Question

x²-[3x+{2x-(x²-1)}+2]
please help me out with this problem
​

Answer 1

Answer:

x²-[3x+{2x-(x²-1)}+2]=(x-1)(2x-3)

Step-by-step explanation:

x²-[3x+{2x-(x²-1)}+2]

=x²-[3x+{2x-x²+1}+2]

=x²-[3x+2x-x²+1+2]

=x²-[5x-x²+3]

=x²-5x+x²-3

=x²+x²-5x-3

=2x²-5x-3

=2x²-2x-3x-3

=(2x²-2x)-(3x-3)

=2x(x-1)-3(x-1)

=(x-1)(2x-3)

Answer 2

Answer:

Required Knowledge

\red{\bigstar}★ Angle sum of a triangle.

The angles of a triangle sum to 180^{\circ}180∘ .

\red{\bigstar}★ The number of diagonals.

In a regular polygon of nn sides, we can draw (n-3)(n−3) diagonals from a point since we cannot connect adjacent sides or itself to get a diagonal. Also, inside is (n-2)(n−2) triangles.

\red{\bigstar}★ Property of regular polygons.

Every single angle is equal in regular polygons.

\large\underline{\text{Solution}}Solution

Since there are (n-2)(n−2) triangles, the angle sum of all the triangles is 180^{\circ}\times(n-2)180∘×(n−2) .

Then since the regular polygon has nn equal angles, one angle will be \dfrac{180^{\circ}\times(n-2)}{n}n180∘×(n−2) .

Now we get the equation.

\implies\dfrac{180^{\circ}\times(n-2)}{n}=144^{\circ}⟹n180∘×(n−2)=144∘

\implies180^{\circ}\times n-360^{\circ}=144^{\circ}\times n⟹180∘×n−360∘=144∘×n

\implies(180^{\circ}-144^{\circ})\times n=360^{\circ}⟹(180∘−144∘)×n=360∘

\implies36^{\circ}\times n=360^{\circ}⟹36∘×n=360∘

\implies n=10⟹n=10

Hence, the shape is a regular 10-sided polygon.