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3 years ago

Four times the square of a certain number increased by 6 times the number equals 108. Find the number

1 answer:

3 0

-6 and 4.5
So to do this you’ll make an equation x will represent the number so 4x^2+6x=108 so we want the equation to equal 0 so we can solve it to do that you have to subtract 108 from both sides so it ends up being 4x^2+6x-108=0 we want to isolate the x onto one side and division property allows us to divide both sides by 2 the reason it’s two is because 2 is the biggest divisible number that every number in the equation is divisible by so once you divide every number by two you get 2x^2+3x-54=0 now you have to factor cause but since you only have 3x and nothing else to factor with you have to write 3x as a difference so you could do 2x^2+12x-9x-54=0 so you are complicating the problem so you can factor out 2x from the expression so 2x(x+6)- 9(x+6)=0 the 54 got factored because it’s divisible by 9 that 6 is in replace of the 54 cause if you solved it 9x6 is still 54 next factor out x+6 so (x+6) x (2x-9) =0 so one of the two have to equal 0 so right the equations separately x+6=0 minus 6 from both sides and you’ll get x to equal -6 and 2x-9=0 add 9 to both sides and you’ll get 2x=9 divide both sides by 9 and you’ll get x to equal 4.5 when you plug 4.5 and -6 for x the equation works out

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Find the nth term of the sequence 7,25,51,85,127

olya-2409 [2.1K]

Let a (n) denote the n-th term of the given sequence.

Check the forward differences, and denote the n-th difference by b (n). That is,

b (n) = a (n + 1) - a (n)

These so-called first differences are

b (1) = a (2) - a (1) = 25 - 7 = 18

b (2) = a (3) - a (2) = 51 - 25 = 26

b (3) = a (4) - a (3) = 85 - 51 = 34

b (4) = a (5) - a (4) = 127 - 85 = 42

Now consider this sequence of differences,

18, 26, 34, 42, …

and notice that the difference between consecutive terms in this sequence b is 8:

26 - 18 = 8

34 - 26 = 8

42 - 34 = 8

and so on. This means b is an arithmetic sequence, and in particular follows the rule

b (n) = 18 + 8 (n - 1) = 8n + 10

for n ≥ 1.

So we have

a (n + 1) - a (n) = 8n + 10

or, replacing n + 1 with n,

a (n) = a (n - 1) + 8 (n - 1) + 10

a (n) = a (n - 1) + 8n + 2

We can solve for a (n) by iteratively substituting:

a (n) = [a (n - 2) + 8 (n - 1) + 2] + 8n + 2

a (n) = a (n - 2) + 8 (n + (n - 1)) + 2×2

a (n) = [a (n - 3) + 8 (n - 2) + 2] + 8 (n + (n - 1)) + 2×2

a (n) = a (n - 3) + 8 (n + (n - 1) + (n - 2)) + 3×2

and so on. The pattern should be clear; we end up with

a (n) = a (1) + 8 (n + (n - 1) + … + 3 + 2) + (n - 1)×2

The middle group is the sum,

$\displaystyle 8\sum_{k=2}^nk=8\sum_{k=1}^nk-8=\frac{8n(n+1)}2-8=4n^2+4n-8$

so that

a (n) = a (1) + (4n ² + 4n - 8) + 2 (n - 1)

a (n) = 4n ² + 6n - 3

4 0

3 years ago

(tanx+cotx)/(secxcscx)=1

Dennis_Churaev [7]

Hello here is a solution:

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3 years ago

Can someone help me with part b on this question please?

Galina-37 [17]

9514 1404 393

Answer:

12.0 cm

Step-by-step explanation:

The Pythagorean theorem applies:

(12 cm)² +b² = (17 cm)²

b² = (289 -144) cm² = 145 cm²

b = √145 cm ≈ 12.04 cm

b ≈ 12.0 cm . . . . rounded to 1 decimal place

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Musya8 [376]

Answer:

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