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Rina8888 [55]
2 years ago
13

I got $15 on cshapp for the right answer

Advanced Placement (AP)
1 answer:
Jobisdone [24]2 years ago
4 0
1) ii x ii, the square would have an i next to every box and ii in every space. Ratio is 100% ii.
2) AA x BB. Across the top both are As and on the side both Bs. All boxes are filled with AB. Ratio is 100% AB.
3) Ai x Bi. Across the top there is a large A and an i. Across the side is a B and an i. At the spot both A and B meet, you have AB, where A and i meet you have Ai, B and i meet at Bi, and i and i meet for ii. Ratio is 25% AB, 25% Bi, 25% Ai, 25% ii.
4) ii x AB. Across the side is i on both, and on top is A and B. Both under A in the boxes would be Ai and both under B would be Bi. Ratio is 50% Ai, 50% Bi.
5) AB x AB. Across the top is A then B and down the side is A and B. Where the As meet is AA, where the Bs meet is BB, and where A and B meet is AB. Ratio is 50% AB, 25% AA, 25% BB.
6) Alice has Ai and Mark has Bi. On the square, across the top first is an A and second is an i, and down first is B second i. In the boxes where A and B combine is AB, where A and i combine is Ai, B and i is BI, and i and i is ii.
7) No, it is not possible, there is no way to give an i phenome. In the square across the top draw A and B on separate rows, and for the ones under A draw a big A and the ones under B a big B.
8) the baby can have type O blood. Ralph’s phenotype is Bi and Rachel is Ai. Across the top put B then i, and on the side A then i. At the intersection of A and B put AB, the intersection between A and i put Ai, between B and i put Bi, and from i and i put ii.

$Jay360b
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Huge help needed ! im lost and need an explanation to the correct answer.
Semmy [17]

Answer:

m<MON = 100°

Explanation:

Given:

Area of shaded sector LOM = 2π cm²

NL = 6 cm

Required:

m<MON

Solution:

m<MON = 180° - m<LOM (angles in a straight line)

We don't know m<LOM. Therefore, let's find m<LOM.

Area of a sector = θ/360 × πr²

Area of sector LOM = 2π cm²

r = 3 cm

θ = m<LOM = ?

Plug in the values

2π = m<LOM/360 × π × 3²

2π = m<LOM/360 × 9π

2π = m<LOM × 9π/360

2π = m<LOM × π/40

Multiply both sides by 40

2π × 40 = m<LOM × π

80π = m<LOM × π

Divide both sides by π

80π/π = m<LOM

80 = m<LOM

m<LOM = 80°

✔️m<MON = 180° - m<LOM (angles in a straight line)

Substitute

m<MON = 180° - 80°

m<MON = 100°

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