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2 years ago

URGENT HELP The nth term of a sequence is given 3n-16. What’s the value of the 9th term?

Mathematics

1 answer:

Ivan2 years ago

7 0

Answer:

a₉ = 11

Step-by-step explanation:

Substitute n = 9 into the nth term formula, that is

a₉ = 3(9) - 16 = 27 - 16 = 11

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Need help to do this step by step please <br> 3x2 - 2x=-1

sdas [7]

Answer:

Solving the expression $3x^2-2x=-1$ we get: $\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}$

Step-by-step explanation:

We need to solve the expression: $3x^2-2x=-1$

This is a quadratic expression and it can be solved using quadratic formula

Solving:

$3x^2-2x=-1\\$

we can write it as:

$3x^2-2x+1=0$

The quadratic formula is: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a = 3, b = -2 and c= 1

Putting values and solving:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(1)}}{2(3)}\\x=\frac{2\pm\sqrt{4-12}}{2(3)}\\x=\frac{2\pm\sqrt{-8}}{6}\\We\:know\:that\:\sqrt{-1}=i\\x=\frac{2\pm\sqrt{8}\sqrt{-1} }{6} \\We\:know\:\sqrt{8}=\sqrt{2\times 2 \times 2}=\sqrt{2^2 \times 2}=2\sqrt{2} \\x=\frac{2\pm2\sqrt{2}i }{6}\\Now,\\x=\frac{2+2\sqrt{2}i }{6}\:or\:x=\frac{2-2\sqrt{2}i }{6}\\x=\frac{2(1+\sqrt{2}i) }{6}\:or\:x=\frac{2(1-\sqrt{2}i) }{6}\\x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}$

So, solving the expression $3x^2-2x=-1$ we get: $\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}$

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AURORKA [14]

Answer:

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Helga [31]

The answer is 12:15 A.M.

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Is 2(x - 6) = -35 equivalent to 2x = -23

rusak2 [61]

Yes, if you just distribute 2(x-6). Then you would have 2x-12 = -35 then add 12 to the -35 and your answer would be 2x = -23 and leads to them being equivalent.

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How many sixes are equal to double 27?

Phantasy [73]

Answer:

its not sixes theres 4 sixes equal 24 so there is no answer

Step-by-step explanation:

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