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2 years ago

PLEASE HELP ASAP!!! will mark brainlest!

Mathematics

1 answer:

N76 [4]2 years ago

3 0

Here are the coordinates, mark the ones on the left as x, and the ones on the right as y, but make sure they are lined up with each other: (-2, 11), (-1, 6), (0, 3), (1, 2), (2, 3), (3, 6). Then plot the points on the graph and connect them.

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Solve this equation algebraically!!

NeTakaya

Answer:

See Explanation

Step-by-step explanation:

$1 - \sqrt{x} \neq \: 0 \\ \\ \therefore \: 1 \neq \sqrt{x} \\ \\ {1}^{2} \neq ({ \sqrt{x}) }^{2} \\ \\ 1 \neq \: x \\$

Hence, x will take values other than 1.

5 0

3 years ago

Is 1.1 less than or greater than or equal to the square root of two

dangina [55]

Answer:

less than

Step-by-step explanation:

the Square root of 2 is 1.4142...

3 0

3 years ago

A triangle has the side lengths of 12, 34, 45. Is it a right triangle?

deff fn [24]

Answer:

Step-by-step explanation:

a^2+b^2=c^2?

12^2+34^2=45^2?

1300≠2025

5 0

3 years ago

PLEASE HELP IM STUCK

kkurt [141]

Answer:

y= 2x + 10

Step-by-step explanation:

The variable depends on the $2 so, the green box should be 2

4 0

1 year ago

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

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

8 0

3 years ago