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2 years ago

Determine the slope of the linear function y= - 2/3x -5

Mathematics

2 answers:

Mamont248 [21]2 years ago

6 0

Hello there!

The slope is -2/3x.

Why? Keep the following formula in mind:

y=mx+b

m is the slope (-2/3x in this case)

b is the y-intercept (-5 in this case)

<h2>Therefore, the slope is -2/3.</h2>

Hope this helps!

~Just a felicitous girlie

#HaveAnAmazingDay

$SilentNature :)$

Anastaziya [24]2 years ago

4 0

Answer: The slope of the linear function y= - 2/3x -5 is:

-2/3

, Hope this helps :)

Have a great day!!

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Write the equation of the line passing through the point (-4 "-7)" and (8 "-17)"

kicyunya [14]

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$y = -\dfrac{5}{6} x -\dfrac{31}{3}$

In order to find equation:

<u>Find slope</u>:

$\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} = \dfrac{rise}{run}$

$\rightarrow \sf slope: \dfrac{-17-(-7)}{8-(-4)}$

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2 years ago

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3 years ago

Read 2 more answers

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