Answer:
0,0
2,-1
-2,4
Step-by-step explanation:
The given expression is ![3b^2*(\sqrt[3]{54a}) + 3*(\sqrt[3]{2ab^6})](https://tex.z-dn.net/?f=%203b%5E2%2A%28%5Csqrt%5B3%5D%7B54a%7D%29%20%2B%203%2A%28%5Csqrt%5B3%5D%7B2ab%5E6%7D%29%20)
This can be simplified as :
= ![3*b^2*(\sqrt[3]{27 *2*a}) + 3*(\sqrt[3]{2*a*b^6})](https://tex.z-dn.net/?f=%203%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B27%20%2A2%2Aa%7D%29%20%2B%203%2A%28%5Csqrt%5B3%5D%7B2%2Aa%2Ab%5E6%7D%29%20)
We know that: ![\sqrt[3]{27} = 3](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B27%7D%20%20%3D%203%20%20%20)
Similarly we also can simplify: ![\sqrt[3]{b^6} = b^2](https://tex.z-dn.net/?f=%20%20%5Csqrt%5B3%5D%7Bb%5E6%7D%20%20%3D%20b%5E2%20)
So our expression will look like this:
= ![3*3*b^2*(\sqrt[3]{2a}) + 3*b^2*(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%203%2A3%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20%2B%203%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
= ![9b^2*(\sqrt[3]{2a}) + 3b^2*(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%209b%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20%2B%203b%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
=![\sqrt[3]{2a}*(9b^2 + 3b^2)](https://tex.z-dn.net/?f=%20%20%5Csqrt%5B3%5D%7B2a%7D%2A%289b%5E2%20%2B%203b%5E2%29%20)
=![\sqrt[3]{2a}*(12b^2)](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2a%7D%2A%2812b%5E2%29%20)
This can also be written as:
![12b^2(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%2012b%5E2%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
So the Answer is Option B
Step-by-step explanation:
Let Bridget, Jim and Krutika's shares be 4x, 5x and x respectively.
According to the given information:
4x = x + 39
4x - x = 39
3x = 39
x = 39/3
x = 13
5x = 5*13 = 65
Hence, Jim got 65 sweets.
Answer:
Multiply 2.8 by 1.6 and that gives you the area of 4.48