Abby can buy an 8 pound bag of dog food for $7.40 or a 4 pound bag of the same dog food for $5.38. Which is the better buy?

Mathematics

2 answers:

Lesechka [4]3 years ago

8 0

Find how much cost one pound of dog food.

If Abby can buy an 8 pound bag of dog food for $7.40, then one pound costs $7.40:8=$0.925;
If Abby can buy a 4 pound bag of dog food for $5.38, then one pound costs $5.38:4=$1.345.

As you can see $0.925<$1.345, then the better buy is an 8 pound bag of dog food for $7.40, because the price for one pound is less.

OLEGan [10]3 years ago

4 0

One way to find out which bag is better to buy is to find out how much is 1 pound worth in both situations. So we have to divide the prices of 8 pound bag by 8 and 4 pound bag by 4:

$7.40 ÷ 8 = $0.925 per pound
$5.38 ÷ 4 = $1.345 per pound

1.345 > 0.925, so it's better to buy 8-pound bag

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I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.