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Svetach [21]
3 years ago
15

A quantity with an initial value of 7600 decays exponentially at a rate of 55% every 7 days. What is the value of the quantity a

fter 2 weeks, to the nearest hundredth?
Mathematics
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

1539

Step-by-step explanation:

We solve the above question using the Exponential decay formula

= A(t) = Ao(1 - r) ^t

Ao = Initial Amount invested = 7600

r = Decay rate = 55% = 0.54

t = time in weeks = 2

Hence:

A(t) = 7600(1 - 0.55)²

A(t) = 7600 × (0.45)²

A(t) = 1539

Therefore, the value of the quantity after 2 weeks is 1539

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zysi [14]

lol

Step-by-step explanation:

4 0
3 years ago
The Ferris wheel has a radius of 70 feet. The distance between the wheel and the ground is 4 feet. The rectangular coordinate sy
Gemiola [76]

Answer:

x² + (y-74)² = 4900

Step-by-step explanation:

Circle equation: (x-h)² + (y-k)² = r²

"h" refers to horizontal shifts while "k" refers to vertical shifts. If the center of the wheel is directly above the origin of the rectangular plane and the entire wheel is 4 ft. from the ground then:

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So plugging those into the equation of circles, we get the answer above.

7 0
3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
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4 0
3 years ago
An even function must always have what characteristic?
Gelneren [198K]
D because d is always right hahahahaha
6 0
3 years ago
5 1/2 divided by 3 1/3
Sophie [7]

Answer:

1 13/20

Step-by-step explanation:

5 1/2 ÷ 3 1/3

11/2 ÷ 10/3
11/2 x 3/10 = 33/20

33/20 = 1 13/20

Hope this helps :)

5 0
2 years ago
Read 2 more answers
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