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2 years ago

5 5/8 - 5 1/8 simplify .

Mathematics

1 answer:

uranmaximum [27]2 years ago

4 0

Answer: 5 : 3 1/8 = 8 : 5

Step-by-step explanation:

Change values to whole numbers.

Convert any mixed numbers to fractions.

Convert 3 1/8

3 1/8 = 25/8

We now have:

5 : 3 1/8 = 5 : 25/8

Convert the whole number 5 to a fraction with 1 in the denominator.

We then have:

5 : 3 1/8 = 5/1 : 25/8

Convert fractions to integers by eliminating the denominators.

Our two fractions have unlike denominators so we find the Least Common Denominator and rewrite our fractions as necessary with the common denominator

LCD(5/1, 25/8) = 8

We now have:

5 : 3 1/8 = 40/8 : 25/8

Our two fractions now have like denominators so we can multiply both by 8 to eliminate the denominators.

We then have:

5 : 3 1/8 = 40 : 25

Try to reduce the ratio further with the greatest common factor (GCF).

The GCF of 40 and 25 is 5

Divide both terms by the GCF, 5:

40 ÷ 5 = 8

25 ÷ 5 = 5

The ratio 40 : 25 can be reduced to lowest terms by dividing both terms by the GCF = 5 :

40 : 25 = 8 : 5

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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: