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2 years ago

Long Division 3 Divided by888

Mathematics

1 answer:

aksik [14]2 years ago

3 0

The answer you will get is 296

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How many solutions does the following equation have? 4x+x+3=4x *

kvv77 [185]

Answer:

1 solution

Step-by-step explanation:

4x + x + 3 = 4x (Subtract 4x from both sides)

x + 3 = 0 (Subtract 3 from both sides)

x = -3

4 0

2 years ago

Rewrite with only sin x and cos x.

Annette [7]

Option A

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

$cos 3x = cos(x + 2x)$

We know that,

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

Therefore,

$\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x$ ---- eqn 1

We know that,

$\sin 2 x=2 \sin x \cos x$

$\cos 2 x=\cos ^{2} x-\sin ^{2} x$

Substituting these values in eqn 1

$\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)$ -------- eqn 2

We know that,

$\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x$

Applying this in above eqn 2, we get

$\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)$

$\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}$

$\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x$

Therefore,

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Option A is correct

7 0

2 years ago

Ship A receives a distress signal from the southwest, and ship B receives a distress signal from the same vessel from the north.

mylen [45]

G=t negative 320+f5 t=b

6 0

2 years ago

Rico ran a race that was 3 km long. how many meters did rico run?

pishuonlain [190]

3000 meters in total.

4 0

2 years ago

Find the following: F(x, y, z) = e^(xy) sin z j + y tan^−1(x/z)k Exercise Find the curl and the divergence of the vector field.

natulia [17]

$\vec F(x,y,z)=e^{xy}\sin z\,\vec\jmath+y\tan^{-1}\dfrac xz\,\vec k$

Divergence is easier to compute:

$\mathrm{div}\vec F=\dfrac{\partial(e^{xy}\sin z)}{\partial y}+\dfrac{\partial\left(y\tan^{-1}\frac xz\right)}{\partial z}$

$\mathrm{div}\vec F=xe^{xy}\sin z-\dfrac{xy}{x^2+z^2}$

Curl is a bit more tedious. Denote by $D_t$ the differential operator, namely the derivative with respect to the variable $t$ . Then

$\mathrm{curl}\vec F=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\D_x&D_y&D_z\\0&e^{xy}\sin z&y\tan^{-1}\frac xz\end{vmatrix}$

$\mathrm{curl}\vec F=\left(D_y\left[y\tan^{-1}\dfrac xz\right]-D_z\left[e^{xy}\sin z\right]\right)\,\vec\imath-D_x\left[y\tan^{-1}\dfrac xz\right]\,\vec\jmath+D_x\left[e^{xy}\sin z}\right]\,\vec k$

$\mathrm{curl}\vec F=\left(\tan^{-1}\dfrac xz-e^{xy}\cos z\right)\,\vec\imath-\dfrac{yz}{x^2+z^2}\,\vec\jmath+ye^{xy}\sin z\,\vec k$

5 0

3 years ago