If y varies inversely as the square of x and y = 8 when x = 2, then what is the value of y when x = -1?

Question

Eva Abramson · Accepted Answer

undefined To find the value of y when x = -1, we can use the inverse variation formula.

The inverse variation formula states that if y varies inversely as the square of x, then y = k/x^2, where k is a constant.

To find the value of k, we can substitute the given values of y and x into the formula.

When y = 8 and x = 2, we have 8 = k/2^2.

Simplifying this equation, we get 8 = k/4.

To solve for k, we can multiply both sides of the equation by 4, giving us 32 = k.

Now that we have the value of k, we can substitute it back into the inverse variation formula to find the value of y when x = -1.

Using y = k/x^2 and substituting k = 32 and x = -1, we get y = 32/(-1)^2.

Simplifying this equation, we have y = 32/1, which means y = 32.

Therefore, when x = -1, the value of y is 32.

Ethan Adkins · Answer

The answer is #32#, based on an inverse proportionality model.

According to the statement, #y prop 1/x^2# This expression takes us to the exact formula: #y = alpha / x^2# where #alpha# is a constant of proportionality. To find it, we need a numerical value: #y(x=2) = 8#. By substituting: #8 = alpha / 2^2 rightarrow alpha = 32# So our formula is: #y = 32 / x^2# Now, let us calculate #y(x=-1)#: #y = 32 / (-1)^2 = 32#

Hazel Anglin · Answer

undefined If $y$ varies inversely as the square of $x$, we can express this relationship using the formula:

$$y = \frac{k}{x^2}$$

Where $k$ is a constant of proportionality.

Given that $y = 8$ when $x = 2$, we can substitute these values into the equation to find the value of $k$:

$$8 = \frac{k}{2^2}$$

$$8 = \frac{k}{4}$$

$$k = 8 \times 4 = 32$$

Now that we know $k = 32$, we can find $y$ when $x = -1$ by substituting this value into the equation:

$$y = \frac{32}{(-1)^2}$$

$$y = \frac{32}{1}$$

$$y = 32$$

So, when $x = -1$, $y$ has a value of $32$.