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1 year ago

Which equation shows a true proportion

Mathematics

1 answer:

Oduvanchick [21]1 year ago

8 0

Answer: the first equation is the answer.

Step-by-step explanation: As you can see the first equation is 5/7=15/21. The are proportional because 5*3=15 and 7*3=21. If they are multiplied or divided by the same number, then they are proportional. It does not work with adding, subtracting or flipping.

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A student has a class with four tests, each worth 100 points, and has earned these

Wewaii [24]

Answer:

average = 52, 100 on fourth test

Step-by-step explanation:

average = 75 + 74 + 71 = 156 / 3 = 52

average = 52

b grade = 75 + 74 + 71 + 100 = 320 / 4 = 80

to get a b grade she must get 100 on her fourth test

4 0

2 years ago

Which is the same as 15.39 + 100?

enot [183]

1.530 divided by 10 is your answer

3 0

2 years ago

Read 2 more answers

5.21 as a decimal rounded

frutty [35]

5.21 would just be 5 or 5.2 depending on what place you need to round.

3 0

2 years ago

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2(x-1) ≥ 10 or 3-4x > 15<br> Solve the compound inequality, and show work!

Veseljchak [2.6K]

Answer:

$x\geq 6\text{ or } x$

Step-by-step explanation:

We have the compound inequality:

$2(x-1)\geq10\text{ or } 3-4x>15$

Let's solve each of them individually first:

We have:

$2(x-1)\geq10$

Divide both sides by 2:

$x-1\geq5$

Add 1 to both sides:

$x\geq6$

We have:

$3-4x>15$

Subtract from both sides:

$-4x>12$

Divide both sides by -4:

$x$

Hence, our solution set is:

$x\geq 6\text{ or } x$

5 0

2 years ago

Read 2 more answers

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