1 year ago

Non homogeneous recurrence relation2fn - 6fn-1 = (n+1)(3^n)

Mathematics

1 answer:

mel-nik [20]1 year ago

3 0

Answer:

Non-Homogeneous Recurrence Relation and Particular Solutions

A recurrence relation is called non-homogeneous if it is in the form. Fn=AFn−1+BFn−2+f(n) where f(n)≠0.

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Juli2301 [7.4K]

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2 years ago

Cameron rent is $1,308 per month .cameron wants to pay $1 per bills .how many $100 bills will Cameron need to pay his rent for t

AnnyKZ [126]

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2 years ago

Read 2 more answers

Someone please help me

exis [7]

V = 1/3 l h w

V = 1/3 (12)(15)(9)
V= = 540

answer 540 m^3

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2 years ago

Is the given value a solution of the inequality

dsp73

Answer:

35 <5m

35/5 < 5m/5

7 < m

m = 8,9,10,11,12,.......

a + 6 $\geq$ 9

a + 6 - 6 $\geq$ 9 -6

a $\geq$ 3

a = 3,4,5,6,7,......

8-g > 5

8 -g +g > 5 +g

8 > 5 +g

8 -5 > 5 -5 +g

3 > g

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Step-by-step explanation:

7 0

2 years ago

Choose the appropriate pattern and use it to find the product: (p4−q4)(p4+q4).

sp2606 [1]

The expression can be solved by expanding the bracket and multiplying out the terms

$(p^4-q^4)(p^4+q^4)$ $\begin{gathered} =p^4(p^4+q^4)-q^4(p^4+q^4) \\ =p^8+p^4q^4-p^4q^4-q^8 \\ =p^8-q^8 \end{gathered}$

Therefore, the expression can be simplified as;

$p^8-q^8$

Alternatively, using the theorem of difference of two squares, which is

$a^2-b^2=(a-b)(a+b)$

Hence,

$p^8-q^8=(p^4)^2-(q^4)^2$

7 0

7 months ago

Non homogeneous recurrence relation2fn - 6fn-1 = (n+1)(3^n)​

Non homogeneous recurrence relation2fn - 6fn-1 = (n+1)(3^n)