Answer:
Part 1) ![\frac{a^4}{4b^2}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E4%7D%7B4b%5E2%7D)
Part 2) ![-\frac{v^9}{w^6}](https://tex.z-dn.net/?f=-%5Cfrac%7Bv%5E9%7D%7Bw%5E6%7D)
Step-by-step explanation:
we know that
When divide exponents (or powers) with the same base, subtract the exponents
Part 1) we have
![\frac{3a^{2}b^{-4}}{12a^{-2}b^{-2}}=(\frac{3}{12})(a^{2+2})(b^{-4+2} )=\frac{1}{4}a^{4}b^{-2}=\frac{a^4}{4b^2}](https://tex.z-dn.net/?f=%5Cfrac%7B3a%5E%7B2%7Db%5E%7B-4%7D%7D%7B12a%5E%7B-2%7Db%5E%7B-2%7D%7D%3D%28%5Cfrac%7B3%7D%7B12%7D%29%28a%5E%7B2%2B2%7D%29%28b%5E%7B-4%2B2%7D%20%29%3D%5Cfrac%7B1%7D%7B4%7Da%5E%7B4%7Db%5E%7B-2%7D%3D%5Cfrac%7Ba%5E4%7D%7B4b%5E2%7D)
Part 2) we have
![\frac{v^3w^{-3}}{-v^{-6} w^3} =-v^{3+6}w^{-3-3}=-v^9w^{-6}=-\frac{v^9}{w^6}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E3w%5E%7B-3%7D%7D%7B-v%5E%7B-6%7D%20w%5E3%7D%20%3D-v%5E%7B3%2B6%7Dw%5E%7B-3-3%7D%3D-v%5E9w%5E%7B-6%7D%3D-%5Cfrac%7Bv%5E9%7D%7Bw%5E6%7D)
Answer:
Step-by-step explanation:
The area would be 25 square centimeter
So... hmmm if you check the first picture below, for 2)
we could use the proportions of those small, medium and large similar triangles like
![\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"} \\\\\\ \cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"} \\\\\\ \cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7Bsmall%7D%7Blarge%7D%5Cqquad%20%5Ccfrac%7Bx%7D%7B12%7D%3D%5Ccfrac%7B6%7D%7Bx%7D%5Cimpliedby%20%5Ctextit%7Bsolve%20for%20%22x%22%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bsmall%7D%7Blarge%7D%5Cqquad%20%5Ccfrac%7Bz%7D%7B18%7D%3D%5Ccfrac%7B6%7D%7Bz%7D%5Cimpliedby%20%5Ctextit%7Bsolve%20for%20%22z%22%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Blarge%7D%7Bmedium%7D%5Cqquad%20%5Ccfrac%7By%7D%7B12%7D%3D%5Ccfrac%7B18%7D%7By%7D%5Cimpliedby%20%5Ctextit%7Bsolve%20for%20%22y%22%7D)
now.. for 3) will be the second picture below
Answer:
![n\geq 23](https://tex.z-dn.net/?f=n%5Cgeq%2023)
Step-by-step explanation:
-For a known standard deviation, the sample size for a desired margin of error is calculated using the formula:
![n\geq (\frac{z\sigma}{ME})^2](https://tex.z-dn.net/?f=n%5Cgeq%20%28%5Cfrac%7Bz%5Csigma%7D%7BME%7D%29%5E2)
Where:
is the standard deviation
is the desired margin of error.
We substitute our given values to calculate the sample size:
![n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23](https://tex.z-dn.net/?f=n%5Cgeq%20%28%5Cfrac%7Bz%5Csigma%7D%7BME%7D%29%5E2%5C%5C%5C%5C%5Cgeq%20%28%5Cfrac%7B1.96%5Ctimes%2012%7D%7B5%7D%29%5E2%5C%5C%5C%5C%5Cgeq%2022.13%5Capprox23)
Hence, the smallest desired sample size is 23
The question as presented is incomplete, here is the complete question with the multiple choice:
The sequence a1 = 6, an = 3an − 1 can also be
written as:
1) an = 6 ⋅ 3^n
2) an = 6 ⋅ 3^(n + 1)
3) an = 2 ⋅ 3^n
4) an = 2 ⋅ 3^(n + 1)
The correct choice is option 3) an = 2⋅3^n.
If we look at the initial sequence an = 3⋅an-1, and
a1 = 3⋅a0 = 6
a0 = 6/3
a0 = 2
We can now look at the sequence.
a0 = 2
a1 = 6
a2 = 18
a3 = 54
etc...
A common factor in each of those numbers is 2, so we can rewrite the sequence by factoring out 2.
a0 = 2⋅1
a1 = 2⋅3
a2 = 2⋅9
a3 = 2⋅27
The numbers being multiplied by 2 are all factors of 3. So we can rewrite the sequence again as:
a0 = 2⋅3^0
a1 = 2⋅3^1
a2 = 2⋅3^2
a3 = 2⋅3^3
This sequence can now be rewritten as an = 2⋅3^n.