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Delicious77 [7]
2 years ago
14

The perimeter of a rectangular garage is 28 m it is 5 m wide how long is it

Mathematics
1 answer:
Ronch [10]2 years ago
3 0

Answer:

The garage is 9 m long.

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PLEASE ANSWER THIS QUESTION RIGHT AND THE ANSWER CHOICES ARE IN THE TOP Elize is scuba diving in plans to leisurely Ascend to
yulyashka [42]

anyone know that answer ?

7 0
3 years ago
If the graph of a line goes from the upper left corner of the graph to the lower right corner what can you say about its slope ?
GREYUIT [131]

Answer: You can say that the slope is negative since the graph is decreasing.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t
Rzqust [24]

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

<h3>What is sequence ?</h3>

Sequence is collection of  numbers with some pattern .

Given sequence

a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20

We can see that

\frac{a_1}{a_2}=\frac{-10}{5}=-2\\

and

\frac{a_2}{a_3}=\frac{20}{-10}=-2\\

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now n^{th}  term of this Geometric progression can be written as

T_{n}= 5\times(-2)^{n-1}

So summation of 15 terms can be written as

\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

To learn more about Geometric progression visit : brainly.com/question/14320920

8 0
2 years ago
The fish population of a lake is decreasing each year. A study is conducted. But, unfortunately some data was lost. The research
Solnce55 [7]

Answer:

F(t) = 18000(0.6666)^{t}

Step-by-step explanation:

The fish population after t years can be modeled by the following equation:

F(t) = F(0)(1-r)^{t}

In which F(0) is the initial population and r is the constant rate of decay.

Year one the fish population was 18000

This means that F(0) = 18000

In year three the fish population was 8000 fish.

Two years later, so F(2) = 8000

F(t) = F(0)(1-r)^{t}

8000 = 18000(1-r)^{2}

(1-r)^{2} = 0.4444

\sqrt{(1-r)^{2}} = \sqrt{0.4444}

1 - r = 0.6666

r = 0.3334

So

F(t) = 18000(0.6666)^{t}

4 0
3 years ago
Robbie was given $80 to spend at the Fair. His admission to the park costs $15.50 and each ride cost $5. He anticipates the cost
Ad libitum [116K]

Answer:

7 or 8

Step-by-step explanation:

Robbie was given $80.

His admission costs $15.50

80 - 15.50 = $64.50

He anticipates the cost of food at the fair will be $25.

64.50 - 25 = $39.50

What is the maximum number of rides he can take at the Fair?

(Total $ after admission and cost of food/ ride cost)

$39.50 / $5 = 7.9

4 0
3 years ago
Read 2 more answers
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