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Mazyrski [523]
3 years ago
12

Solve. Show all your work. a) -3x+2 = 5 b) }(4x – 4) = 7 Help help help please

Mathematics
2 answers:
Doss [256]3 years ago
6 0
A) x = -1
b) x = 11 / 4

a explanation)
-3 x + 2 = 5
-3 x + 2 - 2 = 5 - 2
-3 x = 5 - 2
-3 x = 3
-3 x / -3 = 3 / -3
x = 3 / -3
x = -1

b explanation)
4 x - 4 = 7
4 x - 4 + 4 = 7 + 4
4 x = 7 + 4
4 x = 11
4 x / 4 = 11 / 4
x = 11 / 4
Reptile [31]3 years ago
3 0

Answer:

A) x=-1

B) x=11/4

Step-by-step explanation:

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2 out of 50 were tagged.

Divide 2 by 50:

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6 0
3 years ago
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.




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Answer:

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Answer:

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