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KonstantinChe [14]
3 years ago
14

What is the greatest common factor for 30,48,60 and whats the GCF ??????

Mathematics
2 answers:
Svetach [21]3 years ago
4 0
30 = 2 * 3 * 5 

<span>48 = 2⁴ * 3 
</span>
<span>60 = 2² * 3 * 5 

</span><span>GCF = 2 * 3 = 6</span>
Lesechka [4]3 years ago
3 0
The greatest common factors are
30: 1,2,3,5,6,10,15,30
48: 1,2,3,4,6,8,12,16,24,48
60:1,2,3,4,5,6,10,12,15,20, 30,60
So therefore the GCF is 6
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What shapes does a kite consist of? ​
Oduvanchick [21]

Step-by-step explanation:

A kite is a special case of quadrilateral. In which larger diagonal bisects the smaller diagonal at right angle.

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If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

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∠PCO = 45°

∠POC = 45°

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∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

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Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

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(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

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Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴  ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

Note that DP = 297 - 210 = 87mm

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∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

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