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Ratling [72]
3 years ago
14

What is the unit rate of (8,14) as a decimal

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0
You would need to calculate the unit rate by dividing the y-value by the x-value. 

14/8 = 7/4 

7/4 in decimal form is 1.75
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-37 ≥ n - 20 <br> What could n be??
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Step-by-step explanation:

-78 - 20 = -98 which is less than -37,

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In triangle TRS, TZ = (3x) inches and WZ = (2x - 3) inches. Triangle T R S has centroid Z. Lines are drawn from each point to th
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2 years ago
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Find x <br> Picture Below
Mumz [18]

m∠BEC=(1/2)(mBC + mAD)

110 = (1/2)(x+97)

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2 years ago
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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