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3 years ago

Write the first four terms of the sequence: a = 2 an = a(n-1) + 4

Mathematics

1 answer:

eduard3 years ago

3 0

We have sequence equation $a_n=a(n-1)+4$ .

In this case n is a natural number (1, 2, 3, ...).

So start inserting and computing value of $a_n$ given that you know the value of n and a.

$a_1=2(1-1)+4=4$ (first term)

$a_2=2(2-1)+4=6 \\a_3=2(3-1)+4=8 \\a_4=2(4-1)+4=10$

Hope this helps.

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John plans to buy a ps4 for 249.00. Games cost 49.99 each. Controllers cost 29.99. Write an algerbraic expression if he buys a p

Neporo4naja [7]

Answer:

The algebraic expression is 249 + 29.99(n) where n represents the number of controllers

Step-by-step explanation:

Here in this question, we are interested in writing an algebraic expression which shows the amount spent by John on a ps4 console and any number of controllers;

So the expression will be;

249 + 29.99(n)

Where n represents the number of controllers both

6 0

3 years ago

For a moving object, the force acting on the object varles directly with the object's acceleration. When a force of 25 N acts on

Radda [10]

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

According to the Question,

$\large{\sf{ f \propto \: a}}$

<h3>From the Question,</h3>

Initial Acceleration,a = 5 m/s²

Force at a = 5 m/s²,f = 25 N

Final Acceleration,A = 9 m/s²

Force exerted on the object at a = 9 m/s²

We Know that,

$\huge{\boxed{ \boxed{ \sf{f = ma}}}}$

Firstly,we need to find the mass of the object

Thus,

$\sf{m = \frac{f}{a} } \\ \\ \implies \: \sf{m = \frac{25}{5} } \\ \\ \huge{ \implies \: \sf{m = 5 \: kg}}$

Now,

$\sf{F = mA} \\ \\ \implies \: \sf{F = (5)(9)} \\ \\ \huge{ \implies \: \sf{F = 45 \: N}}$

7 0

3 years ago

Solve the inequality for x<br><br> 19>x-5<br><br> Simplify your answer as much as possible.

Stolb23 [73]

19>x-5

Add 5 on each side giving you
19+5>x
24>x

x<24

Hope this helps!!

7 0

4 years ago

PLEASE HELP ME 40 PTS!!!!!!

SSSSS [86.1K]

Answer:

The sequence converges to infinity.

Step-by-step explanation:

$a_n = \frac{n^3 - n}{n^2 + 5n}\\\\ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^3 - n}{n^2 + 5n}$

$= \lim_{n \to \infty} \frac{n(n^2 - 1)}{n(n + 5)} \\\\= \lim_{n \to \infty} \frac{(n^2 - 1)}{(n + 5)}\\\\= \lim_{n \to \infty} \frac{n(n - \frac{1}{n})}{n(1 + \frac{5}{n})}\\\\= \lim_{n \to \infty} \frac{(n - \frac{1}{n})}{(1 + \frac{5}{n})}$