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Pachacha [2.7K]
3 years ago
8

Consider the sequence 4,-12,36,-108... What is the 9th term in the sequence

Mathematics
2 answers:
JulijaS [17]3 years ago
6 0

Answer:

The 9th term in the sequence is 29

Step-by-step explanation:

can i have brainliest please

kumpel [21]3 years ago
4 0

Answer:

he 9th term in the sequence is 29Step-by-step explanation:

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A rectangle has a length of twenty one cm and a diagonal of twenty nine cm. How wide is the rectangle?
GuDViN [60]

The width of the rectangle become 20 cm.

According to the statement

we have to find the width of the rectangle.

So, For this purpose, we know that the

Rectangle is a parallelogram all of whose angles are right angles.

According to the information:

A rectangle has a length of 21 cm and a diagonal of 29 cm.

Here we use the this formula

d² = l² + w²

So, Substitute the values

29² = 21² + w²

841 = 441 + w²

Now, solve it then

w² = 400

Solve the square root then value becomes

w = 20 cm.

The width of the rectangle become 20 cm.

So, The width of the rectangle become 20 cm.

Learn more about Rectangle here

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3 0
1 year ago
I don't know how to do this
vesna_86 [32]
He could have 30 dimes and 13 quartars
7 0
3 years ago
The equation y=x^2-12x+45 models the number of books y sold in a bookstore x days after an award-winning author appeared at an a
romanna [79]
Answer: The first day the author reaches 100 days is on day 16.

To solve this problem, you could use a graphing calculator to graph the given equation. Then, determine when this line crosses 100. It crosses when x = 15.539. Therefore, we would have to round up to 16 so it is at least 100.

You could use the quadratic equation to solve: 100 = x^2 -12x + 45

Either you will get 16. If you use the quadratic formula, make sure to only use the positive answer.
7 0
3 years ago
In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)
dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about  44622  humpback whales in the year 2018

5 0
3 years ago
5-7:
garri49 [273]
The 3 panes would cost $7.65.
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