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Nuetrik [128]
2 years ago
5

I need help with math​

Mathematics
1 answer:
Doss [256]2 years ago
3 0

Answer:

E) |g-540|

Step-by-step explanation:

E) |g-540|

If it weighed a bit over, say 545, you would get |545-540|=|5|=5

If it weight a little under, say 535, you would get |535-540|=|-5|=5

Either way, it tells you how far away from 540 the package is in grams from the desired weight.

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The annual energy consumption of the town where Camilla lives in creases at a rate that is onal at any time to the energy consum
boyakko [2]

Answer:

6.575 trillion BTUs

Step-by-step explanation:

<em>Let represent the annual energy consumption of the town as E</em>

<em>The rate of annual energy consumption *  energy consumption at time past</em>

<em>dE/dt * E</em>

<em>dE/dt =K</em>

<em>k = the proportionality constant</em>

<em>c= the integration constant</em>

<em>(dE/dt=) kdt</em>

<em>lnE = kt + c</em>

<em>E(t) = e^kt+c ⇒ e^c e^kt  e^c is a constant, and e^c = E₀</em>

<em>E(t) = E₀ e^kt</em>

<em>The initial consumption of energy is E(0)=4.4TBTU</em>

<em>set t = 0 then</em>

<em>4.4 = E₀ e⇒ E₀ (1) </em>

<em>E₀ = 4.4</em>

<em>E (t) = 4.4e^kt</em>

<em>The consumption after 5 years is t = 5, e(5) = 5.5TBTU</em>

<em>so,</em>

<em>E(5) = 5.5 = 4.4e^k(5)</em>

<em>e^5k = 5/4</em>

<em>We now take the log 5kln = ln(5/4)</em>

<em>5k(1) = ln(5/4)</em>

<em>k = 1/5 ln(5/4) = 0.04463</em>

<em>We find  the town's annual energy consumption, after 9 years</em>

<em>we set t=9  </em>

<em>E(9) = 4.4e^0.04463(9)</em>

<em>= 4.4(1.494301) = 6.5749TBTUs</em>

<em>Therefore the annual energy consumption of the town after 9 years is </em>

<em>= 6.575 trillion BTUs</em>

<em />

3 0
3 years ago
Is 2.5(3+x ) equivalent to 2.5x + 7.5
Likurg_2 [28]
Yes, you are correct
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Answer:

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Step-by-step explanation:

2 x - 1/2 = -1

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