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3 years ago

The triangles are similar by the AA Similarity Postulate. Find the value of x.

Mathematics

2 answers:

cricket20 [7]3 years ago

3 0

Line x is the midsegment of the triangle.

5/2= 2.5

I hope this helps!
~kaikers

CaHeK987 [17]3 years ago

3 0

Answer: 2.5

Step-by-step explanation:

Given: The triangles in the picture are similar by the AA Similarity Postulate.

We know that is two triangles are similar triangles then its corresponding sides are proportional [Larger side is proportional to larger side of another triangle or vise versa].

$\Rightarrow\frac{x}{5}=\frac{3}{3+3}\\\\\Rightarrow\frac{x}{5}=\frac{3}{6}\\\\\Rightarrow x=\frac{1}{2}\times5\\\\\Rightarrow x=2.5$

Hence, the value of x is 2.5 .

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Answer:

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Step-by-step explanation:

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3 years ago

Marcus sorts his 85 baseball cards into stacks of nine cards each how many stacks of nine cards can Marcus make?

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Step-by-step explanation:

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3 years ago

What is the slope of the line described by the data in the table below?

bezimeni [28]

Given:

The table of values is:

x y

-4 3

0 8

4 13

8 18

To find:

The slope of the line that described by the data in the table.

Solution:

Consider any two points from the given table.

Let the two points are (0,8) and (4,13). So, the slope of the line is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

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3 years ago

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

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

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

Substitute $x_1=-2$

$f(x_1)=f(-2)=(-2)^3+(-2)+2=-8$

$f'(x_1)=f'(-2)=3(-2)^2+1=13$

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

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

For n=2

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

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

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

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

7 0

3 years ago