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3 years ago

Solve 2m + 4m - 3m squared

Mathematics

1 answer:

vaieri [72.5K]3 years ago

3 0

It might be 6m-3m squared but i’m not sure if you want it simplified or if you want to find what is m

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If x10=235 find x <br> Answer needed ASAP please

nikdorinn [45]

Answer:

23.5 is the answer

Step-by-step explanation:

X10 =235 so x=235÷10 = 23.5

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A.) x (5 + x) = 5x + x^2
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Using English like form write an algorithm to find the area and perimeter of a rectangle

alexdok [17]

Answer:

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Step-by-step explanation:

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3 years ago

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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3 years ago