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bogdanovich [222]
3 years ago
15

If a right triangle has equal legs, and the hypotenuse is 26 ft. how long is each leg (leave answer in radical form)

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
5 0

If a right triangle has equal legs, it is by definition a 45-45-90 triangle. So, the two sides would be x and the hypotenuse would x\sqrt{2}. However, we know the hypotenuse, and we do not see a \sqrt{2} anywhere. That means it has already been multiplied by it, so to get the lengths of the sides we simply divide 26 by \sqrt{2}. If you have to rationalize the denominator, the answer will be \frac{26}{\sqrt{2}}  = \frac{26\sqrt{2}}{2}  = 13\sqrt{2}

Marta_Voda [28]3 years ago
5 0

\boxed {\text{pythagorean theorem :}a^2 + b^2 = c^2}

Let the leg be x.

x² + x² = 26²

-

Solve x:

x² + x² = 26²

<u><em>Combine like terms:</em></u>

2x² = 676

<u><em>Divide both sides by 2:</em></u>

x² = 338

<u><em>Square root both sides:</em></u>

x= √338

<u><em>Put it in radical form:</em></u>

x = 13√2

-

Answer: 13√2

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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
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Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

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x-1

  • divides the remaining fish into three.

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  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

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Refer the attached figure for the graphic representation of the given quadratic equation.

<u>Step-by-step explanation:</u>

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By applying the values, the axis of symmetry of given equation is

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The vertex form of quadratic equation is f(x)=a(x-h)^{2}+k

Where, (h,k) are the vertex.

Convert the quadratic equation into vertex form.

By completing the square,

f(x)=x^{2}-8 x+24

f(x)=\left(x^{2}-2(4) x+4^{2}\right)-4^{2}+24

f(x)=(x-4)^{2}+8

On comparison,

(h , k) = (4 , 8)

Now, plot the equation with vertex (4,8) [refer attached figure].

4 0
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