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-BARSIC- [3]
3 years ago
12

Jessica is baking a cake. The recipe says that she has to mix 96 grams of sugar into the flour. Jessica knows that 1 cup of this

particular sugar has a mass of 128 grams. She added one over two of a cup of sugar to the flour. Should Jessica add more sugar to make the exact recipe, or did she go over and by what amount?
She went over by two over three of a cup

She needs to add two over three of a cup

She went over by one over four of a cup

She needs to add one over four of a cup
Mathematics
2 answers:
maria [59]3 years ago
4 0
One/four
thats the answer
Elden [556K]3 years ago
3 0

Hello and thanks for posting your question:)

The question is asking if Jessica gave to much or to little to the cake recipe. The answer is She needs to add 1/4 of a cup.

I hope this answer helps. Have a great day:)

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Step-by-step explanation:

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2 years ago
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inysia [295]

Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

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Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

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Answer:

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Answer:

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Step-by-step explanation:

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