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Pani-rosa [81]
2 years ago
12

Explain how the distance formula can prove Pythagoras' Theorem. In your

Mathematics
1 answer:
Paul [167]2 years ago
8 0

Answer:

Given a triangle ABC, Pythagoras' Theorem shows that:

c^2=a^2+b^2

Thus,

c = \sqrt{a^2+b^2}

The distance formula, gives an equivalent expression based on two points at the end of the hypotenuse for a triangle.

d^2 = (x_{2} -x_{1})^2 + (y_{2} -y_{1})^2

d = \sqrt{(x_{2}-x_{1})^2 +  (y_{2} -y_{1})^2 }

Therefore when given the hypotenuse with endpoints at

(x_{1}, y_{1})  and {(x_{2}, y_{2})

We know that the third point of the right triangle will be at

(x_{2}, y_{1})

and that the two side lengths will be defined by the absolute values of:

(x_{2} - x_{1}) = a

(y_{2} - y_{1}) = b

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uysha [10]

Answer:  No

Step-by-step explanation:

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zubka84 [21]

PEMDAS.

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

Lets apply that here.

72-[(3x2^3)+48]

So first we start with (3x2^3)

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3 years ago
Read 2 more answers
Write the slope-intercept equation of the line parallel to 5y = 2x + 20 that goes through (-1, 3).
poizon [28]

The first thing I'll do is solve "5y = 2x + 20" for "<span>y=</span>", so that I can find my reference slope:

y = (2/5)x + 4;

So the reference slope from the reference line is <span>m = 2/5;</span>.

Now I need to find two new slopes, and use them with the point they've given me; namely, with the point (-1, 3). They want me to find the line through (4, –1) that is parallel to 5y = 2x + 20; that is, through the given point, they want me to find a line that has the same slope as the reference line. 

Since a parallel line has an identical slope, then the parallel line through (-1, 3) will have slope <span>m = 2/5</span>. Now I have a point and a slope! So I'll use the point-slope form to find the line: y - 3 = (2/5)( x + 1);

Finally, y = (2/5)x + 17/5;


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3 years ago
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