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Mathematics 10 Online
OpenStudy (anonymous):

Solve the variations: 1. If (a) varies inversely as (b), and a=9 when b=4, find a when b=12. 2. What is the value of the constant in the variation in problem 31? 3. x varies directly as y and inversely as z. If x=1 when y=2 and z=6, then what is the value of y when x=12 and z=1and 1/2? 4. What is the value of the constant in the variation in problem 33? How in the world do i do these? Please help.

jimthompson5910 (jim_thompson5910):

# 1 If a varies inversely as b, then a = k/b for some constant k We know that when a = 9, b = 4, so plug these in to get 9 = k/4 From here, solve for k 9 = k/4 9*4 = k 36 = k k = 36 So the equation is a = 36/b Now onto finding the value of a when b = 12. So just plug in b = 12 and simplify a = 36/b a = 36/12 a = 3 So when b = 12, a = 3

jimthompson5910 (jim_thompson5910):

# 2 I don't see problem #31, so I can't answer it

OpenStudy (anonymous):

Number 31? huh?

jimthompson5910 (jim_thompson5910):

# 3 "x varies directly as y and inversely as z" means that x = ky/z We're given x=1, y=2 and z=6, so 1 = k*2/6 1 = 2k/6 1 = k/3 1*3 = k k = 3 Giving the equation x = 3y/z if x=12 and z = 1 1/2, then x = 3y/z 12 = 3y/(3/2) .... Note: 1 and 1/2 = 3/2 12*(3/2) = 3y 36/2 = 3y 18 = 3y 18/3 = y 6 = y y = 6 So when x = 12 and z = 1 1/2, y = 6

jimthompson5910 (jim_thompson5910):

In #2, it says "2. What is the value of the constant in the variation in problem 31?" but I don't see #31

OpenStudy (anonymous):

sorry. typo. its talking about number 1

OpenStudy (anonymous):

& number 4 at the end is suppose to say number 3.

jimthompson5910 (jim_thompson5910):

oh that makes a ton more sense now

OpenStudy (anonymous):

Ha. sorry.

jimthompson5910 (jim_thompson5910):

# 2 the constant of variation is k = 36

jimthompson5910 (jim_thompson5910):

# 4 the constant of variation is k = 3

OpenStudy (anonymous):

Thank you so much. I have one more question and ill let you be. x^2-5x+6 is (x-3) a factor?

jimthompson5910 (jim_thompson5910):

yes, another factor is x-2

OpenStudy (anonymous):

Oh ok. Thank you so much. For all your help. I have so many more questions. I have finals next week. But i dont wanna ask too many tonight.

jimthompson5910 (jim_thompson5910):

alright, you're welcome, good luck on your finals

OpenStudy (anonymous):

Thank you!

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