Serena is making a large meatloaf that contains ground beef and ground pork. the beef is $3.50 per pound and the pork is $2.00 per pound. if she buys 0.75 pounds miri pork than beef and spends 15.25 in total , algebraically model this problem and determine how many pounds of each type of meat was bought.

Hi there! :)

Serena bought 0.75 pounds of pork and about 3.93 pounds of beef.

Step-by-step explanation:

Lets start by laying out the important information we know:

- Beef = $3.50/pound → 1 pound of beef = $3.50

- Pork = $2.00/pound → 1 pound of pork = $2.00

- Buys 0.75 pounds of pork → buys 0.75 of 1 pound of pork

- Buys x pounds of beef

- Spends $15.25 in total → 0.75 pounds of pork + x pounds of beef = $15.25

Once this is done, it's much easier to see what we are dealing with (when it's just a bunch of words it's a lot harder to understand).

Creating your algebraic equation:

Algebraically, your equation needs to be equal to the total amount that Serena spent on pork and beef.

She bought 0.75 of 1 pound of pork, which in other words means that she bought 0.75 of $2.00. The word "of" is the same thing as a multiplication sign. SO, all of this would translate into this: (0.75 × 2)

She then bought beef, but we don't know how much. What we can do is replace the "amount of pounds of beef she bought" with the letter "x". Remember, 1 pound of beef is $3.50. SO, all of this would translate into this: 3.50x

Your algebraic equation should look like this:

(0.75 × 2) + 3.50x = 15.25

Now all you need to do is solve this equation by isolating "x":

(0.75 × 2) + 3.50x = 15.25

1.5 + 3.50x = 15.25

Subtract 1.5 from each side of the equation → 15.25 - 1.5 = 13.75

3.50x = 13.75

Divide each side of the equation by 3.50 → 13.75 ÷ 3.50 = 3.92857143...

x ≈ 3.93

There you go! I really hope this helped, if there's anything just let me know! :)

answer: 7270

step-by-step explanation:

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.