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dsp73
3 years ago
11

Are these all correct?

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0
Problem 1) Line n is the line of symmetry (not line o) because we can fold the lower half to match up with the upper half. The folding line is over line n.

Problem 2) I agree. Nice work on getting the correct answer. The folding line is a vertical line through the center

Problem 3) It's hard to say for sure, but I think the top left corner is NOT reflective over any line of symmetry no matter how you rotate it. So I would uncheck that box. I agree with your other choices though. Great job with those. 
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Answer: The answer is A and D

Step-by-step explanation:

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3 years ago
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Example 2
PIT_PIT [208]

Answer: (a) 314.2cm², (b) 157.1cm², (c) 78.55cm² (e) 6.77

Step-by-step explanation: (a)  Area of the circle with radius of 10 cm = πr²

                                                                          = 3.142 × 10 × 10

                                                                          = 3.142 × 100

                                                                          = 314.2cm²

The formula                                                       = πr²

(b)  Area of the half of a circle known as semicircle

                                                                         = πr²/2

                                                                         = 3.142 ×10 × 10/2

                                                                         = 3.142 × 50

                                                                         = 157.1cm²

The formula                                                      = πr²/2

(c)  A quarter of a circle is called quadrant

                                                            = πr²/4

                                                            = 3.142 × 10 × 10/4

                                                            = 314.2/4

                                                            = 78.55cm²

The formula is written thus = πr²/4, which implies that the circle is divided into 4 unit

(d) The conjecture about how to determine the area of the sector is

Formula of a sector = ∅/360(πr²)

<u>Information</u>

The arc  cant be 60°, therefore information incomplete.

(e) Area of the sector with the angle AOB of 60° = 24.

To find the radius of the angle, make v the subject of the formula from the formula.

Sector area = πr²∅/360°

equate formula to 24.

Therefore πr²∅/360° = 24

Multiply through by360° to make it a linear expression

It now becomes πr²∅ =24× 360°

                                                     r² = 24  x 360/π × ∅°

                                                     r² = 24 × 360° /3.142 × 60°

                                                     r² = 3,640/188.52

                                                     r² = 45.8

To find r , we take the square root of both side by applying laws of indicies

                                    Therefore r = √45 .8

                                                      r = 6.77

(f)   General formula = ∅°/360° × (πr²)

angle substended at centre by the arc = x°

assuming the radius of the circle = ycm, Therefore,  area of the sector = { ∅°/360° × πy² }

                                                     

                                                   

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3.56 as a fraction simplified
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