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
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![\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)
This is a classic example of where you would need to use the process of elimination. Let's start with answer A, 3 classes.
If A were to be correct, 3 classes of 45 minutes each would be 132 greater than the amount he paid, 3 classes worth $12. We could easily express this as a math equation:
45 * 3 = 132 + (12 * 3) Then, all you would have to do is simplify:
135 = 132 + 36
135 = 168
Obviously this equation isn't true, and you can continue with answer choice B.
But wait... there's a better way. Using this starting equation, we can come up with a formula to solve for c, the number of classes. Let's take our starting equation:
45 * 3 = 132 + (12 * 3) Now, let's turn all the 3s into cs.
45c = 132 + 12c Then, using algebra, we can solve this equation.
-12c - 12c
33c = 132
33c/33 = 132/33
c = 4
And finally, we have our answer, B! I hope both methods help you in your further studies.
Answer:
3
Step-by-step explanation:
The scale factor of ABC to DEF is the number you need to multiply a corresponding side of ABC to get one of DBC. We are given the two triangles are similar, so we can say that sides AB and DE are proportional. We are looking for the number we need to multiply AB by to get DE. From this relation, we can get the equation:
AB * x = DE
where x is our scale factor. We can substitute in the values of AB and DE, and solve for x:
5x = 15
x = 3
Therefore, the scale factor is three. This means that you can multiply any side of ABC by 3 to get a side of DEF.
Well the area of the square is is 54.94^2 but the area of the circle is <span>201.06^2. I hope this helped ^^</span>
Answer:
Step-by-step explanation:
Given
Required
The sample mean
This is calculated using:
So, we have:
