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2 years ago

How to solve pythagorean theorem i need an assay someone help

Mathematics

1 answer:

Veronika [31]2 years ago

5 0

Answer:

first write the formula

Step-by-step explanation:

Hypotenuse^2 = Perpendicular ^2 + Base^2

only applicable for a right triangle (one of the sides is of measure 90 degree

Give an example

let is assume in Triangle ABC, angle B = 90 degrees

AB = 3cm

BC = 4 cm

to find AC or hypotenuse

Solution

by using pythagoras formula we get

(AB)^2 + (BC)^2 = (AC)^2

(3)^2 + (4)^2 = (AC) ^2

9 + 16 = (AC)^2

25 = (AC)^2

Root of 25 = AC

5cm =AC

Write about how pythagoras theorem was formed

mathematician who derived it

in which year

Application of Pythagorean theorem

Pythagorean Triplets etc.

I think this willbe enough for an essay

GOOD LUCK FOR THE FUTURE! :)

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Answer: The correct line is

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Step-by-step explanation: We are given the following two sets of quadratic expressions in various forms:

$\textup{Line 1: }x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25,\\\\\textup{Line 2 :}x^2+5x+6=(x+2)(x+3)=(x+2.5)^2+6.25.$

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We can see that there are three different forms of a quadratic expression in each of the lines:

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So, to check which line is correct, we need to calculate the factorised form and the vertex form from the simplified form.

We have

$\textup{Line 1: }\\\\x^2+3x+2\\\\=x^2+2x+x+2\\\\=x(x+2)+1(x+2)\\\\=(x+1)(x+2),$

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$x^2+3x+2\\\\=x^2+2\times x\times 1.5+(1.5)^2-(1.5)^2+2\\\\=(x+1.5)^2-2.25+2\\\\=(x+1.5)^2-0.25.$

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$\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.$

Thus, Line 1 contains three equivalent expressions.

Now,

$\textup{Line 2: }\\\\x^2+5x+6\\\\=x^2+3x+2x+6\\\\=x(x+3)+2(x+3)\\\\=(x+2)(x+3),$

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$x^2+5x+6\\\\=x^2+2\times x\times 2.5+(2.5)^2-(2.5)^2+6\\\\=(x+2.5)^2-6.25+6\\\\=(x+2.5)^2-0.25\neq (x+2.5)^2+6.25.$

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$\textup{Line 2 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2+6.25.$

Thus, Line 2 does not contain three equivalent expressions.

Hence, Line 1 is correct.

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