Answer:
Yes that is the correct answer
Step-by-step explanation:
Solving for X it would be
X=-4-2y
Solving for Y it would be
Y=-2-x/2
Answer: The answer would be 8.5 times 10^-5
Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the
10 If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.
Answer:
$$\sqrt[3]{27x^6y^4}=3 x^2\sqrt[3]{y^4}$$
Step-by-step explanation:
![\sqrt[3]{27x^6y^4}= \sqrt[3]{3^3x^6y^4}\\\sqrt[3]{27x^6y^4}=\sqrt[3]{3^3} \sqrt[3]{x^6} \sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E6y%5E4%7D%3D%20%5Csqrt%5B3%5D%7B3%5E3x%5E6y%5E4%7D%5C%5C%5Csqrt%5B3%5D%7B27x%5E6y%5E4%7D%3D%5Csqrt%5B3%5D%7B3%5E3%7D%20%5Csqrt%5B3%5D%7Bx%5E6%7D%20%5Csqrt%5B3%5D%7By%5E4%7D)
therefore,
![\sqrt[3]{27x^6y^4}=3 x^2\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E6y%5E4%7D%3D3%20x%5E2%5Csqrt%5B3%5D%7By%5E4%7D)
