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sergey [27]
3 years ago
6

Please help! (would really appreciate it)

Mathematics
1 answer:
love history [14]3 years ago
6 0

Divide:

( 8x^4 - 3x^2 + x - 10) by ( x - 1 ) ============================> 8x^5 − 8x^4 − 3x^3 + 4x^2 − 11x +10

Simplify:

(8x^4 − 3x^2 + x − 10)(x − 1)

(8x^4 + −3x^2 + x + −10)(x + −1)

(8x^4)(x) + (8x^4)(−1) +(−3x^2)(x) + (−3x^2)(−1) + (x)(x) + (x)(−1) + (−10)(x) +(−10)(−1)

8x^5 − 8x^4 −3x^3 + 3x^2 + x^2 − x − 10x +10

Hence, Your Answer, =====> 8x^5 − 8x^4 − 3x^3 + 4x^2 − 11x + 10

Hope that helps!!!! : )

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torisob [31]
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Could someone please help me with this? Solution set of lx^2+mx+n=0 is...?
IgorLugansk [536]

Answer:

Simplifying

lx2 + mx + n = 0

Solving

lx2 + mx + n = 0

Solving for variable 'l'.

Move all terms containing l to the left, all other terms to the right.

Add '-1mx' to each side of the equation.

lx2 + mx + -1mx + n = 0 + -1mx

Combine like terms: mx + -1mx = 0

lx2 + 0 + n = 0 + -1mx

lx2 + n = 0 + -1mx

Remove the zero:

lx2 + n = -1mx

Add '-1n' to each side of the equation.

lx2 + n + -1n = -1mx + -1n

Combine like terms: n + -1n = 0

lx2 + 0 = -1mx + -1n

lx2 = -1mx + -1n

Divide each side by 'x2'.

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Simplifying

l = -1mx-1 + -1nx-2

Step-by-step explanation:

Hope this helped you!

6 0
3 years ago
What is the cofunction of cos 2pi/ 9
fiasKO [112]

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4 0
3 years ago
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

3 0
3 years ago
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