Answer:
Options (1), (2), (3) and (7)
Step-by-step explanation:
Given expression is ![\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B8%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%203%7D%20%7D%7B3%5Ctimes2%5E%7B%5Cfrac%7B1%7D%7B9%7D%7D%7D)
.
Now we will solve this expression with the help of law of exponents.
![\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B8%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%203%7D%20%7D%7B3%5Ctimes2%5E%7B%5Cfrac%7B1%7D%7B9%7D%7D%7D%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B%282%5E3%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%203%7D%20%7D%7B3%5Ctimes2%5E%7B%5Cfrac%7B1%7D%7B9%7D%7D%7D)
![=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%5Ctimes%203%7D%20%7D%7B3%5Ctimes2%5E%7B%5Cfrac%7B1%7D%7B9%7D%7D%7D)
[Option 2]
![2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B2%7D%7B9%7D%7D%5Ctimes%203%5E%7B-%5Cfrac%7B2%7D%7B3%7D%20%7D%3D%28%5Csqrt%5B9%5D%7B2%7D%29%5E2%5Ctimes%20%28%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%29%5E2)
[Option 1]
![=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }](https://tex.z-dn.net/?f=%3D%5Csqrt%5B9%5D%7B4%7D%5Ctimes%20%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B9%7D%20%7D)
[Option 3]
![=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B9%5D%7B2%5E2%7D%5Ctimes%20%5Csqrt%5B3%5D%7B3%5E%7B-2%7D%7D)
[Option 7]
Therefore, Options (1), (2), (3) and (7) are the correct options.
I would say A. becuase squares can be similar by looking alike but it doesn't matter what size, big or small.
The answer is b, -11.
plug in 3 as the input and solve.
Answer:
(-1,3)
Step-by-step explanation:
The slope is given by
m = (y2-y1)/(x2-x1)
1 = (2-3)/(-2-x)
Multiply each side by (-2-x)
(-2-x) = -1
Multiply by -1
2+x = 1
Subtract 2 from each side
2+x-2 = 1-2
x = -1
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
