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QveST [7]
2 years ago
5

The height, h, in feet of a piece of cloth tied to a waterwheel in relation to sea level as a function of time, t, in seconds ca

n be modeled by the equation h = 15 cosine (StartFraction pi Over 20 EndFraction t). How long does it take for the waterwheel to complete one turn?
5 seconds
10 seconds
20 seconds
40 seconds
Mathematics
1 answer:
Anni [7]2 years ago
7 0

Considering the period of the cosine function, it is found that it takes 40 seconds for the wheel to complete one turn.

<h3>What is the period of the cosine function?</h3>

The cosine function is defined by:

f(x) = acos(bx + c) + d.

For the period, we have to look at coefficient b, and the period is:

P = 2π/|B|

For this problem, the function is given by:

h(x) = 15 cos(π/20)

Hence B = π/20, and the period is:

P = 2π/|B| = 2π/(π/20) = 2 x 20 = 40 seconds.

Hence it takes 40 seconds for the wheel to complete one turn.

More can be learned about the period of trigonometric functions at brainly.com/question/12502943

#SPJ1

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200 * 1/10 = 20
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4 years ago
The graph of y = x^2 -3 is translated in 4 units to the left of the graph to give the graph A
Vedmedyk [2.9K]

Answer:

a) b = 8, c = 13

b) The equation of graph B is y = -x² + 3

Step-by-step explanation:

* Let us talk about the transformation

  • If the function f(x) reflected across the x-axis, then the new  function g(x) = - f(x)
  • If the function f(x) reflected across the y-axis, then the new  function g(x) = f(-x)
  • If the function f(x) translated horizontally to the right  by h units, then the new function g(x) = f(x - h)
  • If the function f(x) translated horizontally to the left  by h units, then the new function g(x) = f(x + h)

In the given question

∵ y = x² - 3

∵ The graph is translated 4 units to the left

→ That means substitute x by x + 4 as 4th rule above

∴ y = (x + 4)² - 3

→ Solve the bracket to put it in the form of y = ax² + bx + c

∵ (x + 4)² = (x + 4)(x + 4) = (x)(x) + (x)(4) + (4)(x) + (4)(4)

∴ (x + 4)² = x² + 4x + 4x + 16

→ Add the like terms

∴ (x + 4)² = x² + 8x + 16

→ Substitute it in the y above

∴ y = x² + 8x + 16 - 3

→ Add the like terms

∴ y = x² + 8x + 13

∴ b = 8 and c = 13

a) b = 8, c = 13

∵ The graph A is reflected in the x-axis

→ That means y will change to -y as 1st rule above

∴ -y = (x² - 3)

→ Multiply both sides by -1 to make y positive

∴ y = -(x² - 3)

→ Multiply the bracket by the negative sign

∴ y = -x² + 3

b) The equation of graph B is y = -x² + 3

4 0
3 years ago
Whats the y intercept of y=4x -3
AleksAgata [21]

Answer:

-3

Step-by-step explanation:

the y intercept is the number that does not have a variable behind it (3)

the slope is the number that does have a variable behind it (4)

6 0
3 years ago
Read 2 more answers
Classify the triangle​
Bogdan [553]

Answer: B I believe

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
“encontrar la integral indefinida y verificar el resultado mediante derivación”
Oliga [24]

I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C

Para confirmar el resultado:

\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}

I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx

Sustituye:

y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx

\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C

(Te dejaré confirmar por ti mismo.)

I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx

Sustituye:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C

I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}

Sustituye:

u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}

\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C

Podemos hacer que esto se vea un poco mejor:

\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}

\implies I=-\dfrac{(t+1)^4}{4t^4}+C

4 0
4 years ago
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