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2 years ago

10

choose the missing step in the given solution to the inequality −x − 3 < 13 3x. −x − 3 < 13 3x _______________ −4x < 16

x > −4 −4x − 3 > 13 2x − 3 < 18 −4x − 3 < 13 2x − 3 > 18

1 answer:

Ahat [919]2 years ago

5 0

Answer:

$-4x-3 < 13$

Step-by-step explanation:

we have

$-x-3 < 13+3x$

Subtract 3x both sides

$-x-3-3x < 13+3x-3x$

$-4x-3 < 13$

Adds 3 both sides

$-4x-3+3 < 13+3$

$-4x < 16$

Divide by -4 both sides

$x > -4$

therefore

The missing step is

$-4x-3 < 13$

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1 year ago

Read 2 more answers

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,

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Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

$D_uT(x,y)=T_x(x,y)i+T_y(x,y)j$

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

$i=\frac{1}{\sqrt{2} }$

$j=\frac{1}{\sqrt{2} }$

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

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and express T(x,y) as:

$T(x,y)=88*u^{-1}$

The partial derivative with respect to x is:

Using the chain rule:

$\frac{\partial u}{\partial x}=2x$

Hence:

$T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}$

Symplying the expression and replacing the value of u:

$T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}$

The partial derivative with respect to y is:

Using the chain rule:

$\frac{\partial u}{\partial y}=2y$

Hence:

$T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}$

Symplying the expression and replacing the value of u:

$T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}$

Therefore:

$D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})$

Evaluating the point (3,1)

$D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46$

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