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anonymous
 one year ago
Can some one please help me step by step to solve these two problems? Screen shots included. Medal!
anonymous
 one year ago
Can some one please help me step by step to solve these two problems? Screen shots included. Medal!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure... I may get the question wrong. You should probably use this formula A=1/2 BH

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is my first time doing this I need like a step by step on how to do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, I'm sorry.. I dont get it either lol really wish I did.. but it's confusing

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1Use the Pythagorean Theorem \[a^2 + b^2 = c^2 \]

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0use pythagorss theorem

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1For question 1 we know that a = 16 and b = 63 so let's plug that into the formula to find c (hypotenuse) \[16^2 + 63^2 = c^2 \]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand how to find the answer to this problem?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1Good and then find the aquaria root of 4,225 because when c is being squared, it equals to 4225.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you can you wait until I do the second problem so i can see if its right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would the answer be 3.5805.?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1Let's check! \[a^2 + 2.1^2 = 2.9^2\] \[a^2 = 4\] (because 2.9^2  2.1^2) \[a = \sqrt4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay! I just plugged it in wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! I'll contact you if i need help with anything else

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1Sure! and You're welcome :)
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