Mathematics, 13.08.2020 21:01 artursino

HELPPp me asap please I'm hopeless The arithmetic progressions $\{2, 5, 8, 11, \ldots\}$ and $3, 10, 17, 24, \ldots \}$ have some common values. What is the largest value less than 500 that they have in common?

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Answer from: culturedxnat

479

Work Shown:

We have two sequences, which we'll call A and B

A = {2,5,8,11,14, 17 ,20,23,26,29,32,35, 38, 41,44,47,50,53,56, 59 ,62...}B = {3,10, 17 ,24,31, 38 ,45,52, 59 , 66...}

Both sequences are arithmetic. The first sequence starts at 2, and increases by 3 each time. Sequence B starts at 3 and increases by 7 each time.

The terms in bold are what the two sequences have in common.

It turns out that the bold terms follow an arithmetic sequence of their own.

The sequence {17, 38, 59, ...} is arithmetic.

Call this sequence C. This sequence starts at 17 and increases by 21 each time. The nth term for sequence C is

C(n) = a+d(n-1)

C(n) = 17+21(n-1)

C(n) = 17+21n-21

C(n) = 21n-4

Plugging in n = 1 leads to C(n) = 17. Also, plug in n = 2 and you should get C(n) = 38. Plugging in n = 3 leads to C(n) = 59. And so on.

The question is now "what is the largest term in sequence C such that it is less than 500?".

In other words, we want to find the term C(n) where C(n) < 500 and C(n) is as large as possible.

Use a bit of algebra to get

C(n) < 500

21n - 4 < 500

21n < 500+4

21n < 504

n < 504/21

n < 24

So n must be smaller than 24 to get what we want.

Because n is a natural number (positive whole number), we drop to n = 23 to find that

C(n) = 21n-4

C(23) = 21(23)-4

C(23) = 479

The largest value, less than 500, the two original sequences have in common is 479.

If we didn't have the restriction "less than 500", then there would be no largest value as sequence C goes on forever.

Answer from: Quest

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

Answer from: Quest

answer:

total cost for tiles and paints is $924.

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

$\text{area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{area of walls and ceiling}=1200+90+120$

$\text{area of walls and ceiling}=1410$

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

$\text{ the total painting cost}=10*(\frac{1410}{25})$

$\text{ the total painting cost}=10*56.4=564$

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

$\text{area of floor} = 40*3\text{ square meters}$

$\text{area of floor} =120\text{ square meters}$

given: 1m squared floor tiles costs $3. so,

$\text{total cost for tiles} = 3*120 = 360$

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

$\text{combined total cost}= 564+360 = 924$

therefore, the combined total cost of tiles and paint is $924.

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