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lisabon 2012 [21]
3 years ago
10

Pls help me answer the question and if you don't know, pls don't answer. thank you!

Mathematics
1 answer:
Damm [24]3 years ago
6 0
The answer is 162in2 to its orbit
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Rearrange the formula C = 5/9 (F − 32) for F.
shtirl [24]
The answer is (9c/5)+32=F
5 0
3 years ago
Read 2 more answers
Which of the ratios below is equivalent to 5:2? Select all that apply.
Nata [24]

Answer:

D

Step-by-step explanation:

This is because they are both mulitplied by two so they are still equivalent

6 0
3 years ago
Step by step explanation
vodka [1.7K]

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.

4 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)

3 0
3 years ago
Sandy has $200 in her bank account.
Mademuasel [1]
The answer it is b because that is the right answer
3 0
4 years ago
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