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3 years ago

Given the recursive formula shown, what are the first 4 terms of the sequence?

Mathematics

1 answer:

Lorico [155]3 years ago

3 0

Answer:

5,20,80,320

Step-by-step explanation:

a1 = 5

an = 4 an-1

Let n = 2

a2 = 4 * a1 = 4*5 = 20

Let n = 3

a3 = 4 * a2 = 4*20 = 80

Let n = 4

a4 = 4 * a3 = 4*80 = 320

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Need help to do this step by step please 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: $\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}$

8 0

3 years ago

If the gradient of the tangent to

Marizza181 [45]

Answer:

Point A(9, 3)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$