the pressure (p) that a 160-pound person's shoe exerts on the ground when walking varies inversely with area (a) of the sole of the shoe. when the shoes have a sole area of 40 sq. in., the pressure is 4 pounds per sq. in. what is the sole area of the sole when the pressure is 10 pounds per sq. in. ?
To solve this problem, we can use the inverse variation formula:
p = k/a
where p is the pressure, a is the area, and k is the constant of variation.
Given that when the sole area is 40 sq. in., the pressure is 4 pounds per sq. in., we can plug in these values into the formula:
4 = k/40
To determine the value of k, we can multiply both sides of the equation by 40:
4 * 40 = k
k = 160
Now that we have the constant of variation, we can find the sole area (a) when the pressure is 10 pounds per sq. in.:
10 = 160/a
To find the area, we can multiply both sides of the equation by a:
10a = 160
Finally, to isolate a, we divide both sides of the equation by 10:
a = 160/10
a = 16
Therefore, the sole area of the shoe when the pressure is 10 pounds per sq. in. is 16 sq. in.
To solve this problem, we can use the inverse variation equation, which states that the product of the two variables (pressure and area) is constant.
Let's define the variables first:
- pressure (p): the pressure exerted by the shoe on the ground (in pounds per square inch, psi)
- area (a): the area of the sole of the shoe (in square inches)
According to the given information, when the area is 40 sq. in., the pressure is 4 psi. We can write this as an equation:
p * a = k
where k is the constant of variation. We can find the value of k by substituting the given values:
4 * 40 = k
k = 160
Now we have the value of k, so we can use it to find the sole area when the pressure is 10 psi. Let's denote the unknown sole area as x:
10 * x = 160
To solve for x, we can divide both sides of the equation by 10:
x = 160/10
x = 16
Therefore, when the pressure is 10 psi, the sole area of the shoe is 16 square inches.
3x+y=20 x+3y=12
we know that p*a = k
also, 4*40 = k = 160
so, 10*a = 160
. . .