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masha68 [24]
2 years ago
7

There is a row of 15 trees in a garden. Starting from the first tree, a gardener marked every second tree, then, on his way back

, he marked the first tree, and then every third tree. How many trees are left without a mark?
Mathematics
2 answers:
cluponka [151]2 years ago
7 0
<h3>Answer:  5</h3>

=====================================================

Explanation:

The gardener starts at the first tree and then marks every second tree. So they'll skip a tree each time.

The set of trees he marks has the labels {1,3,5,7,9,11,13,15}. Basically it's the set of odd whole numbers between 1 and 15. Or you could notice that we add on 2 each time we need another tree to mark.

Then he goes back to the first tree and marks every third tree. So we have this new set: {1,4,7,10,13}. Start at 1 and add 3 each time to generate a new item.

Union the two sets mentioned and you'll get {1,3,4,5,7,9,10,11,13,15}

We see that the following trees aren't marked: {2,6,8,12,14}

There are <u>  5  </u> items in that last set.

Side note: The trees with labels {1,7,13} have been marked twice. There's a gap of 6 units between each adjacent label.

zzz [600]2 years ago
4 0

Answer:2

Step-by-step explanation:

15/3 = 5

15/2 = 7.5

+1

= 13

15 - 13 = 2

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For the high school's production of "Grease," 490 tickets were sold. The cost of a student
damaskus [11]

<em><u>The system of equations could  be used to determine the number of each type of ticket sold is:</u></em>

s + p + t = 490

8s + 10p + 12t = 4600

s + p = 6t

<em><u>Number of each type of ticket sold is:</u></em>

number of student tickets sold = 220

number of pre-sale non student ticket sold = 200

number of ticket sold at door = 70

<em><u>Solution:</u></em>

Let "s" be the number of student tickets sold

Let "p" be the number of pre-sale non student ticket sold

Let "t" be the number of ticket sold at door

Cost of 1 student ticket = $ 8

Cost of 1 pre-sale non student ticket = $ 10

Cost of 1 ticket sold at door = $ 12

From given information,

<u><em>490 tickets were sold. So we can frame a equation as:</em></u>

number of student tickets sold + number of pre-sale non student ticket sold + number of ticket sold at door = 490

s + p + t = 490 ---------- eqn 1

<em><u>Also given that, The  total income from ticket sales was $4600</u></em>

So we can frame a equation as:

number of student tickets sold x Cost of 1 student ticket + number of pre-sale non student ticket sold x Cost of 1 pre-sale non student ticket + number of ticket sold at door x Cost of 1 ticket sold at door = $ 4600

s \times 8 + p \times 10 + t \times 12 = 4600

8s + 10p + 12t = 4600 ------ eqn 2

<em><u>The combined number of student and pre-sale  tickets was 6 times more than the tickets sold at the door</u></em>

s + p = 6t --------- eqn 3

So eqn 1 eqn 2 and eqn 3 could  be used to determine the number of each type of ticket sold

<em><u>Let us solve eqn 1, eqn 2, eqn 3 to find values of "s" "p" and "t"</u></em>

Substitute eqn 3 in eqn 1

6t + t = 490

7t = 490

<h3>t = 70</h3>

Substitute t = 70 in eqn 2

8s + 10p + 12(70) = 4600

8s + 10p = 3760 ----- eqn 4

Substitute t = 70 in eqn 3

s + p = 6(70)

s + p = 420  ----- eqn 5

Multiply eqn 5 by 8

8s + 8p = 3360  ----- eqn 6

Subtract eqn 6 from eqn 4

8s + 10p = 3760

8s + 8p = 3360

(-) ---------------------

2p = 400

<h3>p = 200</h3>

Substitute p = 200 in eqn 5

s + 200 = 420

<h3>s = 220</h3>

<em><u>Summarizing the results:</u></em>

number of student tickets sold = 220

number of pre-sale non student ticket sold = 200

number of ticket sold at door = 70

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Answer:

1/8 is the missinh part

Step-by-step explanation:

1/8+1/2+3/8= 1 whole

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