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Determine the MOST PRECISE name for the quadrilateral below.
A) rhombus
B) parallelogram
C) square
D) trapezoid
E) kite

Answer: The answer is D

Step-by-step explanation:

Pythagorean theorem: a²+b²=c²

x²+x²=14²

2x²=196

Evaluate...

x=7√2

Answer:

3/4 of a bag of dog food will last 3 days.

The function g(t) is defined as:

g(t) = 1/√(16-t^2)

The function is continuous for all values of t that satisfy the following conditions:

The denominator is non-zero:

The denominator of the function is √(16-t^2). Therefore, the function is undefined when 16-t^2 < 0, or when t is outside the interval [-4,4].

There are no vertical asymptotes:

The function does not have any vertical asymptotes, because the denominator is always positive.

Thus, the function g(t) is continuous on the interval [-4,4].

Samir's new balance is $5,044.17.

To calculate Samir's new balance, add the previous balance, subtract the payment, add the new transaction, and multiply by the interest rate for one period. The following formula can be used to calculate the interest for a single period:

balance * APR / 12 = interest

where APR stands for annual percentage rate and 12 represents the number of months in a year.

When we apply this formula to Samir's balance and APR, we get:

5336.22 * 0.29 / 12 = 128.95 in interest

As a result, the total new balance is:

5336.22 - 607 + 186 + 128.95 = 5044.17

We get the following when we round to the nearest hundredth:

$5,044.17

As a result, Samir now has a balance of $5,044.17.

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The joint probability density function for the random variables Y1 and Y2 E(Y1+Y2) and V(Y1+Y2) is 41/144.

To find E(Y1+Y2), we need to integrate the sum of Y1 and Y2 over their joint probability density function:

E(Y1+Y2) = ∫∫ (y1 + y2) f(y1,y2) dy1 dy2

= ∫∫ (y1 + y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 <=1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex](y1 + y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (2y1 + 1) (1-y1)² dy1

= 5/12

To find V(Y1+Y2), we can use the formula V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]².

First, we need to find E[(Y1+Y2)^2]:

E[(Y1+Y2)²] = ∫∫ (y1+y2)² f(y1,y2) dy1 dy2

= ∫∫ (y1² + y2² + 2y1y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 = 1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex] (y1² + y2² + 2y1y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (1/3)y1³ + (1/2)y1² + (1/2)y1

(1/3)y1 + (1/4) dy1

= 7/12

Next, we need to find [E(Y1+Y2)]²:

[E(Y1+Y2)]² = (5/12)² = 25/144

Therefore, V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]² = (7/12) - (25/144) = 41/144.

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Therefore, the volume of the sphere to the nearest hundredth is 724,775.70 cubic millimeters.

Volume is a measurement of the amount of space occupied by a three-dimensional object. It is often expressed in units such as cubic meters (m³), cubic centimeters (cm³), cubic feet (ft³), or gallons (gal), depending on the context. The volume of a solid object can be calculated by multiplying its length, width, and height or using a specific formula depending on the shape of the object. For example, the volume of a rectangular box can be calculated as length x width x height, while the volume of a cylinder can be calculated as π x radius² x height. In general, volume is an important concept in many fields, including physics, chemistry, engineering, and architecture. It is often used to describe the capacity of containers, the displacement of fluids, and the amount of material used in construction or manufacturing.

Here,

The formula for the volume of a sphere is given as V = (4/3)πr³, where r is the radius of the sphere and π is approximately 3.14.

Substituting the given value of the radius, we get:

V = (4/3) x 3.14 x 48³

V ≈ 724,775.68 cubic millimeters

Rounding this value to the nearest hundredth, we get:

V ≈ 724,775.68 ≈ 724,775.70 cubic millimeters (rounded to two decimal places)

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If λ is an eigenvalue of A, then λ is an eigenvalue of [tex]A^T[/tex]. Show with an example that the eigenvectors of A and [tex]A^T[/tex] are not the same.

The equation Av = λv, where v is a non-zero vector, is satisfied by an eigenvector v and an eigenvalue given a square matrix A. In other words, the eigenvector v is multiplied by the matrix A to produce a scalar multiple of v. Due to their role in illuminating the behaviour of linear transformations and differential equation systems, eigenvectors play a crucial role in many branches of mathematics and science. When the eigenvector v is multiplied by A, the eigenvalue indicates how much it is scaled.

The eigenvalue and eigenvector states that, let v be a non-zero eigenvector of A corresponding to the eigenvalue λ.

Then, we have:

Av = λv

Taking transpose on both sides we have:

[tex]v^T A^T = \lambda v^T[/tex]

The above equations thus relates transpose of vector and transpose of A to λ.

Now, consider a matrix:

[tex]\left[\begin{array}{cc}1&2\\3&4\\\end{array}\right][/tex]

Now, the eigen values of this matrix are λ1 = -0.37 and λ2 = 5.37.

The eigenvectors are:

[tex]v1 = [-0.8246, 0.5658]^T\\v2 = [-0.4159, -0.9094]^T[/tex]

Now, for transpose of A:

[tex]A^T=\left[\begin{array}{cc}1&3\\2&4\\\end{array}\right][/tex]

The eigen vectors are:

[tex]u1 = [-0.7071, -0.7071]^T\\u2 = [0.8944, -0.4472]^T[/tex]

Hence, we see that, if λ is an eigenvalue of A, then λ is an eigenvalue of [tex]A^T[/tex]. Show with an example that the eigenvectors of A and [tex]A^T[/tex] are not the same.

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According to the question. A. No punctuation is missing.

Punctuation is the use of symbols to indicate the structure and organization of written language. It is used to help make the meaning of sentences clearer and to make them easier to read and understand. Punctuation marks can also be used to indicate pauses in speech, to create emphasis, and to indicate the speaker’s attitude. There are many different types of punctuation marks, each with its own purpose. The most commonly used punctuation marks are the period, comma, question mark, exclamation mark, quotation marks, and the apostrophe.

Quotation marks are used to enclose quoted material, while the apostrophe is used to indicate possession or to replace missing letters in a word or phrase. By using punctuation correctly, writers can ensure that their messages are correctly understood by their readers.

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Therefore, the monthly payment for a student loan of $15,000 at a fixed APR of 12% for 20 years is $144.36.

Interest is the cost of borrowing money or the return on investing money. When you borrow money, you usually have to pay back more than you borrowed, and the additional amount you pay is the interest. The interest rate is expressed as a percentage of the borrowed amount, and it can vary depending on factors such as the borrower's credit score, the term of the loan, and the lender's policies.

Given by the question.

Assuming the loan has a fixed interest rate of 12% per annum, the amount of interest charged each year will be:

12% of $15,000 = $1,800

The total interest charged over 20 years will be:

$1,800 x 20 = $36,000

The total amount to be repaid (principal + interest) will be:

$15,000 + $36,000 = $51,000

If the loan is being repaid in equal monthly installments over the 20-year term, the monthly payment can be calculated using the following formula:

M = P * (r[tex](1+r)^{n}[/tex]) / ([tex](1+r)^{n}[/tex]- 1)

Where:

M = Monthly payment

P = Principal amount (in this case, $15,000)

r = Monthly interest rate (12% per annum / 12 months = 1% per month)

n = Total number of payments (20 years x 12 months per year = 240)

Plugging in the values:

M = $15,000 * (0.01[tex](1+0.01)^{240}[/tex]) / ([tex](1+0.01)^{240}[/tex] - 1)

M = $144.36

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The value of x is 5

Triangles with the same shape but different sizes are said to be similar triangles. To be more specific, two triangles are comparable if their respective sides are proportionate and their corresponding angles are congruent.

Two triangles are similar if corresponding angles are congruent and corresponding sides are proportional.

from the below figure, both the triangles are similar, ∆ABC ≈ ∆EFB

By using Thales's theorem, the ratio of the sides of triangles are;

BE/EA = BF/FC

15/30 = x/10    

x = 5

Therefore the value of x is 5

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The ratio of triangle sides can be calculated using Thales' theory, and it is the value of x is 5

Similar triangles are those with the same shape but varying sizes. To be more precise, two triangles are comparable if their matching angles and respective sides are congruent.

If matching sides are proportional and corresponding angles are congruent, two triangles are similar.

Both triangles in the following figure are comparable ∆ABC ≈ ∆EFB

The ratio of triangle sides can be calculated using Thales' theory, and it is;

BE/EA = BF/FC

15/30 = x/10    

x = 5

Therefore the value of x is 5

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Sin 30 =3/x

1/2=3/x

x=6

Step-by-step explanation:

There are many possible solutions to this problem, but one possible set of values for P, Q, R, S, T, and U is:

P = 2

Q = 5

R = 1

S = 8

T = 9

U = 9

With these values, we have:

PQR = 251

STU = 156

And the sum of PQR and STU is indeed 407.

Answer: numbers 1,3 and 4

Step-by-step explanation:

The value of the triple integral[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex] by using spherical coordinates [tex]2\pi(e^{-1}-e^{-9})[/tex].

Given that the triple integral is-

[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex]

E is the region bounded by the spheres which are,

[tex]x^2+y^2+z^2=1\\\\x^2+y^2+z^2=9[/tex]

In spherical coordinates we have,

x = r cosθ sin ∅

y = r sinθ sin∅

z = r cos∅

dV = r²sin∅ dr dθ d∅

E contains two spheres of radius 1 and 3 () respectively, the bounds will be like this,

1 ≤ r ≤ 3

0 ≤ θ ≤ 2π

0 ≤ ∅ ≤ π

Then

[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex]

[tex]\int\int\int _{E} \frac{e^{-r^2}}{r}r^2Sin\phi drd\phi d\theta\\\\2\pi \int_{0}^{\pi} \int_1^3 re^{-r^2} dr d\phi\\\\2\pi \int_1^3 re^{-r^2} dr\\\\2\pi(e^{-1}-e^{-9})[/tex]

The complete question is-

Use spherical coordinates to evaluate the triple integral ∭ee−(x2 y2 z2)x2 y2 z2−−−−−−−−−−√dv, where e is the region bounded by the spheres x2 y2 z2=1 and x2 y2 z2=9.

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Parallelogram (Opposite sides have the same length). Parallelogram (Area is one-half the base times the height). Parallelogram (Opposite sides are parallel). Parallelogram (Angles can be right angles)

According to the parallelogram law, the sum of the squares of a parallelogram's four sides is equal to the sum of the squares of its two diagonals. It is essential for the parallelogram to have equal opposite sides in Euclidean geometry.

A parallelogram is a particular sort of polygon. It is a quadrilateral in which the opposite side pairs are parallel to one another. There are six crucial parallelogram characteristics to be aware of: Congruent sides are those when AB = DC.

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The actual intake of carbohydrates is 138% as compare to recommended intake.

Recommended intake of carbohydrates or any other nutrient are,

Based on the information provided,

Consumed 159 grams of carbohydrates,

Recommended intake is between 115 and 166 grams.

Calculate the percentage of actual intake compared to the recommended intake, use the following formula,

Percentage = (Actual Intake / Recommended Intake) x 100%

Substituting the values in the formula we have,

⇒Percentage = (159 / 115) x 100%

⇒Percentage ≈ 138.3%

Therefore,  the actual intake of carbohydrates is about 138% of the recommended intake, indicating that consumption of more carbohydrates than recommended.

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Answer:

Step-by-step explanation:

i think it B

Answer:

Let's assume that Mr. Nkalle invested an amount of x in Scheme A and (20900 - x) in Scheme B.

The simple interest earned on Scheme A in 2 years would be:

SI(A) = (x * 9 * 2)/100 = 0.18x

The simple interest earned on Scheme B in 2 years would be:

SI(B) = [(20900 - x) * 8 * 2]/100 = (3344 - 0.16x)

The total simple interest earned in 2 years is given as N$3508:

SI(A) + SI(B) = 0.18x + (3344 - 0.16x) = 3508

0.02x = 164

x = 8200

Therefore, Mr. Nkalle invested N$8200 in Scheme A and N$12700 (20900 - 8200) in Scheme B. So the amount invested in Scheme B was N$12700.

On the 7th day after the videο was pοsted, 640 peοple will see the videο.

Statistics is the discipline that cοncerns the cοllectiοn, οrganizatiοn, analysis, interpretatiοn, and presentatiοn οf data.

1 Let's start by finding the pattern in the number οf viewers. We knοw that οn the first day, 5 peοple watched the videο. On the next day, the number οf viewers dοubled tο 5 x 2 = 10. On the third day, the number οf viewers dοubled again tο 10 x 2 = 20. We can see that the number οf viewers is dοubling each day, which means we can write the number οf viewers as:

[tex]V = 5 x 2^n[/tex]

where n is the number οf days after the videο was pοsted.

Nοw we want tο find οn which day the number οf viewers will be 640. Sο we can set V equal tο 640 and sοlve fοr n:

[tex]640 = 5 x 2^n[/tex]

[tex]2^n = 128[/tex]

n = lοg2(128) = 7

2.   Tο find the first day when mοre than 20,000 peοple will see the videο, we can set V equal tο 20,000 and sοlve fοr n:

[tex]20,000 = 5 x 2^n[/tex]

2^n = 4,000

n = lοg2(4,000) ≈ 11.29

Since n represents the number οf days after the videο was pοsted, we can rοund up tο the next whοle number tο find the first day when mοre than 20,000 peοple will see the videο. Therefοre, οn the 12th day after the videο was pοsted, mοre than 20,000 peοple will see the videο if the trend cοntinues

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The value οf the given angle x = 39.8 degree

Trigοnοmetry uses six fundamental trigοnοmetric οperatiοns. Trigοnοmetric ratiοs describe these οperatiοns. The sine functiοn, cοsine functiοn, secant functiοn, cο-secant functiοn, tangent functiοn, and cο-tangent functiοn are the six fundamental trigοnοmetric functiοns.

The ratiο οf sides οf a right-angled triangle is the basis fοr trigοnοmetric functiοns and identities. Using trigοnοmetric fοrmulas, the sine, cοsine, tangent, secant, and cοtangent values are calculated fοr the perpendicular side, hypοtenuse, and base οf a right triangle.

In the figure tanx = p/h

[tex]x = tan^{-1(15/18)}[/tex]

x = 39.8

Hence the value οf the given angle x = 39.8 degree

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Days when change in temperature more than 10° F are Option B)Tuesday and E) Friday.

calculating the difference by deducting the end temperature from the initial temperature. The temperature difference is therefore 75 degrees Celsius - 50 degrees Celsius = 25 if something begins at 50 degrees Celsius and ends at 75 degrees Celsius.

Change in temperature on Monday from High to low

=15-10=5°F

Change in temperature on Tuesday from High to low

=8-(-4)=12°F

Change in temperature on Wednesday from High to low

=-2-(-5)=3°F

Change in temperature on Thursday from High to low

=-3-(-7)=4°F

Change in temperature on Friday from High to low

=-1-(-12)=11° F

Days when change in temperature more than 10° F are Tuesday and Friday.

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The Complete question is attached below:

Answer:

Find f(g(x)) f(x)=7x-8 , g(x)=3x-2. f(x)=7x−8 f ( x ) = 7 x - 8 , g(x)=3x−2 g ( x ) = 3 x - 2. Step 1. Set up the composite result function. f(g(x)) f ( g ...

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Answer:

y = 4 sin(π x) + 5

Step-by-step explanation:

A sine function with a midline of y=5, an amplitude of 4, and a period of 2 can be written in the following form:

y = A sin(2π/ x) +

where A is the amplitude, is the period, is the vertical shift (midline), and x is the independent variable (usually time).

Substituting the given values, we get:

y = 4 sin(2π/2 x) + 5

Simplifying this expression, we get:

y = 4 sin(π x) + 5

Therefore, the sine function with the desired characteristics is:

y = 4 sin(π x) + 5

The answer of the given question is (a) TR = p(40 - (4/3)p), or TR = 40p - (4/3)p² , (b)  TC = 9q + (3/10)q² + 30 , (c) π = 40p - (4/3)p² - 9q - (3/10)q² - 30 , (d) TR ≈ 342.67 , (e) the profit when 20 units are produced is approximately RM188.27 , (f)  the profit when 7 units are produced is approximately -RM24.44, indicating a loss , (g) the output required to obtain a profit of RM100 is approximately 8.78 units.

An equation is  mathematical statement that asserts yhe equality of two expressions. It typically consists of variables, constants, and mathematical operations like addition, subtraction, multiplication, and division, among others. Equations are often used to solve problems, to model real-world phenomena, and to describe mathematical relationships.

(a) The total revenue is given by TR = p x q. Substituting the demand equation q = 40 - (4/3)p, we get TR = p(40 - (4/3)p), or TR = 40p - (4/3)p².

(b) The total cost is given by TC = q x AC. Substituting the given average cost function, we get TC = 9q + (3/10)q² + 30.

(c) The profit is given by π = TR - TC. Substituting the expressions we found in parts (a) and (b), we get π = 40p - (4/3)p² - 9q - (3/10)q² - 30.

(d) At the break even point, the firm earns zero profit, so we set π = 0 and solve for q. Substituting the expression we found in part (a) for p, we get:

0 = 40p - (4/3)p² - 9q - (3/10)q² - 30

0 = 40(40/3 - (3/4)q) - (4/3)(40/3 - (3/4)q)² - 9q - (3/10)q² - 30

0 = 533.33 - 51.25q - 0.22q^2

Solving for q using the quadratic formula, we get:

q = (51.25 ± sqrt(51.25² - 4(-0.22)(533.33))) / 2(-0.22)

q ≈ 22.75 or q ≈ 206.58

We reject the solution q ≈ 206.58 because it is outside the relevant range of output, which is between 0 and 40. Therefore, the total output at the break even point is approximately 22.75 units. To find the total revenue at the break even point, we substitute q = 22.75 into the demand equation from part (a) and get:

p = (40/3) - (3/4)q

p ≈ 15.08

TR = p x q

TR ≈ 342.67

(e) To find the profit when 20 units are produced, we substitute q = 20 into the expression for profit we found in part (c) and get:

π = 40p - (4/3)p² - 9q - (3/10)q² - 30

π ≈ 188.27

Therefore, the profit when 20 units are produced is approximately RM188.27.

(f) To find the profit when 7 units are produced, we substitute q = 7 into the expression for profit we found in part (c) and get:

π = 40p - (4/3)p² - 9q - (3/10)q² - 30

π ≈ -24.44

Therefore, the profit when 7 units are produced is approximately -RM24.44, indicating a loss.

(g) To find the output required to obtain a profit of RM100, we set the profit equation equal to 100 and solve for q:

Profit = TR - TC

100 = pq - ACq

100 = (40-(4/3)p)*q - (9+(3/10)q+30/q)*q

100 = (40-(4/3)p - 9q - 3q²/10)

Multiplying by 10 and rearranging terms, we get a quadratic equation in q:

3q² + 91q - 310 = 0

Solving for q using the quadratic formula, we get:

q = (-91 ± sqrt(91² - 43(-310)))/(2*3)

q ≈ 8.78 or q ≈ -29.44

Since the quantity produced cannot be negative, the output required to obtain a profit of RM100 is approximately 8.78 units.

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4775 g968r648 747474874 483892874 23773259635y84b2375789325 7437594365825 4378574937587 49388959365n 98437858746587 32o4iy548569

Answer:

?

Step-by-step explanation:

Answer: D. Square

Step-by-step explanation:

The shape created by the cross section of the cut through a square pyramid is a square.

To see why, imagine the pyramid sitting on a table with its square base flat against the surface. The cut goes through the vertex, which is the point at the top of the pyramid. Since the cut is perpendicular to the base, it divides the pyramid into two smaller pyramids with congruent, but not identical, bases. Each of these smaller pyramids has a triangular base that is an isosceles right triangle. The two triangles share a common hypotenuse, which is the line of the cut.

The cross section of the cut is the shape formed where the two triangles meet along the hypotenuse. Since both triangles are congruent and the hypotenuse is the same for both, the cross section is a square. The sides of the square are equal to the base of the original pyramid, which is one of the legs of the isosceles right triangles formed by the cut. Therefore, the answer is D, a square.

Answer:

The answer is brand B

Step-by-step explanation:

You divide $14.88 by 24 which equals 68 cents per item.

Then brand B is 60 cents per item which is the better buy!

Yes, there is an analogous ratio between 2:1 and 20:10.

We just cancel by a common factor. So 4:2=2:1 . The simplest representation of the ratio 4 to 2 is the ratio 2 to 1. Also, since each pair of numbers has the same relationship to one another, the ratios are equivalent.

By dividing the terms of each ratio by their greatest common factor, we may simplify both ratios to explain why.

As the greatest common factor for the ratio 2:1 is 1, additional simplification is not necessary.

The greatest common factor for the ratio 20:10 is 10. When we multiply both terms by 10, we get:

20 ÷ 10 : 10 ÷ 10

= 2 : 1

As a result, both ratios have the same reduced form, 2:1, making them equal.

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Answer:

One possible function that models the ocean tide is:

h(t) = A sin(ωt + φ) + B

where:

h(t) represents the height of the tide (in meters) at time t (in hours)

A is the amplitude of the tide (in meters)

ω is the angular frequency of the tide (in radians per hour)

φ is the phase shift of the tide (in radians)

B is the mean sea level (in meters)

This function is a sinusoidal function, which is a common type of function used to model periodic phenomena. The sine function has a natural connection to circles and periodic motion, making it a good choice for modeling the regular rise and fall of ocean tides.

The amplitude A represents the maximum height of the tide above the mean sea level, while B represents the mean sea level. The angular frequency ω determines the rate at which the tide oscillates, with one full cycle (i.e., a high tide and a low tide) occurring every 12 hours. The phase shift φ determines the starting point of the tide cycle, with a value of zero indicating that the tide is at its highest point at time t=0.

To determine specific values for A, ω, φ, and B, we would need to gather data on the tide height at various times and locations. However, typical values for these parameters might be:

1. A = 2 meters (representing a relatively large tidal range)

2. ω = π/6 radians per hour (corresponding to a 12-hour period)

3. φ = 0 radians (assuming that high tide occurs at t=0)

4. B = 0 meters (assuming a mean sea level of zero)

Using these values, we can write the equation for the tide as:

h(t) = 2 sin(π/6 t)

We can evaluate this equation for various values of t to get the height of the tide at different times. For example, at t=0 (the start of the cycle), we have:

h(0) = 2 sin(0) = 0

indicating that the tide is at its lowest point. At t=6 (halfway through the cycle), we have:

h(6) = 2 sin(π/2) = 2

indicating that the tide is at its highest point. We can also graph the function to visualize the rise and fall of the tide over time:

Tide Graph

Overall, this function provides a simple and effective way to model the ocean tide using trigonometric functions.

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