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babunello [35]
2 years ago
12

I need help on this- nothing is right for me

Mathematics
2 answers:
tresset_1 [31]2 years ago
7 0

Answer:

5. Area = 28.5 cm; 6. Area = 544,500 cm

Step-by-step explanation:

Area = Length×Width

5. You have to convert the mm to cm by dividing the mm by 10. Then just multiply the length and width.

6. Same thing; different unit, convert the m to cm (or vice versa, not sure which unit you need to answer in) and multiply the length and width.

kvv77 [185]2 years ago
4 0

Answer:

5. Area = 28.5 cm; 6. Area = 544,500 cm

Step-by-step explanation:

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hodyreva [135]

The trigonometric function that best models the data is determined as y = 1.87sin(–0.29x + 4.79) + 6.13.

<h3>Trigonometric function that models the data</h3>

The trigonometric function that best models the data is determined as follows;

y = 1.87sin(-0.29x + 4.79) + 6.13

Use the above equation to check if its gives you values in the table;

<h3>when x = 1</h3>

y = 1.87sin(-0.29 x 1  + 4.79) + 6.13

y = 1.87sin(4.5 rad) + 6.13

y = 4.3

<h3>when x = 2</h3>

y = 1.87sin(-0.29 x 2  + 4.79) + 6.13

y = 1.87sin(4.21 rad) + 6.13

y = 4.49

<h3>when x = 4</h3>

y = 1.87sin(-0.29 x 4  + 4.79) + 6.13

y = 1.87sin(3.63 rad) + 6.13

y = 5.2

<h3>when x = 7</h3>

y = 1.87sin(-0.29 x 7  + 4.79) + 6.13

y = 1.87sin(2.76 rad) + 6.13

y = 6.8

<h3>when x = 9</h3>

y = 1.87sin(-0.29 x 9  + 4.79) + 6.13

y = 1.87sin(2.18 rad) + 6.13

y =7.7

<h3>when x = 10</h3>

y = 1.87sin(-0.29 x 10  + 4.79) + 6.13

y = 1.87sin(1.89 rad) + 6.13

y = 7.9

<h3>when x = 13</h3>

y = 1.87sin(-0.29 x 13  + 4.79) + 6.13

y = 1.87sin(1.02 rad) + 6.13

y = 7.72

<h3>when x = 17</h3>

y = 1.87sin(-0.29 x 17  + 4.79) + 6.13

y = 1.87sin(-0.14 rad) + 6.13

y = 5.9

<h3>when x = 20</h3>

y = 1.87sin(-0.29 x 20  + 4.79) + 6.13

y = 1.87sin(-1.01 rad) + 6.13

y = 4.6

Thus, the trigonometric function that best models the data is determined as y = 1.87sin(–0.29x + 4.79) + 6.13.

Learn more about trigonometry functions here: brainly.com/question/1143565

#SPJ1

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2 years ago
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