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san4es73 [151]
3 years ago
11

Which theorems or postulates could be used to show ABC =ECD

Mathematics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

?

Step-by-step explanation:

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A company invests $15,000.00 in an account that compounds interest annually. After two years, the account is worth $16,099.44. U
ANEK [815]
Principal = $15000
Amount = $16099.44
Rat of interest = r
Time = 2 years
Then
A = P(1 + r)^t
16099.44 = 15000 (1 + r)^2
(1 + r)^2 = 1.073
(1 + r)^2 = (1.036)^2
Then
1 + r = 1.036
r = 1.036 - 1
  = 0.036
  = 0.036/100
  = 3.6%
From the above deduction, it can be easily concluded that the correct option among all the options that are given in the question is the second option.
6 0
4 years ago
Say what now?! I don’t get this please help!
Nuetrik [128]

Answer:

D)

Step-by-step explanation:

7x + 4 = 3x + 68

7x - 3x = 68 - 4

4x = 64

x = 64/4

x = 16

I Hope I've helped you.

7 0
3 years ago
Read 2 more answers
Mr. Doyle washed
KiRa [710]

Answer:

Mr Doyle washed the most.

4/15 still need to be washed.

Step-by-step explanation:

2/5=6/15  Mr Doyle

1/3=5/15   Son

so Mr Doyle washed the most.

Together they washed 11/15 of the laundry so 4/15 is left to wash.

7 0
3 years 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
65 - 14 = 65-5-______
Fed [463]

Answer: 60? to be hostens i have no clue sorry i hope this help...

Step-by-step explanation:

7 0
3 years ago
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