All categories

3 years ago

8/9-5/6 what is the answer. Need help please

Mathematics

1 answer:

kkurt [141]3 years ago

6 0

Answer:

yes i and the kids were going through a different time of their own food but the food was good but they had to make it very

Step-by-step explanation:

he is a good boy he has to go through a different color he is very good and he’s a good boy he has to go through the

You might be interested in

Day one finding a weeb finish this sentence and I will give you brainiest

Hatshy [7]

Answer:

weeb -_-

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please help me with this translating trigonometry graphs question. Brainliest and Points Available.

andreev551 [17]

According with the definition of translation, we conclude that the equations of graphs M and N are m(x) = f(x - 5) and n(x) = f(x) - 2, respectively.

<h3>How to apply translations on a given function</h3>

Rigid transformations are transformation such that the Euclidean distance of every point of a function is conserved. Translations are a kind of rigid transformations and there are two basic forms of translations:

Horizontal translation

g(x) = f(x - k), k ∈ $\mathbb {R}$ (1)

Where the translation goes rightwards for k > 0.

Vertical translation

g(x) = f(x) + k, k ∈ $\mathbb {R}$ (2)

Where the translation goes upwards for k > 0.

According with the definition of translation, we conclude that the equations of graphs M and N are m(x) = f(x - 5) and n(x) = f(x) - 2, respectively.

To learn more on translations: brainly.com/question/17485121

#SPJ1

3 0

2 years ago

Add the product of 5 and 6 with the product of 3 and 7

xz_007 [3.2K]

It is 51.

5x6=30

3x7=21

30+21=51

Hope that I was of help.

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.

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