Use the expression 5(6 + 4x) to answer the following:

The man is envious because he's jealous of the dog's fur. For he has natural covering and is kept warm because of it. The man has to wear several layers of clothing to be warm while the dog is warm enough with his fur/natural covering.

8 0

3 years ago

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

The sum of two consecutive multiples of 5 is 55.what are the multiples

wariber [46]

Answer:

25 and 30

Step-by-step explanation:

Let the smaller consecutive multiples of 5 be x. Therefore, other consecutive multiples will be x + 5.

Now as per statement the sum of two consecutive multiples of 5 is 55. To find the multiples. Thus

x + x + 5 = 55

2x + 5 = 55

2x = 55 - 5

2x = 50

x = 50/2

x = 25

This the smaller consecutive multiples of 5 is 25, the other consecutuve multiple is x+ 5, 25 + 5 = 30.

The consecutive multiple numbers of 5 are 25 and 30

Answer the two consecutive multiples of 5 are 25 and 30

5 0

3 years ago

Read 2 more answers