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kirill115 [55]
3 years ago
8

Pythagorean theorem..

Mathematics
2 answers:
Paraphin [41]3 years ago
8 0
<h3>By using Pythagoras Theorem:-</h3>

(Hypotenuse)² = (Base)²+ (Perpendicular)²

Taking, Hypotenuse As H

Base As B

Perpendicular As P

_________________________

<h3>Solving The Question</h3>

H²= B²+ P²

15 {}^{2}  = 9 {}^{2}  + p {}^{2}  \\  \\  p {}^{2}  = 15 {}^{2}  + 9 {}^{2}  \\  \\  p {}^{2}  = 225 - 81 \\  \\p {}^{2}  = 144 \\  \\  p =  \sqrt{144}  \\  \\ p = 12

<h3>Hope This Helps You ❤️</h3>
DedPeter [7]3 years ago
4 0

{\huge{\underbrace{\bf{\pink{Answer}}}}}

Length of missing leg = 12 km

Using Pythagoras theorem

a² = 15² - 9²

a = √ 225 - 49

a = √144

a = 12 km

HOPE IT HELPS..

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3 years ago
How do I solve this system using substitution: y = x - 7 ; 2x + y = 8
Rudiy27
Y = x - 7
2x + y = 8

2x + (x - 7) = 8
  2x + x - 7 = 8
        3x - 7 = 8
        <u>    + 7 + 7</u>
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              3      3
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7 0
4 years ago
Read 2 more answers
Review the proof.
andrezito [222]

Answer:

Step-by-step explanation:

A 2-column table with 8 rows. Column 1 is labeled Step with entries 1, 2, 3, 4, 5, 6, 7, 8. Column 2 is labeled Statement with entries cosine (2 x) = 1 minus 2 sine squared (x), let 2 x = theta, then x = StartFraction theta Over 2 EndFraction, cosine (theta) = 1 minus 2 sine squared (StartFraction theta Over 2 EndFraction), negative 1 + cosine (theta) = negative 2 sine squared (StartFraction theta Over 2 EndFraction), 1 + cosine (theta) = 2 sine squared (StartFraction theta Over 2 EndFraction), StartFraction 1 minus cosine (theta) Over 2 EndFraction = sine squared (StartFraction theta Over 2 EndFraction), sine (StartFraction theta Over 2 EndFraction) = plus-or-minus StartRoot StartFraction 1 minus cosine (theta) Over 2 EndFraction EndRoot.

Which step contains an error?

Answer:

 

Step-by-step explanation:

 

7 0
3 years ago
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WINSTONCH [101]

Step-by-step explanation:

({ {a}^{m}) }^{n}  =  {a}^{mn}

({ {5}^{3}) }^{5}

{5}^{5 \times 3}

{5}^{15}

Option A

4 0
3 years ago
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