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GrogVix [38]
2 years ago
14

Colin invests £2100 into his bank account. He receives 5% per year simple interest. How much will Colin have after 3 years? Give

your answer to the nearest penny where appropriate.
Mathematics
1 answer:
GaryK [48]2 years ago
3 0

Answer:£2415

Step-by-step explanation:

2100x0.01= 21 = 1%

21x5= 105 = 5%

105x3= 315 = 5% for 3 years

2100+315=2415

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Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse func
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Answer:

<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>

Step-by-step explanation:

Given the function  j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6}, to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

j(k(x)) = j[(ln x/11.6)]

j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}

j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)

j[(ln x/11.6)] = 11.6 * x/11.6

j[(ln x/11.6)] = x

Hence j[k(x)] = x

Similarly for k[j(x)];

k[j(x)] = k[11.6e^x]

k[11.6e^x] = ln (11.6e^x/11.6)

k[11.6e^x]  = ln(e^x)

exponential function will cancel out the natural logarithm leaving x

k[11.6e^x]  = x

Hence k[j(x)] = x

From the calculations above, it can be seen that j[k(x)] =  k[j(x)]  = x, this shows that the functions j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6} are inverse functions.

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