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Alona [7]
3 years ago
9

Find the inverse of g(n) = - 3 - 1

Mathematics
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

g'(n) = -\frac{n+1}{3}

Step-by-step explanation:

Given

g(n) = -3n - 1

Required

Determine the inverse

g(n) = -3n - 1

Replace g(n) with y

y = -3n - 1

Swap positions of x and y

n = -3y - 1

Make y the subject

3y = -n - 1

3y = -(n + 1)

y = -\frac{n+1}{3}

Replace y with g'(n)

g'(n) = -\frac{n+1}{3}

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The graph shows g(x), which is a translation of f(x) = x². Write the function rule for g(x).
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Answer:

g(x) = (x + 8)^{2}

Step-by-step explanation:

7 0
2 years ago
Rewrite the expression by collecting like terms.<br> 1/2k - 3/8k
pychu [463]

Answer:

1/8k

Step-by-step explanation:

1/2k= 4/8k

4/8k-3/8k=1/8k

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2 years ago
A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 ​min, t
ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

T(t)=T_a+(T_0-T_a)e^{-kt}

where T_a is the ambient temperature, T_0 is the initial temperature, t is the time and k is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say T_0

We know that the ambient temperature is T_a=65, so

T(t)=65+(T_0-65)e^{-kt}

We also know that when t=5 \:min the temperature is T(5)=35 and when t=10 \:min the temperature is T(10)=50 which gives:

T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}

T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}

Rearranging,

35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}

50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}

If we divide these two equations we get

\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}

\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}

Now, that we know the value of k we can use it to find the initial temperature of the beam,

35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5

so the beam started out at 5 °F.

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x = <em>20</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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3 years ago
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