The lengths of two sides of a triangle are 10 inches and 4 inches. Which of the following dimensions is the third side of this triangle? 4 inches 5 inches 6 inches 7 inches
@butterflydreamer
do you know the angles of the triangle?
do u go to centions 8grade
do u go to conenction s8grade
hmm.. is there any other bits of information? :O Like a diagram perhaps?
no im in 7th going to be in 8th on flvs
let me check
depends what triangles we are talking about...
no @butterflydreamer & @Chiko_1278 idek
take a screenshot and post it....
seriously.. O_O...? No other information at all? Can you check to make sure you typed up the question correctly? :)
1.cant do screenshot on my computer 2.im positive i typed it correctly @butterflydreamer & @Chiko_1278
Then i don't think this question can be solved... :S... The sides are 10 and 4 ?
yes 10 and 4 inches i just need to figure out the third side
Is there an angle perhaps?
im guessing we have to do pythagorean theorem
idk it just says a triangle! and no
even by using pythagorean theorem, we don't end up with a nice number/ any of the solutions provided... @Chiko_1278 :S
what i think is that C=10 and A=4...we just have to find the B
yes we do
@butterflydreamer
0.0
let me work it out
ok....
ok chiko
i have another question
nvm is not pytho.....what ever
lol
i got B=√84
how?
lol yeah, i tried pythagoras as well.. :/ No nice numbers in the end so i don't think there isn't enough information to solve your question @lovesvd10 :(
ok thats fine
The figure below shows a shaded circular region inside a larger circle: A shaded circle is shown inside another larger circle. The radius of the smaller circle is labeled as r and the radius of the larger circle is labeled as R. On the right side of the image is written r equal to 2 inches and below r equal to 2 inches is written R equal to 5 inches. What is the probability that a point chosen inside the larger circle is not in the shaded region? 84% 50% 42% 16%
need the pic??????
@Chiko_1278 & @butterflydreamer
where did @butterflydreamer go?
idk
ohhh ok let me check the question......
ok
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