2 years ago

Given that these triangles are similar, what is the perimeter of triangle PQR? Show your work.

Mathematics

1 answer:

dimaraw [331]2 years ago

6 0

Answer:

Step-by-step explanation:

I use the Pythagoras theorem.

$\sqrt{a^{2} +b^{2} } =c^{2}$

$4^{2}{x}6^{2} = c^{2}$

16 x 36= 576

$\sqrt{576} = 24$

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is (-1/2 x + 6/7) - (7/2 x - 4/7) equivalent to -4 x + 2/7 ? use the properties of operations to justify your answer

LuckyWell [14K]

Answer:

The given equation is NOT justified.

Step-by-step explanation:

Here, the given equation is $(\frac{-1}{2}x + \frac{6}{7} ) - (\frac{7}{2}x - \frac{4}{7} )$ = $-4x + \frac{2}{7}$

Now, here Consider the left hand side and simplifying the left side by operating on the like terms,

⇒ $(\frac{-1}{2}x + \frac{6}{7} ) - (\frac{7}{2}x - \frac{4}{7} )$ = $\frac{-1}{2}x + \frac{6}{7} - \frac{7}{2}x + \frac{4}{7} = (\frac{-1}{2}x -\frac{7}{2}x) + (\frac{6}{7} + \frac{4}{7})$

= $(\frac{-1 - 7}{2}x ) + (\frac{6 + 4}{7} ) = \frac{-8x}{2} + \frac{10}{7} \neq -4x + \frac{2}{7}$

Here, Left side of equation ≠ Right side of the equation

Hence, $(\frac{-1}{2}x + \frac{6}{7} ) - (\frac{7}{2}x - \frac{4}{7} )$ ≠ $-4x + \frac{2}{7}$

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