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.
Choose one answer.
a. 260
b. 290
c. 320
d. 360

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.
Choose one answer.
	 a. 260	
	 b. 290	
	 c. 320	
	 d. 360

anonymous · Answer

oh my im so stuck on this question!

anonymous · Answer

a

anonymous · Answer

x went up by 5/3 and n2 went up by 100/16. 5/3 * 100/16 * 25 = 260.42

anonymous · Answer

so...$$y \alpha x ; y \alpha n^2$$  therefore the relation is linear in x

anonymous · Answer

Let K be a constant in $y=k x n^2$.

25 = k 3 16
k = 25/(3 *16)

so, $$y = \dfrac{25}{3*16} * 5 * 100$$

anonymous · Answer

a...using linearity i.e. y/(x*n^2) = const