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3 years ago

Are sin and cos reciprocals of each other

1 answer:

8 0

Hello,

When you compose 2 reciprocals functions, you will find 1.

ex f(x)=√x and g(x)=x² , x>=0
so that g(f(x))=(√x)²=x

sin(cos(0))=sin(1)=0,84147... rad and not 1.
Reciprocal function of sin (x) is arcsin(x).
Reciprocal function of cos (x) is arccos(x).

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A sequence has three terms.

Lunna [17]

Answer: don't know sorry

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8 0

2 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

2 years ago

Write an equation for the parabola whose vertex is at (4, 12) and which passes through (5, 21)

laiz [17]

Answer:

y - 12 = 9(x - 4)

Step-by-step explanation:

The vertex (h, k) is (4, 12) and the point (5, 21) is on the graph. Assuming that this is a vertical parabola, opening up (because the coordinate 21 is greater than the coordinate 12), we insert the knowns into y - k = a(x - h)^2, obtaining

21 - 12 = a(5 - 4), or 9 = a. With a known, we can write the desired equation:

y - 12 = 9(x - 4)

8 0

2 years ago

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yuradex [85]

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4 0

2 years ago

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If f(x) = x2 is vertically stretched by a factor of 4 to g(x) and reflected over the x-axis, what is the equation for g(x)?

alexdok [17]

Step-by-step explanation:

integral fx = gx, gx = fx

gx = -4x2

variable of x2 = replace to left side it mean we got minus when Variable replace to left it mean minus

8 0

2 years ago