Question

y varies jointly with x and the square of n. when y=25, x=3 and n=4. find the value of y when x=5 and n=10. round to the nearest whole number.

Answer 1

\[y = kxn^2 \\ 25 = k(3)(4)^2 \\ 25 = k48 \\ \frac{25}{48} = k\]
So
\[y = (\frac{25}{48})(5)(10)^2 \\ y = (\frac{25}{48})(500) \\ y = \frac{12,500}{48} \\ y = \frac{3,125}{12}\]

Answer 2

thank you soo much, i really appreciate it.