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tresset_1 [31]
3 years ago
8

Why is it important to calculate the distance between two points on a cartesian coordinate plane?

Mathematics
1 answer:
Gwar [14]3 years ago
4 0

Answer:

Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem,

a

2

+

b

2

=

c

2

, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. The relationship of sides

|

x

2

−

x

1

|

and

|

y

2

−

y

1

|

to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example,

|

−

3

|

=

3

. ) The symbols

|

x

2

−

x

1

|

and

|

y

2

−

y

1

|

indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.

c

2

=

a

2

+

b

2

→

c

=

√

a

2

+

b

2

It follows that the distance formula is given as

d

2

=

(

x

2

−

x

1

)

2

+

(

y

2

−

y

1

)

2

→

d

=

√

(

x

2

−

x

1

)

2

+

(

y

2

−

y

1

)

2

We do not have to use the absolute value symbols in this definition because any number squared is positive.

A GENERAL NOTE: THE DISTANCE FORMULA

Given endpoints

(

x

1

,

y

1

)

and

(

x

2

,

y

2

)

, the distance between two points is given by

d

=

√

(

x

2

−

x

1

)

2

+

(

y

2

−

y

1

)

2

Step-by-step explanation:

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Answer:

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The volume of Solid A is 171.5 m and the volume of Solid B is 500 m, If the solids are similar, find the scale factor of Solid A
Bond [772]

Answer:

The scale factor is 1.429

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

so

Let

z----> the scale factor

x----> volume of solid B

y ----> volume of solid A

z^{3}=\frac{x}{y}

we have

x=500\ m^{3}

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substitute

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Suppose M is the midpoint of Segment AB, P is the midpoint of Segment AM, and Q is the midpoint of segment PM.
DerKrebs [107]

The coordinates of M, P and Q in terms of a and b are M = \frac{1}{2}\cdot a + \frac{1}{2}\cdot b, P = \frac{3}{4}\cdot a + \frac{1}{4}\cdot b and Q = \frac{1}{8}\cdot a - \frac{1}{8}\cdot b, respectively.

In this question we are going to use definitions of vectors and product of a vector by a scalar. Based on the information given on statement, we have the following vectorial formulas:

Location of M

\overrightarrow{AM} = \frac{1}{2}\cdot \overrightarrow{AB}

\vec M - \vec A = \frac{1}{2}\cdot \vec B - \frac{1}{2}\cdot \vec A

\vec M = \frac{1}{2}\cdot \vec A +\frac{1}{2}\cdot \vec B

M = \frac{1}{2}\cdot a + \frac{1}{2}\cdot b

Location of P

\overrightarrow{AP} = \frac{1}{2}\cdot \overrightarrow{AM}

\vec P - \vec A = \frac{1}{2}\cdot \vec M - \frac{1}{2}\cdot \vec A

\vec P = \frac{1}{2}\cdot \vec A +\frac{1}{2}\cdot \vec M

\vec P = \frac{3}{4}\cdot \vec A  + \frac{1}{4}\cdot \vec B

P = \frac{3}{4}\cdot a + \frac{1}{4}\cdot b

Location of Q

\overrightarrow{QM} = \frac{1}{2}\cdot \overrightarrow{PM}

\vec M - \vec Q = \frac{1}{2}\cdot \vec M - \frac{1}{2}\cdot \vec P

\vec Q = \frac{1}{2}\cdot \vec P - \frac{1}{2}\cdot \vec M

\vec Q = \frac{1}{2}\cdot \left(\frac{3}{4}\cdot \vec A + \frac{1}{4}\cdot \vec B\right) -\frac{1}{2}\cdot \left(\frac{1}{2}\cdot \vec A + \frac{1}{2}\cdot \vec B\right)

\vec Q = \frac{1}{8}\cdot \vec A -\frac{1}{8}\cdot \vec B

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The coordinates of M, P and Q in terms of a and b are M = \frac{1}{2}\cdot a + \frac{1}{2}\cdot b, P = \frac{3}{4}\cdot a + \frac{1}{4}\cdot b and Q = \frac{1}{8}\cdot a - \frac{1}{8}\cdot b, respectively.

We kindly invite to check this question on midpoints: brainly.com/question/4747771

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