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ahrayia [7]
2 years ago
10

The 3rd degree Taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − (x − π/2) + 1/6 (x − π/2)3 + R3(x). Usin

g this, estimate cos(86°) correct to five decimal places.
Mathematics
1 answer:
Otrada [13]2 years ago
8 0

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at a = \frac{\pi}{2} is:

\cos x \approx -\left(x-\frac{\pi}{2} \right)+\frac{1}{6}\cdot \left(x-\frac{\pi}{2} \right)^{3}

The value of 86º in radians is:

86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}

\cos 86^{\circ} \approx 0.06976

The cosine of 86º is approximately 0.06976.

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Answer:

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

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Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

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