Tom, Lisa, and Jane are competing in a cooking competition. They all used different amounts of sugar from a can containing 4 pounds of sugar.


• Tom used 13% of the total sugar in the can.

• Lisa used 0.6 pound of the total sugar in the can.

• Jane used 0.12 of the total sugar in the can.


Who used the greatest amount of sugar from the can? Show your work and explain your answer in words.

Step-by-step explanation:

Tom used: 0.52 I figured it out by doing this: 0.13x4= 0.52

Lisa used: 2.4 I figured it out by doing this: 0.6x4= 2.4

Jane used: 0.48 I figured it out by doing this: 0.12x4= 0.48

so Lisa used the most

hope it helps!

Answer: Lisa used the most

Step-by-step explanation: cause i got right answer

1.Refer to the system of linear equations shown below. Which of the following statements gives the best choice for multiplying to solve this system using the elimination method?
3x−6y=4

6x+11y=−2

(1 point)

Multiply the second equation by 12, so the x-variables are eliminated.

Multiply the first equation by 2, so the x-variables are eliminated.

Multiply the first equation by −11 and the second equation by 6, so the y-variables are eliminated.

Multiply the first equation by −2, so the x-variables are eliminated.
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4x−y=9

x+3y=12

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13

−4

−13

4
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3x+6y=−18

(1 point)

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(3, −2)

(2, −4)

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4.Solve the system by the elimination method.

3x−5y=4

−9x+3y=−24

(1 point)

(8, 4)

(1, −1)

(3, 1)

(2, −2)
5.Manny says he should multiply the first equation in the system of equations below by 3 and the second equation by 2, then add to eliminate x. Is there a more efficient way to solve this system? Explain your answer.

−2x+y=15

3x+4y=−12

(1 point)

No, there isn't a more efficient way to solve this system.

Yes, a more efficient way is to multiply the first equation by 4, add to eliminate y, then solve for x.

Yes, a more efficient way is to multiply the first equation by −4, add to eliminate y, then solve for x.

Yes, a more efficient way is to multiply the first equation by −4, add to eliminate x, then solve for y.

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