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Snezhnost [94]
3 years ago
13

If an electric kettle uses

Mathematics
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

The energy used in 3.5 hours = 5.25 kilowatts            

Step-by-step explanation:

Energy used by electric kettle:  

  • 2\frac{1}{4}=\frac{9}{4}=2.25 kilowatts

Time taken by electric kettle:

  • 1.5 hours

Electric kettle's rate in kilowatts per hour can be calculated by dividing the spent energy i.e. 2.25 kilowatts by the time taken i.e. 1.5 hours.

Therefore,

Electric kettle's rate = 2.25 kilowatts / 1.5 hour

                                  = 1.5 kilowatts per hour

Thus, the energy used in 3.5 hours = 1.5 × 3.5

                                                           = 5.25 kilowatts

Therefore, the energy used in 3.5 hours = 5.25 kilowatts                        

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Answer:

The mean is 6.2.

Step-by-step explanation:

The "mean" is the same thing as the "average". Essentially, the question is asking for the average of the numbers.

So:

Add up all of the terms. [6 + 11 + 5 + 2 + 7] = 31

You find the average by dividing the sum of the terms (31) by the number of terms (5).

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I hope this helped! :)

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Step-by-step explanation: Hope this helps! :)

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Answer:

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Step-by-step explanation:

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What is the equation of the line that represents the horizontal asymptote of the function f(x)=25,000(1+0.025)^(x)?
posledela

Answer:

The answer is below

Step-by-step explanation:

The horizontal asymptote of a function f(x) is gotten by finding the limit as x ⇒ ∞ or x ⇒ -∞. If the limit gives you a finite value, then your asymptote is at that point.

\lim_{x \to \infty} f(x)=A\\\\or\\\\ \lim_{x \to -\infty} f(x)=A\\\\where\ A\ is\ a\ finite\ value.\\\\Given\ that \ f(x) =25000(1+0.025)^x\\\\ \lim_{x \to \infty} f(x)= \lim_{x \to \infty} [25000(1+0.025)^x]= \lim_{x \to \infty} [25000(1.025)^x]\\=25000 \lim_{x \to \infty} [(1.025)^x]=25000(\infty)=\infty\\\\ \lim_{x \to -\infty} f(x)= \lim_{x \to -\infty} [25000(1+0.025)^x]= \lim_{x \to -\infty} [25000(1.025)^x]\\=25000 \lim_{x \to -\infty} [(1.025)^x]=25000(0)=0\\\\

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