I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$: \begin{array} {}\hfill {}^6{}^439\\ \hfill \times\ 75 \\\hline \hfill {}^1 195 \\ \hfill +\ 273\phantom{0} \\\hline \hfill 2925 \end{array} Since there are two digits to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$.

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!

I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$: \begin{array} {}\hfill {}^6{}^439\\ \hfill \times\ 75 \\\hline \hfill {}^1 195 \\ \hfill +\ 273\phantom{0} \\\hline \hfill 2925 \end{array} Since there are two numbers that are to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$.

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!

I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$. Since there are two digits to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!

I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$: \begin{array} {}\hfill {}^6{}^439\\ \hfill \times\ 75 \\\hline \hfill {}^1 195 \\ \hfill +\ 273\phantom{0} \\\hline \hfill 2925 \end{array} Since there are two digits to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$.

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!

I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$. Since there are two digits to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!

I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different solution methods as possible.

Here is what I have come up with so far:

1. The most common way: multiply the two numbers "vertically", ignoring the decimal, to get $2925$. Since there are two digits to the right of the decimal, place the decimal after the $9$ to get the answer $29.25$

2. Write both numbers as improper fractions: $$3.9= \dfrac{39}{10}$$ and $$7.5=\dfrac{75}{10}$$Then multiply $$\dfrac{39}{10}\cdot\dfrac{75}{10}$$ to get $\dfrac{2925}{100}$ which simplifies to 29.25.

3. Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.

4. Since $3.9$ is very close to $4$, we could instead do $4\cdot7.5=30$ and then subtract $0.1\cdot7.5=0.75$ to get $30 - 0.75=29.25$

5. Similarly, since $7.5$ rounds up to $8$, we can do $3.9\cdot 8=31.2$ and then subtract .$5\cdot 3.9=1.95$ to get $31.2-1.95=29.25$

Are there any other possible methods the students might use? (Note: they are junior college math ed students.) Thanks!