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2 years ago

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Factor out the coefficient of the variable 5/6s+2/3

1 answer:

DochEvi [55]2 years ago

6 0

Okay. so I'm gonna assume you know what factor out means--were gonna take out 5/6. what we need to know is, how does that affect our ending number 2/3? well, we're gonna have 5/6 times some stuff first off, right?

5/6 ( )

so what's in the parenthesis? well, if we distribute in our mind, S would just have to be by itself for it to work out. so we have

5/6 ( s )

but what about 2/3? well, if we're multiplying everything on the inside by 5/6, how can be get it to cancel and leave 2/3 alone? we could divide 2/3 by 5/6, right?

5/6 ( s + (2/3) / (5/6) )

now, when we distribute our 5/6, we'll have just 2/3 at the end. now let's simplify that end term so it looks a little better. if we divide by a fracion, that's the same as multiplying by the flipped version, so let's do that.

2/3 / 5/6 = 2/3 × 6/5 = 2 (6) / 5 (3) = 12/15

we can reduce 12/15 by dividing the top and bottom by three.

= 4/5

so our final answer would be:

5/6 [S + 4/5] you can redistribute to make sure it matches. Anyway, hope that helps!

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2 years ago

Among all right triangles whose hypotenuse has a length of 12 cm, what is the largest possible perimeter?

Veronika [31]

Answer:

Largest perimeter of the triangle =

$P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)$

Step-by-step explanation:

We are given the following information in the question:

Right triangles whose hypotenuse has a length of 12 cm.

Let x and y be the other two sides of the triangle.

Then, by Pythagoras theorem:

$x^2 + y^2 = (12)^2 = 144\\y^2 = 144-x^2\\y = \sqrt{144-x^2}$

Perimeter of Triangle = Side 1 + Side 2 + Hypotenuse.

$P(x) = x + \sqrt{144-x^2} + 12$

where P(x) is a function of the perimeter of the triangle.

First, we differentiate P(x) with respect to x, to get,

$\frac{d(P(x))}{dx} = \frac{d(x + \sqrt{144-x^2} + 12)}{dx} = 1-\displaystyle\frac{x}{\sqrt{144-x^2}}$

Equating the first derivative to zero, we get,

$\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0$

Solving, we get,

$1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0\\\\x = \sqrt{144-x^2}}\\\\x^2 = 144-x^2\\\\x = \sqrt{72} = 6\sqrt{2}$

Again differentiation P(x), with respect to x, using the quotient rule of differentiation.

$\frac{d^2(P(x))}{dx^2} = \displaystyle\frac{-(144-x^2)^{\frac{3}{2}}-x^2}{(144-x)^{\frac{3}{2}}}$

At x = $6\sqrt{2}$ ,

$\frac{d^2(V(x))}{dx^2} < 0$

Then, by double derivative test, the maxima occurs at x = $6\sqrt{2}$

Thus, maxima occurs at x = $6\sqrt{2}$ for P(x).

Thus, largest perimeter of the triangle =

$P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)$

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