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)
.
Answer:
you are a square
Step-by-step explanation:
it says " i have 2 pairs of parallel sides and 4 right angles", and typically 4 right angles make a square
If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
Answer:
You have to use this formula v = ⁴⁄₃πr³ or another way would be by entering the value of the radius in google calculator.
Step-by-step explanation:
Hope this helped :)
3/10 is the answer to 4/5-1/2
