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3 years ago

PLEASE HELP !!!!!!! WILL MARK BRAINLIEST FOR A REAL ANSWER

Mathematics

1 answer:

rjkz [21]3 years ago

3 0

Answer:

A. Find the maximum height of the acrobat.

Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

Given that rectangle LMNO with coordinates L(0,0), M(3,0), N(3,7), O(0,7), P is the midpoint of LM⎯⎯⎯, and Q is the midpoint of

Elina [12.6K]

The midpoint of a line divides the line into equal segments.

The option that proves PQ = LO is (a)

The given parameters are:

$\mathbf{L = (0,0)}$

$\mathbf{M = (3,0)}$

$\mathbf{N = (3,7)}$

$\mathbf{O = (0,7)}$

P is the midpoint of LM.

So, we have:

$\mathbf{P = \frac{LM}{2}}$

$\mathbf{P = (\frac{(0 +3}{2},\frac{0+0}{2})}$

$\mathbf{P = (\frac{3}{2},0)}$

Q is the midpoint of NO.

So, we have:

$\mathbf{Q = \frac{NO}{2}}$

$\mathbf{Q = (\frac{(3 +0}{2},\frac{7+7}{2})}$

$\mathbf{Q = (\frac{3}{2},7)}$

Distance PQ is calculated as follows:

$\mathbf{d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

This gives:

$\mathbf{PQ = \sqrt{(3/2 - 3/2)^2 + (0 - 7)^2}}$

$\mathbf{PQ = \sqrt{ 7^2}}$

$\mathbf{PQ = 7}$

Distance LO is calculated as follows:

$\mathbf{LO = \sqrt{(0 - 0)^2 + (0 - 7)^2}}$

$\mathbf{LO = \sqrt{ 7^2}}$

$\mathbf{LO=7}$

So, we have:

$\mathbf{PQ = 7}$

$\mathbf{LO=7}$

Thus:

$\mathbf{PQ = LO}$

Hence, the correct option is (a)

Read more about distance and midpoints at:

brainly.com/question/11231122

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2 years ago