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4 years ago

What is the tenth term of the sequence 2, 5, 8, 11, 14, … ?

Mathematics

2 answers:

nikitadnepr [17]4 years ago

7 0

The tenth term is 29 :)

ziro4ka [17]4 years ago

4 0

Every term is adding 3 to the previous term. Therefore, the terms are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29 ... The tenth term is 29.

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Where would 3/5 be plotted in a number line?

Mamont248 [21]

Between 0 and 1, add dashes so you have a total of 6 marks including 0 and 1. Then count 3 over and that would be 3/5.

7 0

4 years ago

Read 2 more answers

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Equivalent to (6 + y) × 5?

Sergeeva-Olga [200]

The answer is:
It’s equivalent to 30+5y

6 0

3 years ago

Read 2 more answers

Given that f(–2.4) = -1 and f(-1.9) = -8, approximate f'(-2.4). f'(-2.4)

lora16 [44]

Answer:

f'(-2.4) ≈ -14

General Formulas and Concepts:
Algebra I

Coordinate Planes

Coordinates (x, y)

Slope Formula: $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

Functions

Function Notation

Calculus

Differentiation

Derivatives
Derivative Notation

Step-by-step explanation:

*Note:

The definition of a derivative is the slope of the tangent line.

Step 1: Define

Identify.

f(-2.4) = -1

f(-1.9) = -8

Step 2: Differentiate

Simply plug in the 2 coordinates into the slope formula to find slope m.

[Derivative] Set up [Slope Formula]: $\displaystyle f'(-2.4) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}$
Substitute in coordinates: $\displaystyle f'(-2.4) \approx \frac{-8 - -1}{-1.9 - -2.4}$
Evaluate: $\displaystyle f'(-2.4) \approx -14$