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Marina CMI [18]
3 years ago
8

Which of the quotients are equivalent to -(48/17)

Mathematics
1 answer:
bixtya [17]3 years ago
5 0

Answer:

oiie4546

Step-by-step explanation:

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What is d,the common difference,in the arithmetic sequence 100,98,96,94,92...?
san4es73 [151]

Answer:

-2

Step-by-step explanation:

4 0
3 years ago
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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

3 0
3 years ago
What is the cost of 2 kilograms of flour if 3 kilograms cost $4.86 and the unit price for each packages of flour is the same
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The answer is: $3.24.

\frac{4.86}{3}  =  \frac{x}{2}
\frac{9 .72}{3}  =  \frac{3x}{3}
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8 0
4 years ago
Record your answers as asum of unitfractions.
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1 2/3 = 5/3
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3 years ago
New York City is a popular field trip destination. THis year the senior class at High School A and the senior class at High Scho
maw [93]

Answer:

number of students in a van=12

number of students in a bus=52

Step-by-step explanation:

let number of students in a van=x

number of students in a bus=y

1x+6y=324     ...(1)

9x+3y=264   ...(2)

multiply (2) by 2

18x+6y=528   ...(3)

(3)-(1) gives

17x=204

x=204/17=12

put in (1)

12+6y=324

6y=324-12=312

y=312/6=52

8 0
3 years ago
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