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2 years ago

What is the constant of proportionality represented in the graph?

Mathematics

1 answer:

zheka24 [161]2 years ago

4 0

This constant of proportionality is equal to the slope of the line

If we asume each rectangle side is equal to a unit, then

Point 1 (0,3)
Point 2 (5,-1)

m = (y2 - y1) / (x2-x1)

m = (-1 - 3) / (5-0)

m = (-4) / (5) = -4/5 = k

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Sales tax is a function of the price of an item. The amount of sales tax is 0.08 times the price of the item. Use h to represent

Margaret [11]

Option C:

h(x) = 0.08x

Solution:

Given data:

Sales tax is a function of the price of an item.

Let h be the function.

Take x be the price of an item.

Amount of sales tax = 0.08 times price of the item

= 0.08 × x

Amount of sales tax = 0.08x

⇒ h(x) = 0.08x

Hence the function is h(x) = 0.08x.

Option C is the correct answer.

5 0

1 year ago

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,

-Dominant- [34]

Answer:

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

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:

$u=2+x^2+y^2$

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)

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