Answer :
Sure! Let’s walk through this step-by-step:
1. Identify the values provided:
- We have [tex]\(\pi = \frac{22}{7}\)[/tex].
- The radius of the base [tex]\( r = 3.5 \text{ m} \)[/tex].
- The height [tex]\( h = 5 \text{ m} \)[/tex].
2. Interpret the given expression:
- The expression [tex]\(2 \times \frac{22}{7} \times 3.5 \text{ m} \times 5 \text{ m}\)[/tex] can be seen as finding a certain value, let's call it [tex]\( \text{result} \)[/tex].
3. Break down the calculation into manageable steps:
- Calculate [tex]\(2 \times \frac{22}{7}\)[/tex]:
[tex]\[ 2 \times \frac{22}{7} = 2 \times 3.142857... \approx 6.285714285714286 \][/tex]
- Multiply the result by the radius (3.5 m):
[tex]\[ 6.285714285714286 \times 3.5 = 22 \text{ m} \][/tex]
- Finally, multiply that result by the height (5 m):
[tex]\[ 22 \times 5 = 110 \text{ m}^2 \][/tex]
4. Conclusion:
[tex]\[ \text{result} = 110 \text{ m}^2 \][/tex]
So, when you multiply [tex]\(2 \times \frac{22}{7} \times 3.5 \text{ m} \times 5 \text{ m}\)[/tex], you get [tex]\(110 \text{ m}^2\)[/tex]. This is the final answer.
1. Identify the values provided:
- We have [tex]\(\pi = \frac{22}{7}\)[/tex].
- The radius of the base [tex]\( r = 3.5 \text{ m} \)[/tex].
- The height [tex]\( h = 5 \text{ m} \)[/tex].
2. Interpret the given expression:
- The expression [tex]\(2 \times \frac{22}{7} \times 3.5 \text{ m} \times 5 \text{ m}\)[/tex] can be seen as finding a certain value, let's call it [tex]\( \text{result} \)[/tex].
3. Break down the calculation into manageable steps:
- Calculate [tex]\(2 \times \frac{22}{7}\)[/tex]:
[tex]\[ 2 \times \frac{22}{7} = 2 \times 3.142857... \approx 6.285714285714286 \][/tex]
- Multiply the result by the radius (3.5 m):
[tex]\[ 6.285714285714286 \times 3.5 = 22 \text{ m} \][/tex]
- Finally, multiply that result by the height (5 m):
[tex]\[ 22 \times 5 = 110 \text{ m}^2 \][/tex]
4. Conclusion:
[tex]\[ \text{result} = 110 \text{ m}^2 \][/tex]
So, when you multiply [tex]\(2 \times \frac{22}{7} \times 3.5 \text{ m} \times 5 \text{ m}\)[/tex], you get [tex]\(110 \text{ m}^2\)[/tex]. This is the final answer.