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Snowcat [4.5K]
3 years ago
15

On September 17, 2008, high tide in Portland was at midnight. The water level at high tide was 10.3 feet; later, at low tide, it

was 0.3 feet. Assuming the next high tide is exactly at 12 noon and that the height of the water is given by a sine or cosine curve, find a formula for the water level in Portland as a function of time, where y is the water level in feet, and t is the time measured in hours from midnight.
Mathematics
1 answer:
____ [38]3 years ago
7 0
So the problem ask to calculate and formulate a function that could represent the water level in Portland if y is the water level and t is for time in hour so the possible answer would be that y=cosine(t)-0.3. I hope you are satisfied with my answer and feel free to ask for more
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PLS HELP DUE TODAY
Gekata [30.6K]

Here we go !!

Since it's a regular pentagon, all it's sides are equal so let's solve for x

\hookrightarrow \: 3(x - 1)  = 5x - 6

\hookrightarrow \: 3x - 3 = 5x - 6

\hookrightarrow \:  3x - 5x =  - 6 + 3

\hookrightarrow \:  - 2x =  - 3

\hookrightarrow \: x =  \dfrac{ - 3}{ - 2}

\boxed{\boxed{x =  \dfrac{3}{2} }}

now let's find the measure of each side, i.e

\hookrightarrow \: 5x - 6

\hookrightarrow \: (5 \times  \dfrac{3}{2})  - 6

\hookrightarrow \:  \dfrac{15}{2}   - 6

\hookrightarrow \: 7.5 - 6

\mathrm{\hookrightarrow \: 1.5 \: units}

perimeter = 5 × side length ( for a regular pentagon )

\hookrightarrow \: 1.5 \times 5

\boxed{ \boxed{ \:  \:  \:  \:  \:  \:  \:  \: 7.5 \:  \:  units \:  \:  \:  \:  \:    \:  \:  }}

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3 years ago
A spinner has 7 equal sections which are numbered from 1-7. Which of the following are complementary events? (check all that app
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Spinning an odd number.
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Read 2 more answers
Select all pairs of numbers that have a least common multiple of 30.
ddd [48]

5 and 6  and 3 and 10, the pairs of numbers that have a least common multiple of 30.

<u>Step-by-step explanation:</u>

Case 1: LCM (3, 6)

Prime factorization of 3:  1 \times 3

Prime factorization of 6:  2 \times 3

Using all prime numbers found as often as each occurs most often we take

               2 \times 3=6

Therefore LCM (3, 6) = 6.

Case 2: LCM (5, 6)

Prime factorization of 5:  1 \times 5

Prime factorization of 6:  2 \times 3

Using all prime numbers found as often as each occurs most often we take

                     5 \times 2 \times 3=30

Therefore LCM (5, 6) = 30

Case 3: LCM (3, 10)

Prime factorization of 3:  1 \times 3

Prime factorization of 10:  2 \times 5

Using all prime numbers found as often as each occurs most often we take

                    5 \times 2 \times 3=30

Therefore LCM (3, 10) = 30

Case 4: LCM (5, 10)

Prime factorization of 5:  1 \times 5

Prime factorization of 10: 2 \times 5

Using all prime numbers found as often as each occurs most often we take

                       5 \times 2=10

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So use distributive property
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and the other
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Answer:

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