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timofeeve [1]
3 years ago
9

What function type can have vertical asymptotes?

Mathematics
1 answer:
Sonja [21]3 years ago
8 0
Rational functions of the form
f(x)= polynomial in x / another polynomial in x

See example plot attached. 

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Jessica spent S28 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of the total did she spe
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Answer:

Jessica spent S28 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of the total did she spend on fruit?

Step-by-step explanation:

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2 years ago
A machinist drilled a conical hole into a cube of metal as shown.
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Answer:

512ft³

Step-by-step explanation:

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Order the number 7/4, 1.6, 1 5/8, 1.65 from least to greatest.
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Here are the numbers from least to greatest     
1.6,1.65,15/8,7/4
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3 years ago
Suppose a regular n-gon is inscribed in a circle of radius r. Diagrams are shown for n=6, n=8, and n = 12. n = 6 n=8 n = 12 a a
m_a_m_a [10]

Answer:If an N-gon (polygon with N sides) has perimeter P, then each of the  

N sides has length P/N.  If we connect two adjacent vertices to the  

center, the angle between these two lines is 360/N degrees, or 2*pi/N  

radians.  (Do you understand radian measure well enough to follow me  

this far?)

The two lines I just drew, plus the side of the polygon between them,  

form an isosceles triangle.  Adding the altitude of the isosceles  

triangle makes two right triangles, and we can use one of them to  

derive the equation

 sin(theta/2) =  s/(2R)

where theta is the apex angle (which I said is 2*pi/N radians), R is  

the length of the lines to the center (the radius of the circumscribed  

circle), and s is the length of the side (which I said is P/N).

Putting those values into the equation, we have

 sin(pi/N) = P/(2NR)

so that

 P = 2NR sin(pi/N)

gives the perimeter of the N-gon with circumradius R.

Can we see a connection between this formula and the perimeter of a  

circle?  The perimeter of the circumcircle is 2*pi*R.  As we increase  

N, the perimeter of the polygon should get closer and closer to this  

value.  Comparing the two, we see

 2NR*sin(pi/N) approaches 2*pi*R

 N*sin(pi/N)   approaches pi

You can check this out with a calculator, using big numbers for N such  

as your teacher's N=1000.  If you calculate the sine of an angle in  

degrees rather than radians, the formula will look like

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Step-by-step explanation:

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3 years ago
WHAT IS 659 ROUNDED TO THE NEARST HUNDRED
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Answer: the answer is 700

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Read 2 more answers
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