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NO LINKS AND NO ANSWERING WHAT YOU DON'T KNOW!!! THIS IS NOT A TEST OR ASSESSMENT!!!

Please help and SHOW WORK on these parts.

Unit 1 Assignment- Who's Right?​


NO LINKS AND NO ANSWERING WHAT YOU DONT KNOW THIS IS NOT A TEST OR ASSESSMENTPlease Help And SHOW WORK On These PartsUnit 1 Assignment Whos Right class=

Sagot :

Answer:

Part 1)

1) Sylvia' process was incorrect. In her last step, when she multiplied 0.88 by 10, she also need to divide exponent by 10.

2) Dylan is perfectly correct.

3) Ethan's process was also incorrect. Instead of subtracting the exponents, he added them.

Part 2)

Only Skyler's approach was incorrect. When Skyler acquired (3⁶)ˣ = 9, he made the error of setting the exponent on the left to the value on the right.

Step-by-step explanation:

Part 1)

We want to simplify the expression:

[tex]\displaystyle \frac{3.61\times 10^{-11}}{4.1\times 10^7}[/tex]

We can divide. Recall that xᵃ / xᵇ = xᵃ ⁻ ᵇ. Hence:

[tex]\displaystyle =0.88\times 10^{-11-7}=0.88\times 10^{-18}[/tex]

In scientific notation, the coefficient is always between 1 or 10.

So, we can multiply 0.88 by 10. To keep the equality, we need to divide 10⁻¹⁸ by 10. Hence:

[tex]\displaystyle =0.88(10)\times \frac{10^{-18}}{10}[/tex]

(You can see that the 10s can cancel out, giving us our original expression.)

Simplify. Thus:

[tex]\displaystyle \frac{3.61\times 10^{-11}}{4.1\times 10^7} = 8.8\times 10^{-19}[/tex]

Therefore, as we can see:

1) Sylvia' process was incorrect. In her last step, when she multiplied 0.88 by 10, she also need to divide exponent by 10.

2) Dylan is perfectly correct.

3) Ethan's process was also incorrect. Instead of subtracting the exponents, he added them.

Part 2)

Reviewing their work, we can see that only Skyler's approach was incorrect.

729 is indeed 3⁶. However, when Skyler acquired (3⁶)ˣ = 9, he made the error of setting the exponent on the left to the value on the right.

Both Robert and Kevin are correct, having set each exponent equal to each other after their bases are equivalent and solving for x.