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Zanzabum
2 years ago
5

Determine the points of intersection of the given functions. y = –x2 + 3x + 20 y = –2x + 15 Describe the features of the graphin

g calculator you would choose and why.
Mathematics
2 answers:
Tanzania [10]2 years ago
5 0

After you type in your equations and hit graph you notice that, if you are in the standard window, your parabola is cut off so you have to choose your "window" button to change the viewing window to see the whole graph. Then you would use your 2nd button and "trace" and "intersect" to find the points of intersection of the 2 graphs. The first point is at (-.90901, 16.81812) and the second point is at (5.9090909, 3.1818182). Graphing calculators are quite amazing!

NISA [10]2 years ago
3 0

After you type in your equations and hit graph you notice that, if you are in the standard window, your parabola is cut off so you have to choose your "window" button to change the viewing window to see the whole graph. Then you would use your 2nd button and "trace" and "intersect" to find the points of intersection of the 2 graphs. The first point is at (-.90901, 16.81812) and the second point is at (5.9090909, 3.1818182). Graphing calculators are quite amazing!

Read more on Brainly.com - brainly.com/question/10786940#readmore

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mixas84 [53]
-3000 is correct
Hope this helps :D
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Which statement holds true for a skewed histogram showing a distribution of the weights of students in a class?
Vanyuwa [196]
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A professor wants to determine information about the Internet usage of middle school students in one state. A sample of 500 boys
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2 years ago
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(b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y =
frutty [35]

Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

-4k²Coskt=-25Coskt

Divide through by Coskt and we assume Coskt is not equal to zero

-4k²=-25

k²=-25/-4

k²=25/4

Then, k=√(25/4)

k= ± 5/2

b. Let assume we want to use this

y=ASinkt+BCoskt

Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t).

Then,

-25y

-25(A•Sin(±5/2t)+ B •Cos(±5/2t))

-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt

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