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2 years ago

Consider the diagram to the right below and complete the proof below

Mathematics

1 answer:

kiruha [24]2 years ago

8 0

I don’t know about this question

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B3%7D%20" id="TexFormula1" title="\frac{3}

Scrat [10]

The answer would be 14/15

Find a common denominator that both of the fractions can use. An easy way to do this is simply multiply the denominators together. So for 3/5 you would multiply by 3 to get the fraction 9/15 and then multiply 1/3 by 5 to get 5/15. Add both 9/15 and 5/15 to then get 14/15.

4 0

2 years ago

Read 2 more answers

Question 15<br> Evaluate x = y for x = 3 and y = 12.

nadezda [96]

Answer:

X is four times less/smaller than Y or Y is four times greater than X

Step-by-step explanation:

Not sure what you are asking to do

6 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago

WILL MARK BRAINLYIST IF RIGHT!!I don’t know how to do this

Agata [3.3K]

Answer:

i would just round 0.50 to the nearest tenth

Step-by-step explanation:

5 0

2 years ago

What is the sum of the arithmetic series? ∑[i=1,15,Fn=5i-1]

Helen [10]

$\bf \sum\limits_{i=1}^{15}\ 5i-1\implies 5\sum\limits_{i=1}^{15}\ i-\sum\limits_{i=1}^{15}\ 1\implies 5\left[ \cfrac{15(15+1)}{2} \right]-(15\cdot 1)
\\\\\\
5\cdot (15\cdot 8)-15\implies 585$

8 0

3 years ago