30 over 4 = 15·2 over 2·2 = 15 over 2 = 1 + 7·2 over 2 = 7 1 over 2
Whole number =7
Fraction = 1 over 2
Hope this helps
Answer:
D
Step-by-step explanation:
![f(x)=(x-1)(x^2+2)^3\\f'(x)=(x-1)*3(x^2+2)^2*2x+(x^2+2)^3*1\\f'(x)=6x(x-1)(x^2+2)^2+(x^2+2)^3\\f'(x)=(x^2+2)^2[6x^2-6x+x^2+2]\\f'(x)=(x^2+2)^2(7x^2-6x+2)\\D](https://tex.z-dn.net/?f=f%28x%29%3D%28x-1%29%28x%5E2%2B2%29%5E3%5C%5Cf%27%28x%29%3D%28x-1%29%2A3%28x%5E2%2B2%29%5E2%2A2x%2B%28x%5E2%2B2%29%5E3%2A1%5C%5Cf%27%28x%29%3D6x%28x-1%29%28x%5E2%2B2%29%5E2%2B%28x%5E2%2B2%29%5E3%5C%5Cf%27%28x%29%3D%28x%5E2%2B2%29%5E2%5B6x%5E2-6x%2Bx%5E2%2B2%5D%5C%5Cf%27%28x%29%3D%28x%5E2%2B2%29%5E2%287x%5E2-6x%2B2%29%5C%5CD)
Answer:
6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.
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1/2
When you put a square root into exponential form, the square root turns into an exponential fraction. Since the square root is 2, our fraction would be 1/2
If our root was a cube root, our exponent would be 1/3, etc.
