lb
of sugar. Let 
be the amount of sugar in the tank at time 
. Then 
.
Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of
and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,
Separate variables, integrate, and solve for <em>S</em>.
Use the initial value to solve for <em>C</em> :
The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time
has passed. At this time, we want the tank to contain
(0.5 lb/gal) * (5 gal) = 2.5 lb
of sugar, so we pick <em>P</em> such that
<span>11(23-x)
= 11(23) - 11x
= 253 - 11x</span>
