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Mathematics 17 Online
OpenStudy (anonymous):

medals!!!!! A balloon is released and flies straight up for 6 miles, and then the wind blows it west for 4 miles. Find the distance between the balloon and the site where it was released.

OpenStudy (imstuck):

You will set this up as right triangle problem and what you are solving for is the hypotenuse of the triangle.

OpenStudy (imstuck):

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OpenStudy (anonymous):

ok its a sq root answer

OpenStudy (superhelp101):

yah it is a Pythagorean problem

OpenStudy (superhelp101):

use the formula a^2 + b^2 = c^2

OpenStudy (anonymous):

52

OpenStudy (superhelp101):

so 4^2 + 6^2 = 16 + 36=52 great job!

OpenStudy (anonymous):

but its a sq root answer so 2 sq root 5?

OpenStudy (superhelp101):

what are the other answer choices?

OpenStudy (anonymous):

3 sq root 7

OpenStudy (anonymous):

2 sq root 3

OpenStudy (anonymous):

and 2 sq root 13

OpenStudy (anonymous):

\[A^2+B^2=C^2\] So if C is the Hypotenuse then:\[C = \sqrt{A^2+B^2}\]

OpenStudy (superhelp101):

then @alyygirl you have to do the square root of 52

OpenStudy (anonymous):

So you really have \[C = \sqrt{52}\]

OpenStudy (anonymous):

no 2 sq root 13

OpenStudy (anonymous):

You can break up the inside square root into 4 and 13 because 4 times 13 = 52 and you will have:\[\sqrt{4 \times 13}\] Since 4 is a perfect square of the number 2 and it is multiplied by 13 we can obey the rules of square roots and take the square root of 4 out of the square root sign and multiply it by only square root of 13 like so:\[\sqrt{4 \times 13} = \sqrt{2^2 \times 13} = 2\sqrt{13}\]

OpenStudy (superhelp101):

yep good job! @Johnbc :)

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