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3 years ago

A yearbook printer charges based on the number of pages printed. Here is a table that shows the cost of some recent yearbooks:

1 answer:

5 0

y = 0.012x + 0.65

Number of pages 50.00 100.00 150.00 200.00 
Cost 1.25 1.85 2.45 3.05

x = number of pages
y = cost

intervals of x = 50 pages
intervals of y = 0.60
0.60 / 50 = 0.012

y = 0.012x + 0.65

 x 0.012*x y 
 0.012 50 0.60 0.65 1.25
0.012 100 1.20 0.65 1.85
 0.012 150 1.80 0.65 2.45
 0.012 200 2.40 0.65 3.05

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Find the difference between points M(6,16) and Z(-1,14) to the nearest tenth.

svlad2 [7]

I'm not sure what you mean by difference, so I am guessing you meant to write "distance". If this is not the case, then feel free to report me. Anyway, to find the distance between two points you would use the distance formula.

Distance Formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

Let's substitute our point coordinates into the formula. We'll substitute -1 and 6 for x_2 and x_1, and 14 and 16 for y_2 and y_1. After substituting, your formula will now look like: $d=\sqrt{(-1-6)^2+(14-16)^2}$ .

$d=\sqrt{(-1-6)^2+(14-16)^2}$

Subtract the coordinates.

$d=\sqrt{(-7)^2+(-2)^2}$

Square the subtracted coordinates.

$d=\sqrt{49+4}$

Add 49 and 4.

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3 years ago

The area of the shaded segment is 100cm^2. Calculate the value of r.

Reil [10]

Hello,

The formula for finding the area of a circular region is: $A= \frac{ \alpha *r^{2} }{2}$

then:
$A_{1} = \frac{80*r^{2} }{2}$

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

$cos(40)= \frac{h}{r} \\ \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)$

Now we can find its area:
$A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)$

The subtraction of the two areas is 100cm^2, then:

$A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm$

Answer: r= 1.59cm

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