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DanielleElmas [232]
4 years ago
8

Portia bought 12 bushes and 30 flowers for $240 to plant in her yard. Later, she realized she did not have enough plants and bou

ght 7 more bushes and 10 more flowers for $125. How much did each bush and flower cost?
Mathematics
1 answer:
Anni [7]4 years ago
8 0

Answer: the cost of each bush is $15

the cost of each flower is $2

Step-by-step explanation:

Let x represent the cost of each bush.

Let y represent the cost of each flower.

Portia bought 12 bushes and 30 flowers for $240 to plant in her yard. This is expressed as

12x + 30y = 240- - - - - - - - - - -1

Later, she realized she did not have enough plants and bought 7 more bushes and 10 more flowers for $125. This is expressed as

7x + 10y = 125- - - - - - - - - - -2

Multiplying equation 1 by 7 and equation 2 by 12, it becomes

84x + 210y = 1680

84x + 120y = 1500

Subtracting, it becomes

90y = 180

y = 180/90

y = 2

Substituting y = 2 into equation 1, it becomes

12x + 30 × 2 = 240

12x + 60 = 240

12x = 240 - 60 = 180

x = 180/12

x = 15

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Four universities - 1, 2, 3, and 4 - are participating in a holiday basketball tournament. In first round, 1 will play 2 and 3 w
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Answer:

Part A. {S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Part B. {A} = {1324, 1342, 1423. 1432}

Part C. {B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

Part D.

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Step-by-step explanation:

A. All possible outcomes

There are four teams, each play a semi final where 1 and 2 plays against each other while 3 and 4 plays against each other. Winner of the first semi final can be either 1 or 2 therefore they both can not be in the championship game or in the losers game at the same time same goes for the other semi final.

Using this explanation (1324 denotes: 1 beats 2 and 3 beats 4 in first-round games and then 1 beats 3 and 2 beats 4), All possible outcomes are

{S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

B. Event A in which 1 wins the tournament

From {S} we only have to write the outcomes in which 1 is the first number in 4digit combinations given in part A

{A} = {1324, 1342, 1423. 1432}

C. Event B in which 2 gets into championship game

From {S} we only have to write the outcomes in which 2 is either the first or second digit in 4digit combinations given in part A

{B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

D. Outcomes in (A∪B), (A∩B) and A'

I. (A∪B)

(A∪B) means A union B therefore all we have to do is combine all the members of A and B

(A∪B) = {1324, 1342, 1423, 1432} ∪{2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

II. (A∩B)

(A∩B) means A intersection B in which we have to find the common members of A and B. If there are no common members then the result of (A∩B) is a null set.

(A∩B) =  {1324, 1342, 1423, 1432} ∩ {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

III. {A'}

A' means A compliment, in other words it can be described as all the possible outcomes that are not part of A. So all we do is to subtract outcomes of A from the total possible outcomes S

{A'} =  {S} - {A}

{A'} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231} - {1324, 1342, 1423. 1432}

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

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JA=JB=9 (tangent segments from the same point are congruent)

CK=KB=14 (tangent segments from the same point are congruent)

JK=JB+KB (parts whole postulate)

JK=9+14=23 (substitution, algebra)

LA=LC=10 (tangent segments from the same point are congruent)

LK=LC+CK (parts whole postulate)

LK=10+14=24 (substitution, algebra)

Perimeter of ΔJKL=LK+KL+LJ (perimeter formula for triangles)

Perimeter of ΔJKL=23+24+19=66 (substitution, algebra)

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3 years ago
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