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Can someone help me find the domain
2 answers:
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Answer:
x = - , x = 2
Step-by-step explanation:
To find h(g(x)) substitute x = g(x) into h(x) , that is
h(g(x))
= h(x + 1)
= (x + 1)²
= x² + 2x + 1
For h(g(x)) = 3x² + x - 5 , then
3x² + x - 5 = x² + 2x + 1 ← subtract x² + 2x + 1 from both sides
2x² - x - 6 = 0 ← in standard form
(2x + 3)(x - 2) = 0 ← in factored form
Equate each factor to zero and solve for x
2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = -
x - 2 = 0 ⇒ x = 2
All the theorems are proved as follows.
<h3>What is a Triangle ?</h3>A triangle is a polygon with three sides , three vertices and three angles.
1. The Triangle sum Theorem
According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.
To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.
We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.
180° - ∠A + 180° - ∠B + 180° - ∠C = 360°
- ∠A - ∠B - ∠C = 360° - (180°+ 180°+ 180°) = -180°
∠A + ∠B + ∠C = 180°
(Hence Proved)
2. Isosceles Triangle Theorem
Considering an isosceles triangle ΔABC
with AB = AC, we have by sine rule;
as AB = AC
sin B = sin C
angle B = angle C
3.Converse of the Isosceles theorem
Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;
as ∠B= ∠C ,
AB = AC
4. Midsegment of a triangle theorem
It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.
Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC
In ΔABC and ΔADF
∠A ≅ ∠A
BA = 2 × DA, BC = 2 × FA
Hence;
ΔABC ~ ΔADF (SAS similarity)
BA/DA = BC/FA = DF/AC = 2
Hence AC = 2×DF
5.Concurrency of Medians Theorem
A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and the point of intersection is inside the triangle known as Centroid .
Consider a triangle ABC , X,Y and Z are the midpoints of the sides
Since the medians bisect the segment AB into AZ + ZB
BC into BX + XB
AC into AY + YC
Where:
AZ = ZB
BX = XB
AY = YC
AZ/ZB = BX/XB = AY/YC = 1
AZ/ZB × BX/XB × AY/YC = 1 and
the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).
To know more about Triangle
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