The answer is (9c/5)+32=F
Answer:
D
Step-by-step explanation:
This is because they are both mulitplied by two so they are still equivalent
Answer: x = 4 and y = -3
Step-by-step explanation:
2x + 4y = -4
2x = -4 - 4y
x = -2 - 2y
Now that we have solved the first equation for x, we plug (-2 - 2y) into the other equation in place of x and then solve for y.
3(-2 -2y) + 5y = -3
-6 - 6y + 5y = -3
-y = 3
y = -3
Now that we know y, we plug it into the original equation and solve for x.
2x + 4(-3) = -4
2x + -12 = -4
2x = 8
x = 4
You can check the answer by plugging both x and y into either of the original equations.
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)
The answer it is b because that is the right answer