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cluponka [151]
2 years ago
8

#13 w is the solution of the equation. 3 = b + 3

Mathematics
2 answers:
stealth61 [152]2 years ago
7 0
I'm assuming that you meant what is the solution of the equation.

b would equal 0.

This is because it is a simple one step equation; subtract 3 from both sides which leaves you with 0 = 3.

Hope this helps!
Ulleksa [173]2 years ago
3 0
3-3=b+3-3
solve each side
0=b
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Find an exponential function that passes through each pair of points.<br><br>(2,1.75) &amp; (-2,28)
dusya [7]

Answer:

f(x) = 7\, ((1/2)^{x}).

Step-by-step explanation:

An exponential function is typically in the form f(x) = a\, (b^{x}), where a and b (b > 0) are constants to be found.

In this question:

f(2) = 1.75 means that a\, (b^{2}) = 1.75.

f(-2) = 28 means that a\, (b^{-2}) = 28.

Divide one of the two equations by the other to eliminate a and solve for b.

The number of the right-hand side of the second equation is larger than that of the first equation. Hence, divide the second equation with the first:

\displaystyle \frac{a\, (b^{-2})}{a\, (b^{2})} = \frac{28}{1.75}.

\displaystyle b^{-4} = 16.

b^{-1} = 2.

\displaystyle b = \frac{1}{2}.

Substitute b = (1/2) back into either equation (for example, the first equation) and solve for a:

a\, ((1/2)^{2}) = 1.75.

a = 7.

Substitute a = 7 and b = (1/2) into the other equation. That equation should also be satisfied.

Therefore, this function would be:

f = 7\, ((1 / 2)^{x}).

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Stolb23 [73]

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3 years ago
A sensor output was acquired for 32 seconds at a rate of 200 Hz and spectral analysis was performed using FFT. If the data set w
VashaNatasha [74]

Complete Question

A sensor output was acquired for 32 seconds at a rate of 200 Hz and spectral analysis was performed using FFT. If the data set was split into 5 segments (each 6.4 seconds long), what is the resulting:

a)F minimum

b) F maximum

c) Frequency resolution

Answer:

a) Fmax=100Hz

b) Fmin=0Hz

c) F_r=0.0313

Step-by-step explanation:

From the question we are told that:

Time t=32sec

FrequencyF=200Hz

Segments \mu=5

Generally the equation for Frequency Range is mathematically given by

Fmax=\frac{F}{2}

Fmax=100Hz

Therefore

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b) Fmin=0Hz

c)

Generally the equation for Frequency Resolution is mathematically given by

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Where

N=The Total dat points

N=Sampling Frequency *Time

N=200*32

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Therefore

F_r=\frac{200}{6400}

F_r=0.0313

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