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kirza4 [7]
2 years ago
9

If the instructions for a problem ask you to use the smallest possible domain to completely graph two periods of y = 5 + 3 cos 2

(x - pi divided by three), what should be used for Xmin and Xmax? Explain your answer.
Mathematics
1 answer:
Scrat [10]2 years ago
4 0

The correct given equation is:

y = 5 + 3 cos 2 (x – π/3)

<span> To answer this problem, we must remember that:</span>

y = A cos [w(x - π)], the period is 2 π / w and the phase shift is π (the beginning of one period)

Rewriting the given equation to this form,

y = 5 + 3cos [2(x - π/3)] 

Therefore,

period = 2 π / 2 = π<span>
phase shift = π /3 </span>

So to create two periods, you need an interval of 2 π <span>
If you start at π /3, you'll start and end at the max of the cos function.</span>

Therefore,

 [Xmin , Xmax] = [π /3 , 7 π /3] would create a complete graph of 2 periods.

 

ANSWER: [π /3 , 7 π<span> /3]</span>

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Stephanie has $45,000. Part or all of which she wants to invest into a combination of municipal bonds and corporate bonds. She w
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Answer:

Step-by-step explanation:

Let x be the amount invested in municipal bonds, and let y be the amount invested in corporate bonds.

If Stephanie has $45,000 and she wants to invest part or all into a combination of municipal bonds and corporate bonds, the inequality for this statement would be

x + y ≤ 45000

She wants to invest no less than $20,000 into municipal bonds. It means that

x ≥ 20000

Also, she wants to invest at least four times as much into municipal bonds than into corporate bonds. This means that

x ≥ 4y

x/4 ≥ y

y ≤ x/4

The inequalities are

x + y ≤ 45000

x ≥ 20000

y ≤ x/4

8 0
2 years ago
write a number from 20 to 50 that has both tens and ones. use pictures and words to show the tens and ones.
slavikrds [6]
The answer is 35 because it in beetween 20 and 50

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1 year ago
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Anna bought 8 tetras and 2 rainbow fish for her aquarium. The rainbow fish cost​ $6 more than the tetras. She paid a total of​ $
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Answer:

C, D, and E

Step-by-step explanation:

When trying to solve for the cost of each item you will find a rainbow fish is $8.50 and a tetra is $2.50. If you use that information to solve the other answer choices C, D, and E would be correct

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2 years ago
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Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 50 km/h. After six hours,
Eva8 [605]
Let:
x = hours of travel
y = velocity

slope= rise/run slope=(y2-y1)/(x2-x1) 
(x1,y1) = (2,50) (x2,y2) = (6,54) 
sub values back into the equation m = (54-50)/(6-2) m = 1 
POINT SLOPE FORMy-y1 = m(x-x1) y-50= 1(x-2) y = x -2 +50
y = x + 48

B) 

the graph within the first seven hours can be obtained at point B

x = 7

y = 7+48 = 55
B(7,55)
4 0
2 years ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
2 years ago
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