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3 years ago

PLEASE HELP ME I need help so please help me

Mathematics

2 answers:

Y_Kistochka [10]3 years ago

8 0

Answer:

2/7

Step-by-step explanation:

This is because the slope is positive so that gets rid of both of the negatives, and the line is not straight but if the line was horizontally straight then your slope would be zero. So I just did process of elimination.

Anastaziya [24]3 years ago

4 0

Answer:

The answer is -2/7

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Write an equation of the line passing through the points (4,6) and (-2, - 18).

olga_2 [115]

Answer:

y = 4x – 10

Step-by-step explanation:

slope formula: $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

point-slope formula: $y-y_{1}=m\left(x-x_{1}\right)$

Use slope formula and plug in your numbers: $m = \frac{-18-6}{-2-4 }$
Solve for slope $m = 4$
Use the point-slope formula to solve for the equation: $y-6=4\left(x-4\right)$
Solve and find equation: $y=4x-10$

Note: When given two points you only need to pick one for the point-slope formula! Best to choose the positive one so you don't have to deal with the negatives!

Hope this helps! Please give Brainliest!

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2 years ago

3 1/2 divided by 2 1/4

liberstina [14]

3 1/2 divided by 2 1/4 = 1.55555556

1.55555556 Rounded is 1.55

8 0

3 years ago

Write down 1 16 as (a) a decimal, [1] (b) a percentage.

Ede4ka [16]

Answer:

(a) 1.0625

(b) 6.25% percentage

Step-by-step explanation:

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2 years ago

QUICK HELP!!! How manny differences can you spot? 8 6 7 4 3 5

Eduardwww [97]

Answer:

i can only see 3

Step-by-step explanation:

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3 years ago

Read 2 more answers

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago