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faltersainse [42]
3 years ago
8

A baker made two cakes of the same size. • At the end of the day, there was 23 of a chocolate cake left. • There was 56 of a str

awberry cake left. • The baker divided the remaining chocolate cake into 2 equal pieces and the remaining strawberry cake into 3 equal pieces. Which cake flavor had larger pieces and by how much? A chocolate by 16 of a cake B strawberry by 16 of a cake C chocolate by 118 of a cake D strawberry by 118 of a cake
asap show your work
Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0
This is the concept of algebra, To solve the question  we proceed as follows;
Number of chocolate cake left=23 
The number of pieces that was left after the cake was divided into 2 equal pieces will be:
23*2
=46

Number of strawberry cake left = 56
The number of pieces that was left after the cake was divided into 3 equal pieces will be:
56*3
=168

Comparing the two fraction, the flavor that had larger pieces was strawberry cake;
This cake was larger compared to chocolate by:
168-46
=122
The answer is Strawberry by 122 of a cake
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Given:

The expressions are

(c) \left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3

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(c)

We have,

\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3

We know that, zero to the power of a non-zero number is always 1. So, \left(\dfrac{2^4\times 3^6}{12^2}\right)^0=1

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(d)

We have,

\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}

It can be written as

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\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13\times 13\times 13}{(65\times 49)(65\times 49)}

\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{(5\times 49)(5\times 49)}

\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}

\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}

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