1)Rewrite the table:
70, 49, 34.3, 24.01, 11.807 {The original size of the wound =70}
2) write the quotient of each number by the number before & notice the value:
49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^⁽ⁿ⁻¹⁾
3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
or f(n) = y = 70(0.7)ⁿ / 0.7==> f(n) = [(0.7)ⁿ ]/ 100.
This is a decreasing exponential function where the coefficient
raised to n is < 1.
The domain is for all n>= 0.
When n→∞, f(n)→0; For n=0==>f(n) =70. So the range of f(n) is:<=70
<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>
81/4
Step-by-step explanation:
9=4x/9
81=4×
x=81/4
adding logs means elements are multiplied, subtracting means divide. number in front of log is raised to power
base of log is raised to power of element after =
element in log is then equal to log base raised to power
solve resulting equation
The main property is as follow
(A^p)^q=A^pxq
so only Sam's work is correct because
(8^4)^5 x(7^3)^9=(8^4x5)x(7^3x9)=8^20x 7^27
