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alukav5142 [94]
3 years ago
12

Fifteen dozen eggs were needed for baking four wedding cakes. The first cake

Mathematics
1 answer:
Mazyrski [523]3 years ago
7 0

Answer:

96 eggs

Step-by-step explanation:

A dozen is equal to 12 eggs, so 15 dozen is equal to 180 eggs

(Because 15*12 = 180)

We already know how many eggs are required for the 1st cake: 12 eggs.

Then it says "each successive cake needs twice as many eggs as the previos cake".

(Successive means the cake directly after the previous cake)

Here's how we find the number of eggs needed for the 2nd cake:

The 1st cake needed 12 eggs, and because the 2nd cake is directly after the 1st cake, we are going to need two times the amount of 12 eggs.

This equation represents the above scenario:

12*2 = 24

So we need 24 eggs for the 2nd cake.

Now we repeat this process for the 3rd cake, finding twice the amount of eggs from the 2nd cake to find the amount of eggs needed for the 3rd cake:

24*2 = 48

And we repeat it once more for the 4th cake, using the eggs from the 3rd cake:

48*2 = 96

So here's the list of how many eggs are required for each of the cakes:

1st cake: 12

2nd cake: 24

3rd cake: 48

4th cake: 96

If you add all the eggs from each of the cakes, you will get 180, which is the number of eggs needed for all four cakes. So our answer is correct.

Hope this helps (●'◡'●)

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3

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2

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18

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⎥

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The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

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​3

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and

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1

1

8

0

1

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⎥

⎦

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12

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1

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3

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3

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⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

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x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

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1

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⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

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3

→

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⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

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−

y

+

z

=

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y

−

12

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=

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