Determine if there is a proportional relationship between the length of the diagonals and the perimeters of the squares.

Mathematics

2 answers:

RoseWind [281]3 years ago

6 0

Answer:

where is picture for this question??

garri49 [273]3 years ago

5 0

Answer:

that polygon. Since the ratio P/s is constant, hence we can say that the perimeter of a square is directly proportional to the length of a side. Hence the constant of proportionality for the perimeter of a square and length of it's side is 4.08

Step-by-step explanation:

where's the question and picture

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The slope of the line below is 2. Which of the following is the the point-slope form of the line?

Jobisdone [24]

Answer: y + 1 = 2 (x -1)

Step-by-step explanation:

Point slope form is:

y - y1 = m (x - x1)

Point on the line is (1, -1) and slope is 2

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y1 = -1

x1 = 1

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5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B2%7D-1%7D%7B2-5a%7D%20times%20%5Cfrac%7B15a-6%7D%7Ba%5E%7B2%7D%2B5a-6%7D" id=

stepan [7]

$\bf \cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\\\\
-----------------------------\\\\
recall\quad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
thus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)
\\\\\\
now\quad a^2+5a-6\implies (a+6)(a-1)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}\times \cfrac{3(5a-2)}{(a+6)(a-1)}\\\\
-----------------------------\\\\
$

$\bf now\quad 3(5a-2) \iff -3(2-5a)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}\times \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}$

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3 years ago

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vlabodo [156]

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3 years ago

Read 2 more answers

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Marianna [84]

Answer:

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