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scZoUnD [109]
3 years ago
5

Plz I need help !!!!!

Mathematics
1 answer:
Ad libitum [116K]3 years ago
3 0
So how you do scientific notation is, however many places you move the decimal point to the left is the number that you put in the power. For example, Mercury's mass is 330,000,000,000,000,000,000,000. You move the decimal point 23 places to the left to get 3.3 x 10^23. ( ^ means power. The little number that is next to the 10)

Now according to the question, the mass of Jupiter is 5.8 x 10^3 multiplied by 3.3 x 10^23 (5,800 x 330,000,000,000,000,000,000,000).

The answer to part A is,
1,941,000,000,000,000,000,000,000,000

Then, 3.2 x 10^2(X)=5.8 x 10^3 x 3.3 x 10^23

I don't know if you know how to do algebra, but this is very simple to solve. All you do is divide bolth sides of the equation by 3.2 x 10^2 leaving X (Earths mass) on the left.

For the right side need to divide 320 by 1,941,000,000,000,000,000,000,000,000. This will be the answer to part B

Hopefully this was helpful enough for you to show your work. Good luck : )
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Answer:

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Step-by-step explanation:

multiply 180 by the decimal form of 40%

180 x 0.40 = 72

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3 years ago
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Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (
kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2}) is given as continuous function, so there exist lim_{(x,y)\rightarrow(0,0)}f(x,y) and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})

we know that cos^2A+sin^2A=1, so we have that

f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)

lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19

So f(0,0)=ln19.

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