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3 years ago

~Please help me out~

Mathematics

2 answers:

shtirl [24]3 years ago

8 0

Answer:

C) 54

And thanks I really needed it!

Oxana [17]3 years ago

4 0

Answer:

pretty sure the answer is C) 54

Step-by-step explanation:

I'm in 10th grade and I been in advanced math for my entire life

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What is 24% of 1350 I will give 75 points

nika2105 [10]

<h3>Answer:</h3><h2>324.</h2><h3>Step-by-step explanation:</h3>

To find 24% of 1,350, you must multiply 24% by 1,350.

To do this problem, we must turn the numbers to fractions.

Twenty-Four hundredths, 24 / 100 is your fraction for 24%.

Put 1,350 over 1 since 1,350 is a whole number.

So:

24 / 100 x 1,350 / 1.

1350 x 24 = 32400, 100 x 1 = 100.

Your answer is 32,400 / 100.

Get rid of the 2 zeroes in 400 and 100.

It should be 324 / 1.

Since 1 is just the number 1, the answer is 324.

If you have any questions, feel free to comment below.

Merry Christmas!

8 0

3 years ago

Read 2 more answers

Sum one please help need answerssssss pleaseeeeee

erma4kov [3.2K]

Answer:

Step-by-step explanation:

look it up

3 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.

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