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Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.
\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}
​

Step 1
Step 2
Step 3
Step 4
Step 5
​

93
×51
​

3
90
150
+4,500
​

4,743
​

1×3 ones
1×9 tens
50×3 ones
50×9 tens
​

Respuesta :

abidemiokin

Since molly's solution tally's with the given solution, hence Molly's solution is correct.

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

[tex]\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}​[/tex]

This partial product can also be solved as shown below:

[tex]93 \times 51 = (90+3)\times (50+1)[/tex]

Applying the distributive law:

[tex]93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743[/tex]

Since molly's solution tally's with the given solution, hence Molly solution is correct.

Learn more about partial product at: https://brainly.com/question/24716925