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Leto [7]
3 years ago
11

Long division 12p^5+20p^4+18p^2+45p+25/3p+5

Mathematics
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

4p^4+6p+5

Step-by-step explanation:

Given

\frac{12p^5+20p^4+18p^2+45^p+25}{3p+5}

Step 1:

Now Dividing the above Equation our first quotient will be 4p^4 and First remainder will be 18p^2+45p

Step 2:

Now Dividing the First remainder 18p^2+45p with 3p+5

now our second Quotient will be 4p^4+6p and Second remainder will be 15p+25

Step 3:

Now Dividing Second remainder 15p+25 with 3p+5

now our third Quotient will be 4p^4+6p +5 and Remainder will be 0.

\therefore Our Final Answer is 4p^4+6p +5 with remainder 0

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write the equation for the nth term of the sequence and then find the value of the 24th term: 4, 13, 22, 31, ...
julia-pushkina [17]

Answer :

nth term

an = a1 + (n-1) d

an = 4 + (9-1) 9

an = 4 + (8)9

an = 4 + 72

an = 82

24th term

an = 4 + (24-1) 9

an = 4 + (23) 9

an = 4 + 207

an = 211 is the answer

Hope it helped

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3 years ago
8-2x=5<br> Please help in the next ten minutes
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8-2x=5\\&#10;-2x=-3\\&#10;x=\dfrac{3}{2}
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Which of the following is a complex number?<br> -7<br> 2 + StartRoot 3 EndRoot<br> 4 + 9i<br> Pi
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Second to last one is my guess
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4 years ago
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Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
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Pachacha [2.7K]

Answer:

y = -4x - 12

Step-by-step explanation:

y + 4 = -4(x + 2)

y = -4x -8 - 4

y = -4x - 12

6 0
3 years ago
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