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3 years ago

Find the difference: 3.45 - 0.12 =

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

6 0

Answer:

3.33

Step-by-step explanation:

3.45

- 0.12

______

3.33

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A 1-unit by 4-unit rectangle is shown below

barxatty [35]

Answer:

Step-by-step explanation:

You multiply the height times the width

5 0

4 years ago

Read 2 more answers

Find the value of 3x/2 - 7 if x =8

erastova [34]

Answer:

( ( 3(8) ) /2 ) - 7 = 5

Step-by-step explanation:

Eliminate the denominator by reducing the fraction by 2

3(4) - 7

Solve:

12 - 7

= 5

4 0

3 years ago

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What is 5h-9= -16+6h

valkas [14]

5h - 9 = -16 + 6h

Combine like terms

5h subtracted from both sides.

-9+16 = -16 + h
+16

-9+16 = 7

h = 7

Answer: h = 7

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3 years ago

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1.- Don Francisco reparte 4/5 de su herencia a sus hijos Luis, Pedro y Alondra. ¿Cuánto le toca a cada uno? Datos Operación Resu

guajiro [1.7K]

Respuesta:

4/15

2/15

Explicación paso a paso:

1.)

Fracción de herencia distribuida = 4/5

Número de personas entre las que se distribuyó la herencia = 3

Si cada uno recibió la misma cantidad de herencia;

La cantidad de herencia que tiene cada uno es:

Fracción de la herencia distribuida ÷ número de personas

4/5 ÷ 3

4/5 * 1/3 = (4 * 1) / (5 * 3) = 4/15

Cantidad de herencia que cada uno recibe = 4/15

2.)

Cantidad de pan elaborado por 5 trabajadores = 2/3

Si cada trabajador trabaja al mismo ritmo;

Cantidad de pan elaborado por cada trabajador:

2/3 ÷ 5

2/3 * 1/5 = (2 * 1) / (3 * 5) = 2/15

3 0

3 years ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

3 years ago