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Orlov [11]
3 years ago
15

How do you find a meadian?

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0
You first order a set of number smallest to largest. Then you find the middle number. If it is odd, there will be 1 middle number. If it is an even set of numbers, then there would be two middle numbers. You would then find the middle of those 2 numbers.
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Answer:

it is linear bc it has a constant rate of change

Step-by-step explanation:

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Solve this equation, 2x^2-10=0
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What is the solution set of -|-x| = -12
ozzi

Answer:

{-12,12}

Step-by-step explanation:

-|-x| = -12 multiply both sides by -1

|-x| = 12 apply rule for solving absolute value equations

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x = -12 and x =12

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PLZZ HELP ASAP!!!!!!!!!!!
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4 years ago
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Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE
inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to 20\ units

Part 3) The area is equal to 25\ units^{2}

Step-by-step explanation:

we have

A(5,0), B(2,4), C(-2,1),D(1,-3)

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

A(5,0),B(2,4)

substitute in the formula

d=\sqrt{(4-0)^{2}+(2-5)^{2}}

d=\sqrt{(4)^{2}+(-3)^{2}}

d=\sqrt{25}

AB=5\ units

Find the distance BC

B(2,4), C(-2,1)

substitute in the formula

d=\sqrt{(1-4)^{2}+(-2-2)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}

BC=5\ units

Find the distance CD

C(-2,1),D(1,-3)

substitute in the formula

d=\sqrt{(-3-1)^{2}+(1+2)^{2}}

d=\sqrt{(-4)^{2}+(3)^{2}}

d=\sqrt{25}

CD=5\ units

Find the distance AD

A(5,0),D(1,-3)

substitute in the formula

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

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}        

AD=5\ units

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

B(2,4),D(1,-3)

substitute in the formula

d=\sqrt{(-3-4)^{2}+(1-2)^{2}}

d=\sqrt{(-7)^{2}+(-1)^{2}}

BD=\sqrt{50}\ units        

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

BD^{2}=AD^{2}+AB^{2}

substitute

(\sqrt{50})^{2}=5^{2}+5^{2}

50=50 -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

A=b^{2}

we have

b=5\ units

A=5^{2}=25\ units^{2}

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

P=4b

we have

b=5\ units

P=4(5)=20\ units

6 0
3 years ago
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