Algebra II question? please help
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Algebra II question? please help

[From: ] [author: ] [Date: 12-02-11] [Hit: ]
k = 25/48, and so y (25/48)(5)(100) = 12500/48.......
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.

thank you!

-
First write an equation:
y = k*x*n^2
- "varies jointly" means that one variable relates to two or more other variables ( y relates to x and n)
- the "and" in a problem that varies jointly means to multiply
- our two terms on opposite y are given by "x and the square of n"
- we know that n is squared

* we're looking for some way to modify x to make the equation true

Pug in your known variables and solve for k:
25 = k * (3) * (4)^2
25 = k * (3) * 16
25/16 = k * (3)
[(5/4)^2]/3 = k
(25/48) = k

Now use our value of k and the given values of x and n to solve for y.

y = k * x * n^2
y = (25/48) * 5 * 10^2
y = (25/48) * 5 * 100
y = (25/48) * 500
y = 3125/12
y = 260.416

Rounded to the nearest whole number: y = 260

-
y = kxn^2
25 = k(3)(16)
k = 25/48, and so y (25/48)(5)(100) = 12500/48. U simplify this
1
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