What’s he hypotenuse, x, if ( 20, 21,x) is a Pythagorean triple?

Mathematics

1 answer:

ololo11 [35]3 years ago

7 0

If x is the hypotenuse, that means that sqrt(20^2+21^2)=x=29 since sqrt(a^2+b^2)=c or a^2+b^2=c^2

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Evaluate 2xsquared by 2 − 5y over 3x −y when x = 2 and y = -4.

Arisa [49]

(2x^2 -5y)/(3x-y)
((2*2)^2-(5*-4))/((3*-4)-(-4))
(4^2+20)/(-12+4)
36/-8
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2 years ago

F to the second time F to the Third

miskamm [114]

Answer:

$F^{5}$

Step-by-step explanation:

$F^2 \times F^3 =$

$= F^{2+3}$

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4 0

3 years ago

Read 2 more answers

One of our brainliest, Konrad509, made this:

Reil [10]

$\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x$

A=10, B=11, C=12, etc.

$\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2$

$121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0$

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that $x\in\mathbb{N}$ . Moreover we can deduce that $x\geq19$ since there is $J$ and $J=19$ .