A pizza shop makes large pizzas with a target diameter of 16 inches. A pizza is acceptable if its diameter of a pizza is within 3 (times) 2 (to the power of) -2 in of the target diameter. Let d represent the diameter of a pizza. Write an inequality for the range of acceptable large pizza diameters in inches. (Where do I even start?)

Question

anonymous · Answer

$$
|d_{target}-d_{pizza}|\leq 3\cdot2^{-2}
$$

anonymous · Answer

In this case $d=d_{pizza}$

anonymous · Answer

While $d_{target} = 16$

anonymous · Answer

So is the final equation 
$$d \le3\times2^{-2}$$

anonymous · Answer

Nope...

anonymous · Answer

You have $$
|16-d|\leq 3\times 2^{-2}
$$

anonymous · Answer

I mean shoot. One sec.

anonymous · Answer

$$d \le3\times1/4$$

anonymous · Answer

Is it that? (This is zero and negative exponents is the lesson)

anonymous · Answer

You can't ignoree the $|16-d|$

anonymous · Answer

But you got the exponent right.

anonymous · Answer

How come it is  |16−d| What does it mean? Why is it in absolut value?

anonymous · Answer

It's in the absolute value because we want the difference to be positive.

anonymous · Answer

Because we need it to 16 or less. Right?

anonymous · Answer

For example, if $d=18$ we don't want $16-18=-2$,  we want $|16-18|=|-2|=2$

anonymous · Answer

But it has to be within the targeted area of 16. Going over 16 wouldn't work.

anonymous · Answer

$$
| \dots |\leq a \quad\to\quad  -a\leq \dots \leq a
$$

anonymous · Answer

That doesn't seem like it would be an algerbra lesson. I'm in algebra right now.

anonymous · Answer

Well within could mean  |dw:1362004378487:dw|