In the binomial development, the main problem is calculation of binomial coefficients.
If we want to get term a∧8*b∧2 we see that this is the third member in binomial development (n 2) a∧n-2*b∧2
The given binomial is ((1/3)a∧2 - 3b)∧6, the first element is (1/3)a∧2, the second element is (-3b) and n=6 when we replace this in the formula we get
(6 2) * ((1/3)a∧2)∧(6-2) * (-3b)2 = (6*5)/2 * ((1/3)a∧2)∧4 *9b∧2= 15*(1/81)*9 *(a∧8b∧2) =
= 15*9* a∧8b∧2 = 135*a∧8b∧2
We finally get numerical coefficient 135
Good luck!!!
Answer:
6... if the serving size is 1/6 of the package, that means it will serve 6 people
Step-by-step explanation:
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.
Group terms that have factorable coefficients (like 2&4 and then 9&18)
(4x³ + 2x²) + (-18x - 9)
factor out the like terms
2x²(2x + 1) - 9(2x +1)
parenthesis are the same so you can factor further.. {Like 2x+3x = (2+3)x}
(2x² - 9)(2x+1)
Factored too far they only want Answer C. is the correct one for grouping
