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Trava [24]
4 years ago
14

Last year the girl's basketball team had 8 fifth-grade students and 7 sixth-grade students.What was the ratio of sixth-grade stu

dents to fifth-grade students on the team?
Mathematics
1 answer:
Rashid [163]4 years ago
4 0

Answer:

The ratio of sixth-grade students to fifth-grade students on the team was <u>7 : 8</u>.

Step-by-step explanation:

Given:

The girl's basketball team had 8 fifth-grade students and 7 sixth-grade students.

Now, to find the ratio of sixth-grade students to fifth-grade students on the team.

<em>Number of fifth-grade students = 8.</em>

<em>Number of sixth-grade students = 7.</em>

Now, to get the ratio of sixth-grade students to fifth-grade students on the team :

Number\ of\ sixth-grade\ students:Number\ of\ fifth-grade\ students

=7:8.

Therefore, the ratio of sixth-grade students to fifth-grade students on the team was 7 : 8.

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The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Addition

   c) AdditionSum

   d) in this case we have two possibilities

       * If we move to the right the addition

       * If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

        -3 -1 = -4

        -4 -1 = -5

        -7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) 0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}  ...

In this case we can see more clearly the sequence when writing in this way

      0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}

each term is found by adding 1 to the current term,

      \sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}

Therefore the mathematical operation is the addition

c)   ... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...

      The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

  • To move to the right add 1

           -\frac{-10}{2} + 1 = \frac{-10}{2}  -   \frac{2}{2}  = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}

  • To move left subtract 1

         \frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}

         

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Sum

   c) Sum

   d) This case we have two possibilities

  •  If we move to the right the sum
  •  If we move to the left we subtract

Learn more here: brainly.com/question/4626313

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Answer:

The answer to your question is:th first option is correct.

Step-by-step explanation:

Here we have and hyperbola with center (0, 1), and the hyperbola is horizontal because x² is positive.

Equation

y - k = ±\frac{b}{a} (x - h)

Process

Find a, b

                a² = 9    

               a = 3

                b² = 5

              b = √5

              h = 0    and k = 1

Substitution

              y - 1 = ±\frac{√5}{3} (x - 0)

Equation 1

            y = \frac{\sqrt{5} }{3} x + 1

Equation 2

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