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Mathematics

1 answer:

Alla [95]3 years ago

8 0

Answer:

a) -2n+14

b) -5n+30

Step-by-step explanation:

a) 12 to 10 to 8 etc. is -2 each time.

Since this is linear, so the first part is therefore -2n. To get from -2 to 12, you have to add 14. So the expression is -2n+14. Now, to prove this, we can input 5 where n is to find the 5th term, which should be 4. -2x5 = -10. -10+14 = 4. The expression is proven correct.

b) 25 to 20 to 15 etc. is -5 each time.

This is also linear, so the first part is therefore -5n. To get from -5 to 25, you have to add 30. So the expression is -5n+30. We can prove this similar to question A. We input 5 where n is to find the 5th term which should be 5. -5x5 = -25. -25+30 = 5. The expression is proven correct.

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Given that B=(1,2,3,4) how many subsets have exactly two elements?

balandron [24]

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3 years ago

Read 2 more answers

Could you have a polynomial with solutions 2i and 5i? If not explain why not. If yes, then construct the polynomial.

baherus [9]

If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.

So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.

(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.

(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)

= x^4 + 29x^2 + 100

So the equation would be x^4 + 29x^2 + 100 = 0.

Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.

5 0

3 years ago

Can someone answer the question?

fomenos

Answer:

$\frac{\sqrt{3} }{3} }(cos(\frac{19\pi}{12})+isin(\frac{19\pi}{12}))$

Step-by-step explanation:

Division of complex numbers in polar form is $z_1/z_2=\frac{r_1}{r_2}cis(\theta_1-\theta_2)$ where $z_1$ and $z_2$ are the complex numbers being divided, $r_1$ and $r_2$ are the moduli, $\theta_1$ and $\theta_2$ are the arguments, and $cis$ is shorthand for $cos\theta+isin\theta$ . Therefore: