First: solve for k. Then, go back to the original equation c=k(d-3) and solve for c when d is ten.
Second: Area=k*s^2
Put in the fist conditions, solve for k.
Then, knowing k, solve for A when is eight.
direct variation
1)if c varies directly as d-3, and if c=16 when d=7, find c when d=10
c=k(d-3) then 16=k(4)...how do u do this?
2)the surface area of a cube varies directly as the square of the length of one edge. if the surface area of a cube is 150 cm squared when the length of an edge is 5 cm, what is the surface area of a cube with an edge of length 8 cm?
3x4=3x-3
To solve for k in the first equation, you can substitute the given values into the equation and solve for k.
1) Given: c = 16, d = 7
Substitute these values into the equation c = k(d-3):
16 = k(7-3)
16 = k(4)
Now, to solve for k, divide both sides of the equation by 4:
k = 16/4
k = 4
So, the value of k is 4.
Next, use the value of k to find c when d is 10.
c = k(d-3)
Substitute k = 4 and d = 10 into the equation:
c = 4(10-3)
c = 4(7)
c = 28
Therefore, when d = 10, c is 28.
Now let's move on to the second problem:
2) Given: Area (A) = 150 cm^2, length of one edge (s) = 5 cm
The equation representing the direct variation relationship is A = k*s^2.
Substitute the given values into the equation:
150 = k*(5^2)
150 = k*(25)
To solve for k, divide both sides of the equation by 25:
k = 150/25
k = 6
So, the value of k is 6.
Now, use the value of k to find A when s is 8 cm.
A = k*s^2
Substitute k = 6 and s = 8 into the equation:
A = 6*(8^2)
A = 6*(64)
A = 384
Therefore, when the length of an edge is 8 cm, the surface area of the cube is 384 cm^2.