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SVEN [57.7K]
3 years ago
15

What is the equation of the line? 1 x - 4 y = 2x - 4 3x + 2 y = 2x + 2

Mathematics
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

4

Step-by-step explanation:

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The area of a square garden is 98 m2. How long is the diagonal?
stiks02 [169]

Answer:

The diagonal  is about 14 m

Step-by-step explanation:

A = s^2

98 = s^2

so

s = √98

diagonal c^2 = 98 + 98

c^2 = 196

c = √196

c  ≈ 14

Answer:

The diagonal  is about 14 m

8 0
3 years ago
Read 2 more answers
Line segment JK measures 17cm. Segments JK was translated 4 units right and 8 units up for for J'K'. What is the measurement of
castortr0y [4]

Answer:

c

Step-by-step explanation:

8 0
2 years ago
A hose fills an empty fish tank with 24 gallons of water in 8 minutes. At that same rate, how many minutes will it take the hose
kolezko [41]

Answer:

\frac{8}{24} = \frac{x}{60}

24x = 8*60

24x = 480

x = 20

Step-by-step explanation:

3 0
2 years ago
if mya had 20 candy and ellie had 100000000000000000000000000 candy how many dose ellie and mya have togeather
jolli1 [7]

Answer:

100000000000000000000000020

Step-by-step explanation:

There you go

3 0
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
8 0
3 years ago
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