geometry please
Question?
okay @Cacciatore_J
find the length of each hypotenuse such that knm=lmn
I HAVE a quiz will you help me with them i need to grduate
So, the hypotenuse is the side opposite the right angle, right? SO, in this case, 8x-7 is equal to 5x+3y
okay? so i need to solve it now?
This is a bit tricky. I can straight up give you the answers if you want b/c it's hard to write it all out on the computer.
The difficult bit is that there are two variables, in addition to the unknown side MN that is shared by both
i know i need help please god bless u guys
okay 2 of them are fraction a is 14 over 3 and b is 76 over 3 and c. 49 and d, 188 @kevin4762
Again, there are two variables, so while I can assume that those answer choices are the length of the sides, I can't figure out where they might go. Is there any other information that you have yet to mention? Because otherwise, I'd start by plugging and chugging those numbers into the place of x
is it b?
and no there isnt
|dw:1410586749833:dw|
i really dont understand can you please give me the answer i am being timed and have alot of questions
I don't know if you can find a definite answer. There are too may unknown variables. It's either a or b,likely
>
??
Because, by equating the two hypotenuse equations, it becomes \[x-7/3=y\]
idk:'''(
Neither do I :'(
what would be your guess
a, but only because it's smaller and over 3
ok
Which of the following statements would not provide enough information to prove YAZ=AYB using the HL theorem?
THats the next one
Okay. Give me the answer choices, and I'll help you wade through them
okay there wierd so wait ill draw them out
Can yo snip a picture and post that? It might be easier
thats it
D, then, because that one solely relates to one triangle, whereas the other choices relate to both triangles
yay will you hlep with more pleaseeeeeeee <3
Sure
these questions are werid i dont know how to show
???
Bink, post your questions
Kevin, what do you need?
i dont know howww
Do you have a snipping tool on your device? Then take pictures and place them here
Well look. This is my train of thought for the first problem. I assumed that the triangle was a 3,4,5 triangle. So what I did was set the hypotenuse 8x-7 equal to 5 and the first leg, 3x-y, equal to 3. I found that both x and y is 3/2. For the second triangle I did the same and found that x = 0 and y = 5/2. I'm not sure where to go after this.
????
Is that a safe assumption though? And besides, that wasn't one of the answer choices
You don't know anything about either triangle,only that MN is equal for both triangles and that the two hypotenuse are equal to make the triangle the same. If you assume that they're three 3-4-5 triangles you could be very wrong. We only know that they are right and congruent
Yeah what I'm trying to figure out is how to relate it to each other so that I get one final result.
Personally, I don't think you can solve it, the first one. There are too many variables and not enough information. Bink, are you on a computer?
Alright I got the answer. So what I did was I set the legs equal to each other so that x+2y = 3x-y and I got 2x-3y. Then I did the same for the hypotenuse and got 5x+3 = 8x-7 and I got 3x-10. Then I set them up as a system of equations and got y = 20/9 and x = 10/3. but none of those are the answer choices. I mean if you plug in 10/3 for x you get the same hypotenuse so I'm stumped as to how it can't be 10/3.
and when you plug in 10/3 for x you get 59/3.
so yeah if the hypotenuse isn't 59/3 then I dunno what to tell you because it makes sense.
unless I have to square it.
darn I made a mistake. it's 5x + 3y.
okay so then 3x-3y-7 = 2x-3y and I got x = 7 and y = 14/3. The answer is C. LOL wow sorry it took me so long it really wasn't complicated at all.
@Miracrown will you help me with geomtry
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