Answer:
The correct answer is option D. Dilation by a scale factor of 2 followed by reflection about the x-axis
The variable is b, the coefinet is 1/3
remember
ab-ac=a(b-c)
1/3b-1/3=1/3(b-1)
there, factored
The first question asks who sold more rolls. So start with figuring out how many Christie sold.
5 total - 1 2/3 left = 3 1/3 sold
you can convert the numbers to improper fractions with the same denominator. Like this:
5 x (3/3) - (3+2)/3
15/3 - 5/3 = 10/3
10/3 = 3 1/3
So now we know Christie sold more because 3 1/3 dozen is more than 2 1/2 dozen.
The part asks how many more.. Subtract the amounts the two girls sold.
3 1/3 - 2 1/2
10/3 x (2/2) - 5/2 x (3/3)
20/6 - 15/6 = 5/6
Christie sold 5/6 dozen more rolls. A dozen is 12 rolls so if you wanted to go further you just multiply 12 x 5/6 = 10 rolls
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
