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Lemur [1.5K]
3 years ago
10

Which of the following is the complete factorization of 10x - 3 - 3x 2?

Mathematics
1 answer:
andrew11 [14]3 years ago
5 0
-3x^2+10x-3  

This is of the form ax^2+bx+c.  We need to find two values, j and k, such that:

jk=ac=9 and j+k=b=10, so j and k must be 1 and 9.

Now replace bx with jx and kx in the original equation...

-3x^2+x+9x-3  now factor 1st and 2nd pair of terms

-x(3x-1)+3(3x-1)

(-x+3)(3x-1) or equivalently...

-(x-3)(3x-1) or as they put it:

-(3x-1)(x-3)
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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

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3 years ago
Express the recurring decimal 0.042 as a fraction in its simplest form.
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Answer:

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Step-by-step explanation:

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This will be the numerator  42

Now ,the denominator:

For each repeating decimal we write a 9 (in this case 2 nines)

For each non repeating decimal we write a zero

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