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nikdorinn [45]
3 years ago
6

What are the points of discontinuity? Are they all removable? Please show your work.

Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0
A) By definition, you have that the period of the y=aCos(bx) is:
 Period=2π/|b|
 b=2π/3
 Period=2π/(2π/3)
 Period=3
 b) The amplitude of y=aCos(bx) is:
 Amplitude=|a|
 Amplitude=1/4
 c) The mideline represents the horintal line that passes between the maximum and minimum points. In the equation y=aCos(bx) the mideline is: y=0
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Write an equation in point-slope form for the line that passes through the point with the given slope. (-5, -8), m = -2
Arada [10]

Answer:

Y= -2x - 18

Step-by-step explanation:

Hope it works

4 0
2 years ago
What’s 12 rounded to the whole number
marin [14]

Answer:

Step-by-step explanation:

12.45

3 0
3 years ago
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Hi, can someone help with these questions? Thanks in advance !!
andreev551 [17]
5: 25 degrees
8: 48 degrees

Still working on the other ones

Explanation
- for 5 we know that the straight line has a measurement of 180 degrees so we just add 125 and 30 to get 155 and 155 plus 25 is 180 degrees

- For 8 we know that the measurements of AED are 89 degrees and we know two other angles are 29 and 12 which make 41 if you add them and we know that 41 + 48= 89 which is what we wanted to get to
6 0
2 years ago
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What is the area of this figure?
nalin [4]

Answer:

<em>290.5 square miles</em>

Step-by-step explanation:

Consider splitting this figure into two mini rectangles and a triangle, each of given lengths;

Dimensions of Rectangle 1 - Height ; 5 mi, Length ; 7 mi,\\Area of Rectangle 1 - Height * Length = 5 mi * 7 mi = 35 mi^2\\\\Dimensions of Rectangle 2 - Height ; 18 mi, Length ; 8 mi,\\Area of Rectangle 2 - Height * Length = 23 mi * 8 mi = 184 mi^2\\\\Dimensions of Mini Right Triangle - Height ; 11 mi, Base ; 8 + 5 = 13 mi,\\Area of Mini Right Triangle - 1 / 2 * Base * Height = 1 / 2 * 13 mi * 11 mi = 71.5 mi^2,\\\\Area of Figure = 35 mi^2 + 184 mi^2 + 71.5 mi^2 = 290.5 square miles

<em>Solution; 290.5 square miles</em>

4 0
3 years ago
Solve 0= 2t^3-21t^2+ 40t
sukhopar [10]
0=2t^3-21^2+40t
switch sides
2t^3-21t^2+40t=0
solve factoring 
t(t-8)(2t-5)=0
using the zero factor principle
t=0&#10;
solve
t-8=0:t=8
solve
2t-5=0:t= \frac{5}{2}
t=0,t=8,t= \frac{5}{2}
Hope this helps

4 0
3 years ago
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