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3 years ago

The following steps show how the right side of 1 – cos(2x) = tan(x)sin(2x) can be rewritten to show it is an identity.

Mathematics

2 answers:

damaskus [11]3 years ago

7 0

Answer:

Step-by-step explanation:

Brums [2.3K]3 years ago

4 0

Double Angle Identity, Quotient Identity for Tangent, Simplify, Double Angle Identity

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Which set of equations would have infinite solutions?

vaieri [72.5K]

The answer is b the two equations are equal to each other

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2 years ago

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How did changes to the voting process affect voter participation?

Jobisdone [24]

Answer:

A) more people voted

Step-by-step explanation:

I did it on edge

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3 years ago

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A football field is 1000 yards long.Each if the two end zones adds an extra 10 yards . The field is 160 feet wide . What is the

Kobotan [32]

Answer:

The area of the whole field is 54400yd^2

Step-by-step explanation:

says that I add 10 yards in each zone, that means that 10 yards is added twice at 1000 yards

1000yd + 2*10yd = 1020yd

we have to know that one yard equals 3 feet

so to have the length in feet we only multiply it by 3

1020 * 3 = 3060ft

Now that we know the length and width of the court, we can calculate the area by multiplying them

a = b * h

a = 3060ft * 160ft

a = 489600 ft^2

the result is in square feet

if we want it in square yards we just have to divide 2 times by 3

489600/(3*3) = 54400yd^2

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3 years ago

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What is the solution to the equation 3/4y+1/4=1+y , given the replacement set {−4, −3, −2} ?

ivann1987 [24]

Answer:

It's -3 mate

Step-by-step explanation:

I just did the test

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago