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3 years ago

What is the sixth term in the sequence 3, 6, 11, 18

Mathematics

1 answer:

zlopas [31]3 years ago

5 0

Answer:

Step-by-step explanation:

3 to 6 is an add by 3 then 6 to 11 is an add by 5 then 11 to 18 is an add by 7 each add is increased by 2 so the next add will be 9 giving you 27 then you will add 11 because it's increased by two giving you 38

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Solve:m+ 2/3=1/2 A. 1 1/6 B. 1/6 C.-1 D. - 1/6

natita [175]

Answer:

D. - 1/6

Step-by-step explanation:

M + 2/3 = 1/2

M = 1/2 - 2/3

$\frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}$

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2 years ago

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Complete the following statements using m(x) = x² and n(x) = x – 3.

larisa [96]

Answer: x-3, 3, is not equal to

Step-by-step explanation:

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3 years ago

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3 years ago

Mia had $500 in her account after 5 years. If she got an

irina [24]

Answer:

The initial amount in the account is $333.33

Step-by-step explanation:

Here, we want to calculate the initial amount in the account.

Let the initial amount be $x

Mathematically;

Simple interest = PRT/100

where P is the amount deposited = $x

R is the rate = 10%

T is the time = 5 years

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By substituting;

500-x = (x * 10 * 5)/100

100(500-x) = 50x

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3 years ago

find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on

enyata [817]

Let $f(x)=\tan^{-1}x$ . Then $f'(x)=\frac1{1+x^2}$ . Note that $f(0)=0$ .

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n$

This means that for $|-x^2|=|x|^2$ , or $-1$ , we have

$\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Integrate the series to get

$f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

$\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

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3 years ago