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Mathematics

1 answer:

xz_007 [3.2K]3 years ago

4 0

Answer:

Step-by-step explanation:

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Find the equation of the line passes through (-3,-4) and whose slope is 3/2 in slope intercept form. (Algebra 2 or Geometry)

ioda

Hey there! I'm happy to help!

Slope intercept form is y=mx+b, where m is the slope and b is the y-intercept. We already have the slope, so our equation so far is y=3/2x+b. We just need to find the b for it to be complete.

To do this, we plug in a point on the line and solve for b. We have a point (-3,-4), so let's use it and solve.

-4=3/2(-3)+b

Undo parentheses.

-4=-9/2+b

Add 9/2 to both sides

b=1/2

Therefore, our equation is y=3/2x+1/2.

Have a wonderful day! :D

4 0

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

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