DIRECT VARIATION
1. a is directly proportional to b.If a=15,then b=9
2. m varies directly as n.If n=2/3,then m=1/4
3. T varies as the square root of L. If L=81,then T=10
4.r varies directly as the square of t. If t=16,then r=6.
5. y varies as the cube foot of x. If x=27,then y=2.
6. y is directly proportional to x. If y=15 when x=4,find y when x=5.
7. m varies directly as n. If m= -6 when n=2,find m when n=-3
8. The square of x varies directly as the cube of y. When x=2,y=4,find y when x=8
What do you want for #1 - 5 ?
Are you finding the constant of variation or what?
I will do one of them, #4
r varies directly as the square of t. ---> r = kt^2
when t=16, r=6
6 = k(16)^2
6 = 256k
k = 6/256 = 3/128
so r = (3/128)t^2
#8.
The square of x varies directly as the cube of y. ---> x^2 = ky^3
when x=2, y=4
4 = k(64)
k = 4/64 = 1/16
so x^2 = (1/16)y^3
when x = 8
64 = (1/16)y^3
y^3 = 1024
y = 1024^(1/3) = 8cuberoot(2)
Do the others the same way.
let me know what you got for the rest.
a is directly proportional to b.If a=15,then b=9
I DON'T KNOW THE ANSWER PLEASE TELL OF
To solve these direct variation problems, you need to use the concept of direct variation. In direct variation, two variables are said to vary directly if their ratio remains constant. Mathematically, this relationship can be expressed as:
y = kx
Where:
- y and x are the variables that vary directly.
- k is the constant of variation.
Let's solve each problem step by step:
1. a is directly proportional to b. If a = 15, then b = 9.
To find the constant of variation (k), we can divide a by b: k = a/b = 15/9 = 5/3.
Now, we can find b when a = 15: b = k * a = (5/3) * 15 = 25.
2. m varies directly as n. If n = 2/3, then m = 1/4.
Using the same formula, we can find the constant of variation:
k = m/n = (1/4) / (2/3) = (1/4) * (3/2) = 3/8.
Now, we can find m when n = 2/3: m = k * n = (3/8) * (2/3) = 1/4.
3. T varies as the square root of L. If L = 81, then T = 10.
In this case, the relationship is T = k*√L.
To find the constant of variation, we square both sides: T^2 = k*L.
Now, substitute the given values: (10)^2 = k * 81.
Simplifying, we get k = (10^2)/81 = 100/81.
To find T when L = 81, T = k * √L = (100/81) * √81 = 10.
4. r varies directly as the square of t. If t = 16, then r = 6.
The relationship is r = kt^2.
To find the constant of variation, we divide r by t^2: k = r / (t^2) = 6 / (16^2) = 6/256 = 3/128.
Now, we can find r when t = 16: r = k * t^2 = (3/128) * (16^2) = 6.
5. y varies as the cube foot of x. If x = 27, then y = 2.
The relationship can be written as y = k * x^(1/3).
To find the constant of variation, we divide y by x^(1/3): k = y / (x^(1/3)) = 2 / (27^(1/3)).
Simplifying, k = 2 / 3.
Now, to find y when x = 27: y = k * x^(1/3) = (2/3) * (27^(1/3)) = 2.
6. y is directly proportional to x. If y = 15 when x = 4, find y when x = 5.
The relationship is y = k * x.
To find the constant of variation, divide y by x: k = y / x = 15 / 4.
Now, we can find y when x = 5: y = k * x = (15/4) * 5 = 75/4.
7. m varies directly as n. If m = -6 when n = 2, find m when n = -3.
The relationship is m = k * n.
To find the constant of variation, divide m by n: k = m / n = -6 / 2 = -3.
Now, we can find m when n = -3: m = k * n = -3 * (-3) = 9.
8. The square of x varies directly as the cube of y. When x = 2, y = 4, find y when x = 8.
The relationship is (x^2) = k * (y^3).
To find the constant of variation, divide (x^2) by (y^3): k = (x^2) / (y^3) = (2^2) / (4^3) = 4/64 = 1/16.
Now, we can find y when x = 8: (8^2) = (1/16) * (y^3).
Simplifying, 64 = y^3/16.
Multiplying both sides by 16, 1024 = y^3.
Taking the cube root of both sides, y = ∛1024 = 8.
By using the concept of direct variation and the given information, we were able to find the specific values of the variables in each of the problems.