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OLEGan [10]
2 years ago
15

I need help with number 6 please provide a screenshot of the Desmos and any work on paper

Mathematics
1 answer:
riadik2000 [5.3K]2 years ago
3 0

Answer:

See below

Step-by-step explanation:

Graphing all three equations can be done on one DESMOS plot, but it gets very busy and cluttered.  The attached image demonstrates an approach that simplifies the graph.

Enter the three equations and three inequalities separately into six DESMOS input boxes.  Then I clicked off the color circles of the ones I didn't want to see, leaving only two visible at one time.  Each pair represents one line of the function and limit for that function.  Then find the intersection point for the values indicated in the second half of the equation.  

One can also calculate the answers:

a)  (-21,442)

b)  (-4,-1)

c)  (18,15.5)

d) (20,17)

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Answer: area is 286.16 ft²

Step-by-step explanation:

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L= 29.2 ft

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To find out the area, all we have to do is multiply the lenght by the width:

And do not forget that to calculate the area, all units must be the same and for the area, it is always the unit squared, in this case, feet square or simply (ft²).

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What is the missing reason in the proof? Statement Reason 1. m || n and p is a transversal 1. given 2. ∠2 ≅ ∠3 2. ver. ∠s theore
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Match the function in the left column with its period in the right column.
kiruha [24]

The results for the matching between function and its period are:

  • Option 1 - Letter D
  • Option 2 - Letter A
  • Option 3 - Letter C
  • Option 4 - Letter B

<h3>What is a Period of a Function?</h3>

If a given function presents repetitions, you can define the period as the smallest part of this repetition. As an example of periodic functions, you have: sin(x) and cos(x).

\mathrm{Period\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}

The period of sin(x) and cos(x) is 2π.

For solving this question, you should analyze each option to find its period.

1) Option 1

\mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}\\ \\ \mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{2\pi }{\frac{1}{2} }=4\pi

Thus, the option 1 matches with the letter D.

2) Option 2

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}\\ \\ \mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{4} =\frac{\pi }{2}

Thus, the option 2 matches with the letter A.

3) Option 3

\mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}\\ \\ \mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{2} =\pi

Thus, the option 3 matches with the letter C.

4) Option 4

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}\\ \\ \mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{8} =\frac{\pi }{4}

Thus, the option 4 matches with the letter B.

Read more about the period of a trigonometric function here:

brainly.com/question/9718162

#SPJ1

6 0
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