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2 years ago

A line that includes the point (-8,-3) has a slope of -1/9. What is it’s equation in point-slope form?

Mathematics

1 answer:

yulyashka [42]2 years ago

4 0

Y+3=-1/9(X+8) is the equation

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Simplify -10x - x plzzzz help

Marta_Voda [28]

hi <3

this is a lot simpler if you just ignore the variable (basically the letter)

now you have -10 - 1

which gives you -11

now just add back on the variable to give you your answer of:

-11x

hope this helps :)

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2 years ago

13u - v + 8 Coefficients.

Aleks04 [339]

Answer:

the coefficient in the equation is 13

Step-by-step explanation:

a coefficient is a number that is in front of a variable. 13 is in front of the variable u so it is the coefficient.

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3 years ago

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Angie earns $55000 per year. Each year she revives a $1700 increase in annual salary. What will Angie's annual salary be for the

brilliants [131]

If she starts with 55000 per year, and she get 1700 per year extra, after fifteen years this extra will be 1700*15, or 25500. Add this to her original salary of 55000 per year to get $80500.

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2 years ago

What is the answer for x ?

Bingel [31]

I think its probaly going to be 6

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2 years ago

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Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

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3 years ago