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3 years ago

A rectangular box is 6 inches long, 5 inches wide and 7 inches tall. What is the approximate length of its longest diagonal?

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

7 0

Answer:

10.5 inches

Step-by-step explanation:

The length of the longest diagonal is across two opposite points of the solid, namely

L = sqrt ( 6^2 + 5^2 + 7^2)

= sqrt (36+25+49)

= sqrt(110)

= 10.5"

V125BC [204]3 years ago

3 0

Answer:

im sorry i need points plz dont hate me for not answering and wasting you time

Step-by-step explanation:

truly sorry

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Use Newton's method to find the second and third approximation of a root of x3+x+2=0 starting with x1=−1 as the initial approxim

adoni [48]

Answer:

The third approximation is $x_3=-1.08259$

Step-by-step explanation:

We are given that

$f(x)=x^3+x+2=0$

$x_1=-2$

We have to find the second and third approximation of a root of given equation by using Newton's method.

We know that Newton's method , if nth approximation is given $x_n$ and $f'(x_n)\neq 0$ then, the next approximation is given by

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

$f'(x)=3x^2+1$

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Substitute the value n=1 then, we get

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}$

Substitute the values then , we get the second approximation

$x_2=-2-\frac{-8}{13}=-\frac{18}{13}$

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$f(x_2)=f(-\frac{18}{13})=(-\frac{18}{13})^3-\frac{18}{13}+2=-2.03914$

$f'(x_2)=f'(-\frac{18}{13})=3(-\frac{18}{13})^2+1=6.75148$

$x_3=x_2-\frac{f(x_2)}{f'(x_2)}$

$x_3=-\frac{18}{13}-\frac{-2.03914}{6.75148}=-1.08259$

Hence, the third approximation is $x_3=-1.08259$

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