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3 years ago

. Duke's photography pays $9 for a 5x7 photograph.

Mathematics

1 answer:

g100num [7]3 years ago

5 0

Answer:

the markup percentage is 66.67%

Step-by-step explanation:

The computation of the percent of markup based on cost is shown below:

= (Selling price - paid price) ÷ (paid price)

= ($15 - $9) ÷ ($9)

= 66.67%

By taking the difference of the selling price & paid price and then divided it by paid price we can get the percentage of markup

Hence, the markup percentage is 66.67%

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Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$ . Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$ . So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$ . So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_3%7Bx%5E%7B2%7D-x%20%7D%20%5C%2C%20dx" id="TexFormula1" title="\int\limits

melisa1 [442]

Answer:

$\frac{53}{6}$

Step-by-step explanation:

$\int\limits^4_3 {x^2-x} \, dx$

= [ $\frac{x^3}{3}$ - $\frac{x^2}{2}$ ] ← evaluate for upper limit - lower limit

= ( $\frac{64}{3}$ - $\frac{16}{2}$ ) - ( $\frac{27}{3}$ - $\frac{9}{2}$ )

= $\frac{64}{3}$ - 8 - 9 + $\frac{9}{2}$

= $\frac{155}{6}$ - 17

= $\frac{53}{6}$

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