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2 years ago

What is the formula for the Pythagorean theorem?

Mathematics

2 answers:

pochemuha2 years ago

8 0

Answer:

$\Longrightarrow: \boxed{\sf{A^2+B^2=C^2}}$

Step-by-step explanation:

In the Pythagorean theorem, the formula is:

$\underline{\text{PYTHAGOREAN THEOREM FORMULA:}}}$

$\Longrightarrow: \sf{A^2+B^2=C^2}$

Thus, the Pythagorean theorem's formula is "a²+b²=c²".

I hope this helps. Let me know if you have any questions.

maksim [4K]2 years ago

7 0

Answer:

a^2 + b^2 = c^2

Step-by-step explanation:

a^2 + b^2 = c^2

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What is the prime factorization for 50 using exponents?

galina1969 [7]

Answer:

Either way, it gives us the prime factorization for 50 that is 5 x 5 x 2. The prime factorization of 50 can be expressed using exponents as 5^2 x 2.

Step-by-step explanation:

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3 years ago

Read 2 more answers

Describe the steps required to determine the equation of a quadratic function given its zeros and a point.

faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

$y = a\cdot x^{2}+b\cdot x + c$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$a$ , $b$ , $c$ - Coefficients.

A value of $x$ is a zero of the quadratic function if and only if $y = 0$ . By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: $A(x,y) = (r_{1}, 0)$ , $B(x,y) = (r_{2},0)$ and $C(x,y) = (x,y)$

Based on such information, we form the following system of linear equations:

$a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0$ (2)

$a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0$ (3)

$a\cdot x^{2} + b\cdot x + c = y$ (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

$a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

And finally we obtain the equation of the quadratic function given two zeroes and a point.

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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