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dem82 [27]
3 years ago
10

Rewrite the following as a decimal -5 1/11

Mathematics
1 answer:
Maru [420]3 years ago
3 0

Answer:

-5.09

Step-by-step explanation:

Convert 5 1/11 to an improper fraction.

Exact Form:

-56/11

Decimal Form:

-5.09

Mixed Number Form:

-5 1/11

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Sarah is carrying out a series of experiments which involve using mcreasing amounts of a chemical. In the
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Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the  first experiment she uses 6g of the chemical and in the second experiment she uses 7.8 g of the chemical

(i)Given that the amounts of the chemical used form an arithmetic progression find the total amount of  chemical used in the first 30 experiments

(ii)Instead it is given that the amounts of the chemical used for a geometric progression. Sarah has a  total of 1800 g of the chemical available. Show that the greatest number of experiments possible satisfies the inequality: 1.3^N \leq 91 and use logarithms to calculate the value of N.

Answer:

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(b)N=17

Step-by-step explanation:

(a)

In the first experiment, Sarah uses 6g of the chemical

In the second experiment, Sarah uses 7.8g of the chemical

If this forms an arithmetic progression:

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Total Amount of  chemical used in the first 30 experiments

S_n=\dfrac{n}{2}[2a+(n-1)d] \\S_{30}=\dfrac{30}{2}[2*6+(30-1)1.8] \\=15[12+29*1.8]\\=15[12+52.2]\\=15*64.2\\=963$ grams

Sarah uses 963 grams in the first 30 experiments.

(b) If the increase is geometric

First Term, a=6g

Common ratio, r =7.8/6 =1.3

Sarah has a total of 1800 g

Therefore:

Sum of a geometric sequence

S_n=\dfrac{a(r^N-1)}{r-1} \\1800=\dfrac{6(1.3^N-1)}{1.3-1} \\1800=\dfrac{6(1.3^N-1)}{0.3}\\$Cross multiply\\1800*0.3=6(1.3^N-1)\\6(1.3^N-1)=540\\1.3^N-1=540\div 6\\1.3^N-1=90\\1.3^N=90+1\\1.3^N=91

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1.3^N \leq 91

Next, we solve for N

Changing 1.3^N \leq 91 to logarithm form, we obtain:

N \leq log_{1.3}91\\N \leq \dfrac{log 91}{log 1.3}\\ N \leq  17.19

Therefore, the number of possible experiments, N=17

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