No, your answer are wrong (I assume that the question asks to simplify the roots). You can only remove (their square is placed in from of the root and multiplied by it) the numbers from the root that are squares, the lest must stay, like this:
no roots are among the factors, nothing can be removed
nothing can be removed
nothing can be removed
nothing can be removed
nothing can be removed
nothing can be removed
nothing can be removed
nothing can be removed
(47 is prime) nothing can be removed
In short, to convert two fractions to have the same denominator, we simply multiply one by the denominator of the other, so in this case, we'll multiply 1/3 by 5, top and bottom, and 1/5 by 3, top and bottom, thus
![\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \qquad \qquad \sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%20%5Cimplies%20%20%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%20%0A%5Cqquad%20%5Cqquad%0A%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%5Cimplies%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \cfrac{15x^{\frac{1}{3}}}{y^{\frac{1}{5}}}\qquad \begin{cases} \frac{1}{3}=\frac{1\cdot 5}{3\cdot 5}\\ \qquad \frac{5}{15}\\\\ \frac{1}{5}=\frac{1\cdot 3}{5\cdot 3}\\ \qquad \frac{3}{15} \end{cases}\implies \cfrac{15x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{\sqrt[15]{x^5}}{\sqrt[15]{y^3}} \\\\\\ 15\sqrt[15]{\frac{x^5}{y^3}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7By%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B1%5Ccdot%205%7D%7B3%5Ccdot%205%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B5%7D%7B15%7D%5C%5C%5C%5C%0A%5Cfrac%7B1%7D%7B5%7D%3D%5Cfrac%7B1%5Ccdot%203%7D%7B5%5Ccdot%203%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B3%7D%7B15%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B15%5D%7Bx%5E5%7D%7D%7B%5Csqrt%5B15%5D%7By%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A15%5Csqrt%5B15%5D%7B%5Cfrac%7Bx%5E5%7D%7By%5E3%7D%7D)
Answer : The number of bottles should the pharmacy technician prepare to provide the required dose are 4 bottles.
Step-by-step explanation :
As we are given that a patient needs a total dose of 50 grams of albumin and the pharmacy has 50 mL bottles of 25% Albumin in stock.
Now we have to determine the number of bottles should the pharmacy technician prepare to provide the required dose.
As,
The pharmacy has stock albumin = 
The pharmacy has stock albumin = 12.5 mL
So,
Number of bottles should the pharmacy technician prepare to provide the required dose = 
Number of bottles should the pharmacy technician prepare to provide the required dose = 4
Therefore, the number of bottles should the pharmacy technician prepare to provide the required dose are 4 bottles.
Answer:
2
Step-by-step explanation:
Answer:
I believe it will be 55.9 or 56 along the roadway
Step-by-step explanation:
Hope this Helped
