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Alina [70]
2 years ago
11

Tell me everything you know about Pythagorean Theorem. Some things to include, but not limited to: Why is it that in some exampl

es you have to subtract the square of a leg from the hypotenuse? Can this be used on every triangle? When might someone use this in real life as an adult?
Mathematics
2 answers:
svlad2 [7]2 years ago
6 0

Answer: Check below

Step-by-step explanation:

Okay so the Pythagorean Theorem was made by some old guy and the formula is a^2+b^2=c^2. You ALWAYS have to subtract the leg of the triangle when your finding a leg. You squareroot the square of the two legs if your finding the hypotnuse. The Theorem can be used to find any side of the triangle ONLY IF its a right triangle. An adult might use it in real life like when your an architect trying to find how much material you should use to make something thats in the shape of a right triangle.

Ps. Sorry if there's a bunch of grammar mistakes since this is kinda rushed

Setler [38]2 years ago
5 0

Answer: if you work in stem, it can be used.

Step-by-step explanation:

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Find for x<br><br> 1/5x + 1/3= 3(2/3x -2)
topjm [15]

Answer:

The value of x  for the given expression is  (95/27)

Step-by-step explanation:

Here, the given expression is:

\frac{1}{5} x + \frac{1}{3}   = 3(\frac{2}{3}x -2)

Now here  solving for the value of x , we get

\frac{1}{5} x + \frac{1}{3}   = 3(\frac{2}{3}x) -6\\\implies\frac{1}{5} x + \frac{1}{3}   = 2x -6

Now, taking the variable terms on 1 side, we get

\frac{1}{5} x   - 2x   = -6 -  \frac{1}{3} \\\implies\frac{x  - 10x}{5}   = -(\frac{6(3)  +1}{3} )

or, \frac{-9x}{5}   = -\frac{19}{3}   \implies x  = \frac{19}{3}  \times\frac{5}{9}   = \frac{95}{27}

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3 years ago
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Step-by-step explanation:

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Let f(x)=3x+1 and g(x)=2x2−1<br> Find f(g(2))<br><br><br> 10<br><br> 22<br><br> 97<br><br> 28
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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x + 1

g(x) = 2x² - 1

<u>Step 2: Find g(2)</u>

  1. Substitute:                    g(2) = 2(2)² - 1
  2. Evaluate:                       g(2) = 2(4) - 1
  3. Multiply:                        g(2) = 8 - 1
  4. Subtract:                       g(2) = 7

<u>Step 3: Find f(g(2))</u>

  1. Define:                    g(2) = 7
  2. Rewrite:                   f(7)
  3. Substitute:               f(7) = 3(7) + 1
  4. Multiply:                   f(7) = 21 + 1
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And we have our final answer!

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