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pshichka [43]
3 years ago
12

Can you write this equation in point-slope form? y = ½x + 4?

Mathematics
1 answer:
Svet_ta [14]3 years ago
4 0
Yes.  This equation given:
______________________________
   " y = (½)x + 4 " ;  in point-slope form; also known as: "slope-intercept form" ; is:
______________________________________
    " y = (½)x + 4 " .
______________________________________
In other words, the equation given is ALREADY written in "point-slope form" ; or, "slope-intercept form".
______________________________________
Note:   An equation that is written in "point-slope form"
             (or, "slope-intercept form"), is written in the format of:
______________________________________
" y = mx + b " ;_________________
in which:_________________
"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:
"m" is the slope of the line; which is what we want to solve for;
"b" is the "y-intercept"; or more precisely, the value of "x"
(that is; the "x-coordinate") of the point at which "y = 0";
that is, the value of "x" ; or the "x-coordinate" of the point at which
the graph of the equation crosses the "x-axis".
______________________________________
Note that in our given equation, which is written in "point-slope form" (or, "slope-intercept form" — that is:  " y = mx + b "  ;
_______________________________________
which is:  " y = (½)x + 4 " ;
_______________________________________
we have:
_______________________________________
  "y" isolated as "stand-alone" variable on the "left-hand side" of the equation;

  m = ½ ;
  b =  4 .
_______________________________________

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Initial length of electric cable needed, L=550\ ft.

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Wire needed is given by :

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Step-by-step explanation:

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A line that includes the point (2,10) has a slope of 9. What is its equation in slope-intercept form
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For each of the following vector fields
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(A)

\dfrac{\partial f}{\partial x}=-16x+2y

\implies f(x,y)=-8x^2+2xy+g(y)

\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y

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(B)

\dfrac{\partial f}{\partial x}=-8y

\implies f(x,y)=-8xy+g(y)

\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x

\implies \dfrac{\mathrm dg}{\mathrm dy}=x

But we assume g(y) is a function of y alone, so there is not potential function here.

(C)

\dfrac{\partial f}{\partial x}=-8\sin y

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\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y

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