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yuradex [85]
2 years ago
15

Write a program that accepts the lengths of three sides of a triangle as inputs. the program output should indicate whether or n

ot the triangle is a right triangle. recall from the pythagorean theorem that in a right triangle, the square of one side equals the sum of the squares of the other two sides. use the triangle is a right triangle. and the triangle is not a right triangle. as your final outputs.
Computers and Technology
1 answer:
omeli [17]2 years ago
4 0

The code will have to obey the Pythagorean theorem that says square of the hypotenuse side is equals to the sum of the squares of the other legs.

<h3>How to write a code that check if a triangle is a right angle by using Pythagoras theorem?</h3>

The code is written in python.

def right_triangle(x, y, z):

    if x**2 + y**2 == z**2 or y**2 + z**2 == x**2 or z**2 + x**2 == y**2:

         print("it is a right angle triangle")

    else:

         print("it is not a right angle triangle")

right_triangle(6, 10, 8)

<h3>Code explanation</h3>
  • we defined as function named "right_triangle". x, y and z are argument which are the length of the triangle.
  • Then we check if the sides obeys Pythagoras theorem.
  • If it does we print a positive statement else we print a negative statement.

Learn more about python at: brainly.com/question/21437082

#SPJ4

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Answer:

Therefore the inverse function of  f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

Explanation:

We need to find the inverse of function f(x)=1+\sqrt{1+2x}

Function Inverse definition :

\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}

y=1+\sqrt{1+2x}\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y

x=1+\sqrt{1+2y}

\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+\sqrt{1+2y}-1=x-1

Simplify

\sqrt{1+2y}=x-1

\mathrm{Square\:both\:sides}

\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2

\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y

\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1

1+2y=x^2-2x+1

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+2y-1=x^2-2x+1-1

\mathrm{Simplify}

2y=x^2-2x

\mathrm{Divide\:both\:sides\:by\:}2

\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}

\mathrm{Simplify}

y=\frac{x^2-2x}{2}

Therefore the inverse function of f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

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