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Mazyrski [523]
3 years ago
10

Coconuts were on sale, 5 for $4.45. Pineapples were on sale, 4 for $3.40. Which of the following is true?

Mathematics
2 answers:
melomori [17]3 years ago
3 0

Answer: D

Step-by-step explanation: If you do 4/4.45 you get a number approximate to 89 cents. Answers A and B you can cross of because coconuts are 5 cents more than pineapples

bulgar [2K]3 years ago
3 0
I think the answer is d
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masya89 [10]
The length of the diagonal of the square is the square root of 32 which equals approximately 5.6569.
The length from the corner to the center of the square is half of the diagonal which is 2.8284. I hope this helps!
7 0
3 years ago
If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

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

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

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)

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I can give you the link for the answers
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PLEASE IM IN CLASS AND I REALLY NEED AN ANSWER!! 20 PTS & BRAINLIEST!!
makkiz [27]

Answer:

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Answer:

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