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earnstyle [38]
3 years ago
14

The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is ?

Mathematics
1 answer:
Dahasolnce [82]3 years ago
3 0
I Believe The Answer Is 288.
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Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by
ivanzaharov [21]

Step-by-step explanation:

T=2\pi\sqrt{\frac{l}{g}}

T+\Delta T=2\pi\sqrt{\frac{(l+\Delta l)}{g}} =2\pi\sqrt{\frac{l}{g}}\sqrt{1+\frac{\Delta l}{l}}=2\pi\sqrt{\frac{l}{g}}(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))=T(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So, \Delta T=T\frac{1}{2}\frac{\Delta l}{l}

b. For an increase of 2%, that is:

\frac{\Delta l}{l}=0.02

\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%

7 0
3 years ago
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Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

3 0
3 years ago
What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?
Marrrta [24]

C

the sum to infinity of a geometric series  = \frac{a}{1-r}

where a is the first term and r the common ratio

= \frac{81}{\frac{1}{3} } = 81 × 3 = 243



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3 years ago
EFGH is translated 3 units to the left and 7 units up.<br> What are the coordinates of H'?
Nikitich [7]
Where was H originally?
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3 years ago
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A recipe calls for 5 deciliters of water and 236 milliliters of milk. How many milliliters of liquid does the recipe call for in
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The recipe calls for 736 milliliters of liquid in total.
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3 years ago
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