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erik [133]
3 years ago
10

What is 120/1,000 simplified??????????????????????????????????????????????????????????/

Mathematics
2 answers:
Nikolay [14]3 years ago
5 0
0.12 is your answer!
dedylja [7]3 years ago
5 0
The answer is 3/25.

Hope this helps!
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Step-by-step explanation:

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Please help me ill mark you brainiest
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55.560

Step-by-step explanation:

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Given P(A)=0.5P(A)=0.5, P(B)=0.71P(B)=0.71 and P(A\cap B)=0.415P(A∩B)=0.415, find the value of P(A|B)P(A∣B), rounding to the nea
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0.585

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Solve the inequality 3x - 5
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Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
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