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notka56 [123]
3 years ago
9

Help ASAP Please). Show your work: It takes the earth 24 h to complete a full rotation. It takes Mercury approximately 58 days,

15 h, and 30 min to complete a full rotation. How many hours does it take Mercury to complete a full rotation? Show your work using the correct conversion factors. ( Only answer if you actually know how to work this problem) Will Mark Brainliest. Need 2 answers to mark Brainliest.​
Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

1,407.5 h

Step-by-step explanation:

Rotational period of mercury

= 58 days, 15 h, and 30 min

= 58*24 h + 15 h + 30/60 h

= 1,392 h + 15 h + 0.5 h

= 1,407.5 h

Westkost [7]3 years ago
3 0

58 days * 24 hrs / day = 1392 hrs

30 mins / 60 mins = 0.5

1392 hrs + 15 hrs = 1407 hrs

1407 hrs + 30 mins = 1407.5 hrs

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Which equation represents the graph of the linear function?
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I think the answer is D

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Find an equation for the line perpendicular to y=-1/2x+4 and goes through the point (-1,10)
Artemon [7]

The slope of the perpendicular line will be 2. Remember, a perpendicular slope is the negative reciprocal.

So, we can make the equation y=2x+b. We need to find b, and put b in.

To find b, put in the numbers of y and x.

10=2(-1)+b

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5 0
4 years ago
So I keep getting 188 or 88 but I got the reminder right I’m not sure what I did wrong
9966 [12]

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3 0
4 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)

3 0
3 years ago
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