Answer:
4 ways
Step-by-step explanation:
1 gummy 4 chocolates
2 gummy 3 chocolates
3 gummy 2 chocolates
4 gummy 1 chocolates
This is what i think the answer is hope it helps
4 sq.units
area of a rectangle = length × breadth
length = 4units
breadth = 1units
A = l × b
= 4 × 1
= 4 sq.units
<h3>
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em> </em><em>U</em><em> </em></h3>
<em>IF</em><em> </em><em>YES</em><em> </em>
<em>MARK</em><em> </em><em>ME</em><em> </em><em>AS</em><em> </em><em>THE</em><em> </em><em>BRAINLIEST</em>
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)
Answer:
0.01
can I have brainliest please
Step-by-step explanation:
Answer:
The 6th term is first integer value of n greater then 50 in the 2n^2 sequence.
Step-by-step explanation:
2n^2 sequence:
2,8,18,32,50,72