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katrin2010 [14]
3 years ago
12

Which is the correct formula to find the distance between two points (a, b) and (c, d)?

Mathematics
1 answer:
stepan [7]3 years ago
3 0

Answer:

Learn how to find the distance between two points by using the distance formula, which is an application of the Pythagorean theorem. We can rewrite the Pythagorean theorem as d=√((x_2-x_1)²+(y_2-y_1)²) to find the distance between any two points. Created by Sal Khan and CK-12 Foundation.

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Hi can you help me please?)
AlladinOne [14]

<em>*To solve for a specified variable, you need to isolate that variable onto one side.</em>

<h3>11.</h3>

Firstly, subtract 5r on both sides: 2p=q-5r

Lastly, divide both sides by 2 and <u>your answer will be p=\frac{1}{2}q-\frac{5}{2}r</u>

<h3>12.</h3>

First, subtract z on both sides of the equation: -10-z=xy

Next, divide both sides by y and <u>your answer will be \frac{-10-z}{y}=x</u>

<h3>13.</h3>

Firstly, multiply both sides by b: a=cb

Next, divide both sides by c and <u>your answer will be \frac{a}{c}=b</u>

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3 years ago
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3 years ago
Find the value of the variable.
nalin [4]

Answer:

The variable, y is 11°

Step-by-step explanation:

The given parameters are;

in triangle ΔABC;          {}              in triangle ΔFGH;

Segment \overline {AB} = 14         {}               Segment \overline {FG} = 14

Segment \overline {BC} = 27         {}              Segment \overline {GH} = 19

Segment \overline {AC} = 19         {}               Segment \overline {FH} = 2·y + 5

∡A = 32°                       {}                ∡G = 32°

∡A = ∠BAC which is the angle formed by segments \overline {AB} = 14 and \overline {AC} = 19

Therefore, segment \overline {BC} = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments \overline {FG} = 14 and \overline {GH} = 19

Therefore, segment \overline {FH} = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;

\overline {FH} ≅ \overline {BC} by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴ \overline {FH} = \overline {BC} = 27° y definition of congruency

\overline {FH} = 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°

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