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3 years ago

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The slope of the line below is -2. Use the coordinates of the labeled point to find the point-slope equation of the line. (4, -2

)

1 answer:

natka813 [3]3 years ago

7 0

Point slope form : y - y1 = m ( x - x1)

m = slope = -2 and Substitute the corresponding values ...

x = 4 and y = - 2

POINT SLOPE FORM : y + 2 = -2 ( x - 4 )

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Answer:

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Step-by-step explanation:

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3 years ago

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soldier1979 [14.2K]

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2 years ago

Coleman uses his graphing calculator to find all of the solutions to the equation shown below.

Phoenix [80]

Answer:

The correct option is;

D. x = -1.38 and 0.82

Please find attached the combined function chart

Step-by-step explanation:

The given equation is x³ + 3 = -x⁴ + 4

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f(x) = x³ + 3, h(x) = -x⁴ + 4

x f(x) h(x)

-1.4 0.256 0.1584

-1.39 0.314381 0.26699

-1.38 0.371928 0.373261

-1.37 0.428647 0.477246

-1.36 0.484544 0.57898

Which shows an intersection at the point around -1.38

x f(x) h(x)

0.77 3.456533 3.64847

0.78 3.474552 3.629849

0.79 3.493039 3.610499

0.8 3.512 3.5904

0.81 3.531441 3.569533

0.82 3.551368 3.547878

0.83 3.571787 3.525417

Which shows the intersection point around 0.82

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From the graphing calculator the intersection point is given as

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2 years ago

10^(-10log3) How to solve this???

Olenka [21]

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3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

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

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

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

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(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)

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

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

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

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

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3 years ago