y squared=( y/2) squared + 15 squared can u help solve for y?
Is the equation:\[y ^{2} = (y/2)^{2} + 15^2\]
If yes, you have y^2 = y^2/4 + 225, or 3y^2 = 900, so y^2 = 300 and y = \[10\sqrt{3}\]
uuumm yes the first one is what i ment, but i dont follow the equation is what i got for triangles this well help me solve all three sides
So, (y/2)^2 = y^2/4.
15^2 is 225.
You have \[ y^2=y^2/4+225\]
So, multiply by 4 and you get 4y^2=y^2+900.
Then, subtract y^2 from both sides for 3y^2=900
no, this is just basic geometry, y squared not^
^ Means to the power of.
So y^2 is "y to the power of 2", or "y squared".
Then what is?
ok um it cant be 900 because its triangle and im trying to find the side lenghts all i have is 15^2 and need to find y
\[y^2=y^2/4+225 \implies 3x^2/4-225=0 \]\[\implies x^2-300=0 \implies x=\pm \sqrt{300}=\pm 10\sqrt3\]
Don't make the problem harder than it is.
OOPS, went from y to x as the variable halfway thru.... you see what I did?
umm im trying state that its not hard its simple and i just need a little help with it i have no idea as to how u got that and i dont think that it correct becuase triangles dont have side lenghts up to 300
\[y^2=y^2/4+225\implies 3x^2/4−225=0\]\[\implies y^2−300=0\implies y=±\sqrt{300}=±10\sqrt3\]If the answer is the side of a triangle, we need it to be positive.
The answer is 10 times the square root of three. That's only about 17.
so if ur saying that y=10\[\sqrt{3}\] then plug that into the equation i gave and tell me that iur right and im wrong.
Well, let's substitute.....\[y^2=y^2/4+225 \implies (10\sqrt3)^2=(10\sqrt3)^2/4+225 \implies \frac{900}{4}=225\]Which is true.
what are u talking i dont understand at all! where did u get y/4 and 225?
When you square y/2, it becomes y^2/4, since both the numerator and denominator are squared.
ok i follow that but what about 225
That is 15^2.
ok
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