I have an equation in the format of y=ax^2
y=38
x=360
I have it set up as follows: 38=a(180)^2
I am stuck at 38/32400=19/16200 is this the lowest?
Then i get 19/16200x^2
Is this correct?

Question

I have an equation in the format of y=ax^2
y=38
x=360
I have it set up as follows: 38=a(180)^2
I am stuck at 38/32400=19/16200 is this the lowest?
Then i get 19/16200x^2
Is this correct?

anonymous · Answer

y=ax^2
38=a times 360^2

anonymous · Answer

can you figure the rest out?

anonymous · Answer

38= a times 129600

anonymous · Answer

if you want to find a, a= 38/129600

anonymous · Answer

hasteeerfani00 I must have multiplied wrong

anonymous · Answer

Would it not be 180^2 which is 32400?

anonymous · Answer

180^2 means 180 squared. so it would be 32400

anonymous · Answer

the question says x=360

anonymous · Answer

yes and  38/32400 gives me a decimal I think it has to be in fraction form so I can put it like this 1/675x^2 (different example)

anonymous · Answer

i can reduce it to 19/16200 but thats it.  Is this the answer? 19/16200x^2

anonymous · Answer

well i beileve that you would have to have plugged x in

anonymous · Answer

what is this for? geometry, algebra etc

anonymous · Answer

algebra

anonymous · Answer

one or two

anonymous · Answer

I had another problem and i did nto get decimals. it was 48=a(180)^2
Then I got 48/32400=1/675
Therefore I got y=1/675x^2
But thid problem is giving me a decimal

anonymous · Answer

college algebra