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2 years ago

In the arthmetic sequence {-9, -4, 1, 6, 11,…}, what is the common difference ?

2 answers:

7 0

The common difference is 5 each time. Take for example -9 and -4 it’s goes up 5, and then 1 and 6 it goes up five and so on,hope this helped!

GuDViN [60]2 years ago

6 0

<h3>Answer: 5</h3>

Explanation:

Pick any term and subtract off the previous one to get the common difference.

For instance, let's say we picked 11

11-6 = 5

Or we could pick 6

6-1 = 5

Each time we get 5 as the common difference. It says that to get each next term, we add on 5.

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If amy sends 20emalis in 5minutes what ia the init rate

Maksim231197 [3]

I'm pretty sure it would be 4 20÷5=4

7 0

3 years ago

Read 2 more answers

The cost of a ticket to the circus is $19.00 for

DENIUS [597]

Answer:

The number of tickets bought is 1000 adults tickets and 700 children were present

Explanation:

Let the number of children and adults present in the circus be x and y respectively

According to question,

Then, x + y = 1700 ……………..equation (1)

And, cost of ticket to circus for children and adult be 19, 42 respectively

Total cost of ticket;

for adults = 42y; for children = 19x

19x + 42y = 55300 …………………equation (2)

Multiplying equation (1) by 19

19x + 19y = 32300 ……………… eqn (3)

Subtracting equation 1 and 3, we get

23y = 23000

y = 1000

putting this value in eq (1), we get x = 700

Therefore, 1000 adults and 700 children were present .

5 0

3 years ago

A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$