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alexgriva [62]
3 years ago
11

How can you add eight 8's to get the number 1,000? (only use addition)

Mathematics
1 answer:
Fudgin [204]3 years ago
6 0
The answer is 888+88+8+8+8
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Solve for x. x÷(-2)=1.4​
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x=-2.8

Step-by-step explanation:

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the value of painting increases each year. To find the value of the painting for the next year, the art dealer multiplies the cu
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the value of painting increases each year. To find the value of the painting for the next year, the art dealer multiplies the current value by 1.6. if the original value of the painting is 100, what is the value of the painting next yearStep-by-step explanation:


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2 years ago
Reciprocal of. 0×7/11​
Tpy6a [65]

Answer:

it doesn't exist

Step-by-step explanation:

the expression 0×7/11​ is equivalent to 0. 1/0 isn't possible, so its reciprocal doesn't exist.

8 0
2 years ago
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
2 years ago
For 30 minutes you do a combination of walking and jogging. At the end of your workout your pedometer displays a total of 2.5 mi
yKpoI14uk [10]
Walking for 10 minutes
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3 years ago
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