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3 years ago

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The sets of numbers 6, 8, 10 and 5, 12, 13 are Pythagorean triples. Use what you know about the Pythagorean Theorem and explain

or show why they are Pythagorean triples. Be sure to show your work for each set of triples!

1 answer:

allochka39001 [22]3 years ago

7 0

We know that the Pythagorean Theorem is a² + b² = c². First lets start with 6, 8, and 10. There is a simple way to know that these are Pythagorean triples. They make up what is called a 3-4-5 triangle. 3 x 2 is 6, 4 x 2 is 8, and 5 x 2 is 10. Next 5, 12, and 13 are Pythagorean triplets because 5² + 12² = 13². Because both of these form right triangles, we know that they are Pythagorean triplets.
Hope this helps :-)

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What is the equation for the plane illustrated below?

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Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

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