Answer:
2) The vertex is (1, 2), the domain is all real numbers, and the range is y ≥ 2.
Step-by-step explanation:
Look at the photo below for the graph.
:)
Point P is in the interior of ∠OZQ. If m∠OZQ = 125° and m∠OZP = 62°, what is m∠PZQ?
2 answers:
You might be interested in
Answer:
The correct option is 1.
Step-by-step explanation:
The given parent function is
1. Domain of the function is all positive real number including 0.
2. Range of the function is all positive real number including 0.
3. It is an increasing function. It increases at decreasing rate.
First graph increase at decreasing rate and it starts from (-4,-1), therefore the required function is
Therefore graph 1 is an example of a function whose parent graph is of the form y = √x.
Second graph is a parabola, so it is the graph of a quadratic function.
Third graph is a rectangular hyperbola, so it is the graph of a rational functions.
Fourth graph is increasing at increasing rate, so it is the graph of an exponential function.
Therefore options 2, 3 and 4 are incorrect.
Answer:
The diagram for the question is missing, but I found an appropriate diagram fo the question:
Proof:
since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle
∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
∠DOP = 22.5°
∠PDO = 67.5°
∠ADO = 22.5°
∠AOD = 67.5°
Step-by-step explanation:
Given:
AB = CD = 297 mm
AD = BC = 210 mm
BCPO is a square
∴ BC = OP = CP = OB = 210mm
Solving for OC
OCB is a right anlgled triangle
using Pythagoras theorem
(Hypotenuse)² = Sum of square of the other two sides
(OC)² = (OB)² + (BC)²
(OC)² = 210² + 210²
(OC)² = 44100 + 44100
OC = √(88200
OC = 296.98 = 297
OC = 297mm
An isosceless tringle is a triangle that has two equal sides
Therefore for △OCD
CD = OC = 297mm; Hence, △OCD is an isosceless triangle.
The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles
Since BC = OB = 210mm
∠BCO = ∠BOC
since sum of angles in a triangle = 180°
∠BCO + ∠BOC + 90 = 180
(∠BCO + ∠BOC) = 180 - 90
(∠BCO + ∠BOC) = 90°
since ∠BCO = ∠BOC
∴ ∠BCO = ∠BOC = 90/2 = 45
∴ ∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
For ΔOPD
Note that DP = 297 - 210 = 87mm
∠PDO + ∠DOP + 90 = 180
∠PDO + 22.5 + 90 = 180
∠PDO = 180 - 90 - 22.5
∠PDO = 67.5°
∠ADO = 22.5° (alternate to ∠DOP)
∠AOD = 67.5° (Alternate to ∠PDO)