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3 years ago

Can anyone help me with this the last answer choice is wrong

Mathematics

2 answers:

umka21 [38]3 years ago

8 0

Answer:

It's the 2nd one

Step-by-step explanation:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

Step-by-step explanation:

for all x>6

x-6>0

$\frac{6x-x^2}{x^2-4x-12} \\=\frac{-x(x-6)}{x^2-6x+2x-12} \\=\frac{-x(x-6)}{x(x-6)+2(x-6)} \\=\frac{-x(x-6)}{(x-6)(x+2)} \\=\frac{-x}{x+2} \\by~dividing~by~x-6$

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Please solve these questions for me. i am having a difficult time understanding.

s2008m [1.1K]

Answer:

1) AD=BC(corresponding parts of congruent triangles)

2)The value of x and y are 65 ° and 77.5° respectively

Step-by-step explanation:

Given : AD||BC

AC bisects BD

So, AE=EC and BE=ED

We need to prove AD = BC

In ΔAED and ΔBEC

AE=EC (Given)

$\angle AED = \angel BEC$ ( Vertically opposite angles)

BE=ED (Given)

So, ΔAED ≅ ΔBEC (By SAS)

So, AD=BC(corresponding parts of congruent triangles)

Hence Proved

Refer the attached figure

$\angle ABC = 90^{\circ}$

In ΔDBC

BC=DC (Given)

So, $\angle CDB=\angle DBC$ (Opposite angles of equal sides are equal)

So, $\angle CDB=\angle DBC=x$

So, $\angle CDB+\angle DBC+\angle BCD = 180^{\circ}$ (Angle sum property)

x+x+50=180

2x+50=180

2x=130

x=65

So, $\angle CDB=\angle DBC=x = 65^{\circ}$

Now,

$\angle ABC = 90^{\circ}\\\angle ABC=\angle ABD+\angle DBC=\angle ABD+x=90$

So, $\angle ABD=90-x=90-65=25^{\circ}$

In ΔABD

AB = BD (Given)

So, $\angle BAD=\angle BDA$ (Opposite angles of equal sides are equal)

So, $\angle BAD=\angle BDA=y$

So, $\angle BAD+\angle BDA+\angle ABD = 180^{\circ}$ (Angle Sum property)

y+y+25=180

2y=180-25

2y=155

y=77.5

So, The value of x and y are 65 ° and 77.5° respectively

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3 years ago