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3 years ago

Li Na is going to plant 63tomato plants and 81 rhubarb plants.

Mathematics

1 answer:

Ivan3 years ago

6 0

The greatest number of rows Li Na can plant is 9

<h3><u>Solution:</u></h3>

Given that Li Na is going to plant 63 tomato plants and 81 rhubarb plants

Li Na would like to plant the plants in rows where each row has the same number of tomato plants and each row has the same number of rhubarb plants

<em><u>To find: greatest number of rows Li Na can plant</u></em>

To find the greatest number of rows of plants we have to list the factors of 63 and 81 and find the greatest common one

The factors of 63 are: 1, 3, 7, 9, 21, 63

The factors of 81 are: 1, 3, 9, 27, 81

Then the greatest common factor of 63 and 81 is 9

Thus greatest number of rows Li Na can plant is 9

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The midpoint of y and 31 is -5. what is y

pochemuha

Y = -41.
EXPLANATION.
if -5 is the midpoint, it must be -5 + 36.
Because there is 36 numbers between 31 and -5.
So -5 + 36 = -41.

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3 years ago

Whats the domain of this function?

nika2105 [10]

Answer:

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3 years ago

Alberto ran a race at a speed of 9 meters per second. Omar had a 6 meter head start and ran at a speed of 7 meters per second. B

zloy xaker [14]

Answer: Alberto run 27 meters.

Time taken By Alberto=3 seconds

Step-by-step explanation:

Let d be the distance Alberto ran to catch Omar.

Then, distance ran by Omar=x-6

Also, $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

Since, both started running at the same time.

Time taken By Alberto= $\frac{x}{9}$

Time taken By Omar= $\frac{x-6}{7}$

Now, at the point they meet time taken by both is equal.

$\frac{x}{9}=\frac{x-6}{7}\\\Rightarrow\ 7x=9(x-6)\\\Rightarrow7x=9x-54\\\Rightarrow9x-7x=54\\\Rightarrow\ 2x=54\\\Rightarrow\ x=27$

Hence, Alberto run 27 meters.

Time taken By Alberto= $\frac{27}{9}$ =3 seconds

6 0

3 years ago

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$