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3 years ago

Who can helppp ?????!

Mathematics

1 answer:

Elina [12.6K]3 years ago

3 0

Answer:

well i do not know and would not know what those ovals are or whatever they are but the dude before is right .

Step-by-step explanation:

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The volume of a cube decreases at a rate of 0.5ft cubed per minute. What is the rate of change of the side length when the side

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The solution to the problem is as follows:

a cube has sides s

V = s^3 ---> differentiating

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3 years ago

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Step-by-step explanation:

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Read 2 more answers

Sarah is carrying out a series of experiments which involve using mcreasing amounts of a chemical. In the

Marianna [84]

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:

(a)963 grams

(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:

First term, a =6g

Common difference. d= 7.8 -6 =1.8 g

Therefore:

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$

Therefore, the greatest possible number of experiments satisfies the inequality

$1.3^N \leq 91$

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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3 years ago