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Shalnov [3]
3 years ago
9

Mr.Ramirez teaches dance. He has 18 sixth-grade students and 24 seventh-grade students. He wants to put the students in equal gr

oups. Each group will have students of only one grade level. How many students should be in each group? How many of each group will be there?
Mathematics
2 answers:
kifflom [539]3 years ago
7 0
He could have groups of 6, and have three groups in sixth grade and 4 in 7th grade.
Sever21 [200]3 years ago
6 0

Answer:

groups of 6 students.

Step-by-step explanation:

Mr. Ramirez has 18 sixth grade students and 24 seventh grade students.

He wants to put the students in equal group of only one grade students.

So we calculate the HCF of the numbers.

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

= 2 × 3

HCF = 6

Hence there would be 6 students group for each grade

6th grade = 6 + 6 + 6  = 18 students

7th grade = 6 + 6 + 6 + 6  = 24 students

Total 7 groups of 6 students would be made.

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Mike and Danielle each improve their yards by planting hospice and shrubs. They bought their supplies from the same store. Mike
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Answer: each hospice costs $4 and each shrub costs $9

Step-by-step explanation:

Let x represent the cost of one hospice.

Let y represent the cost of one shrub.

Mike spent $71 on four hospice and seven Travis. This means that

4x + 7y = 71- - - - - - - - - - - - -1

Danielle spent $53 on 13 hospice and three shrubs. This means that

13x + 3y = 53- - - - - - - - - - - 2

Multiplying equation 1 by 13 and equation 2 by 4, it becomes

52x + 91y = 923

52x + 12y = 212

Subtracting, it becomes

79y = 711

y = 711/79

y = 9

Substituting y = 9 into equation 1, it becomes

4x + 7 × 9 = 71

4x + 63 = 71

4x = 71 - 63

4x = 8

x = 8/4 = 2

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Step-by-step explanation:

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Can someone help me solve (a) and (c) pls.<br>Thanks ​
Alex777 [14]

Answer:

Area of ABCD = 959.93 units²

Step-by-step explanation:

a). By applying Sine rule in the ΔABD,

  \frac{\text{SinA}}{46}=\frac{\text{Sin}\angle{DBA}}{35}

  \frac{\text{Sin110}}{46}=\frac{\text{Sin}\angle{DBA}}{35}

 Sin∠DBA = \frac{35\times \text{Sin}(110)}{46}

 m∠DBA = \text{Sin}^{-1}(0.714983)

 m∠DBA = 45.64°

 Therefore, m∠ADB = 180° - (110° + 45.64°) = 24.36°

                  m∠ADB = 24.36°

c). Area of ABCD = Area of ΔABD + Area of ΔBCD

  Area of ΔABD = AD×BD×Sin(\frac{24.36}{2})

                           = 35×46Sin(12.18)

                           = 339.68 units²

   Area of ΔBCD = BD×BC×Sin(\frac{59.92}{2})°

                            = 46×27×(0.4994)

                            = 620.25 units²

   Area of ABCD = 339.68 + 620.25

                            = 959.93 units²

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