Given:
The expression is
![\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%20%5C_%5C_%7D%3D)
To find:
The simplified form of the expression.
Solution:
We have,
The expression ![\sqrt[3]{48}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D)
can be written as
![\sqrt[3]{48}=\sqrt[3]{8\cdot 6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%206%7D)
![\sqrt[3]{48}=\sqrt[3]{8}\cdot \sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%7D%5Ccdot%20%5Csqrt%5B3%5D%7B6%7D)
![[\because \sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5B3%5D%7Bab%7D%3D%5Csqrt%5B3%5D%7Ba%7D%5Csqrt%5B3%5D%7Bb%7D%5D)
![\sqrt[3]{48}=2\cdot \sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Ccdot%20%5Csqrt%5B3%5D%7B6%7D)
![\sqrt[3]{48}=2\sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Csqrt%5B3%5D%7B6%7D)
Therefore, ![\sqrt[3]{48}=\sqrt[3]{8\cdot 6}=2\sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%206%7D%3D2%5Csqrt%5B3%5D%7B6%7D)
.
<h3>
Answer: 7/10</h3>
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Explanation:
There are 30 days in April. Since it rained 9 of those days, the empirical probability of it raining in April is 9/30 = (3*3)/(3*10) = 3/10.
If we assume that the same conditions (ie weather patterns) hold for May, then the empirical probability of it raining in May is also 3/10. By "raining in May", I mean specifically raining on a certain day of that month.
The empirical probability of it not raining on the first of May is therefore...
1 - (probability it rains)
1 - (3/10)
(10/10) - (3/10)
(10-3)/10
7/10
We can think of it like if we had a 10 day period, and 3 of those days it rains while the remaining 7 it does not rain.
P=2(L+W)
if given one side and the perimiter
(P/2)-L=W
(P/2)-W=L
Answer:
$44.04
Step-by-step explanation:
14.68 * 3 = 44.04
Isosceles triangle I think
