End behavior: f. As x -> 2, f(x) -> ∞; As x -> ∞, f(x) -> -∞
x-intercept: a. (3, 0)
Range: p. (-∞, ∞)
The range is the set of all possible y-values
Asymptote: x = 2
Transformation: l. right 2
with respect to the next parent function:
Domain: g. x > 2
The domain is the set of all possible x-values
550×10 I think how do you want it to be explained
8:8,16,24,32,40,48,56,64,
12:12,24,36,48
Least common multiple is 24 because it's less then the other common multiples such as 48
Answer:
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Step-by-step explanation:
for
I= ∫x^n . e^ax dx
then using integration by parts we can define u and dv such that
I= ∫(x^n) . (e^ax dx) = ∫u . dv
where
u= x^n → du = n*x^(n-1) dx
dv= e^ax dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)
then we know that
I= ∫u . dv = u*v - ∫v . du + C
( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =
(u*v) - ∫v*du + C )
therefore
I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
The answer comes from using proportions. 3lbs to 14 people. Turn that into a fraction, then make a second fraction with 55 as the denominator. Find a common denominator for both denominator values, and multiply the numerators to find the missing value.
