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adell [148]
2 years ago
14

a square dance set requires 4 couples 8 dancers with each couple standing on one side of a square. There are 250 people at a squ

are dance. What is the greatest number of sets possible at the dance?
Mathematics
1 answer:
sveta [45]2 years ago
7 0

Given:

1 set requires 4 couples 8 dancers.

Total number of people at a square dance = 250.

To find:

The greatest number of sets possible at the dance.

Solution:

We have,

Total people = 250

1 set = 8 people.

\text{Number of possible sets}=\dfrac{\text{Total people}}{\text{People required for 1 set}}

\text{Number of possible sets}=\dfrac{250}{8}

\text{Number of possible sets}=31.25

Number of possible sets cannot be a decimal or fraction value. So, approx. the value to the preceding integer.

\text{Number of possible sets}\approx 31

Therefore, the number of possible sets at the dance is 31.

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3 years ago
What is the value of x?
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ΔTRQ is an isosceles right triangle, so if we find the value of RT then the value of 'x' will be the same

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3 0
3 years ago
Please help me with this
Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

4 0
3 years ago
Read 2 more answers
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