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Virty [35]
3 years ago
6

Tyler went to the supermarket to buy food for a food pantry. He has $36, and can carry up to 20 pounds of food in his backpack.

Mathematics
1 answer:
krok68 [10]3 years ago
7 0

Answer:

option A: x + 1.5y ≤ 20

option C: x + 3y ≤ 36

option D: x ≥ 0

option H: y ≥ 0

Step-by-step explanation:

It says to assume x = the number of packages of pasta and y = the number of jars of pasta sauce.

Since the number of items can not be a negative number, we have x ≥ 0 and y ≥ 0.

Condition: He has $36, and can carry up to 20 pounds of food in his backpack.

Pasta costs $1 for a 1-pound package. Pasta sauce costs $3 for a 1.5 pound jar.

If he buys 'x' packs of pasta and 'y' jars of pasta sauce, then:-

Total cost = 1x + 3y ≤ 36 dollars.

and Total weight = 1x + 1.5y ≤ 20 pounds.

Hence, we have four inequalities:-

option A: x + 1.5y ≤ 20

option C: x + 3y ≤ 36

option D: x ≥ 0

option H: y ≥ 0

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Answer:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            2x-2& \quad x \geq 5/2 \\            -2x+8 & \quad x < 5/2         \end{array}    \right.

Step-by-step explanation:

We are given the function:

f(x)=3+|2x-5|

Remember that by the definition of absolute value:

\displaystyle |x|= \left\{        \begin{array}{ll}            x & \quad x \geq 0 \\        -    x & \quad x < 0        \end{array}    \right.

Our absolute value is:

|2x-5|

First, we will find when it becomes 0. Set the equation equal to 0:

2x-5=0

Solve for <em>x: </em>

<em />x=5/2<em />

<em />

So, we can see that for all values greater than <em>x </em>= 5/2, 2x - 5 is positive.

For all values less than <em>x </em>= 5/2, 2x - 5 is negative.

Therefore (the positive case go above, and the negative case go below):

|2x-5|= \left\{        \begin{array}{ll}            2x-5 & \quad x \geq 5/2 \\           -(2x-5) & \quad x < 5/2         \end{array}    \right.

Finally, we can add a three:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            3+(2x-5) & \quad x \geq 5/2 \\            3+(-(2x-5)) & \quad x < 5/2        \end{array}    \right.

Simplify if desired:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            2x-2& \quad x \geq 5/2 \\            -2x+8 & \quad x < 5/2         \end{array}    \right.

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