When it hatches from its egg, the shell of certain crab is 1cm across. When fully grown the shell is approximately 10cm across.

Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space consists of all potential outcomes for each experiment:

a) The sample encompasses the outcomes resulting from tossing a coin (heads or tails) and throwing a die (a number between 1 and 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can settle in one of three different positions: on its top, bottom, or side:

S={top,bottom,side}

c) The entire set of cards within a standard deck constitutes the sample space:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Answer:

Tan(b°) = 3/4, which is equivalent to three fourths (C).

Step-by-step explanation:

Triangle JKL is a right triangle with angle K being the right angle and angle L equal to b°.

We will employ SOHCAHTOA principles from trigonometry to calculate the sides' values.

For triangle ∆JKL:

Sin(b°) = opposite/hypotenuse

Sin(b°) = 3/5

Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4.

From the earlier information, we have the values for each side of triangle ∆JKL.

Please refer to the attached diagram for the triangle (1).

Triangle ∆JKL is scaled with a factor of 2.

This implies multiplying each side of ∆JKL by 2, resulting in:

Opposite side = 2(3) = 6

Adjacent side = 2(4) = 8

Hypotenuse = 2(5) = 10.

To find tan(b°) for the new triangle, we use the tangent ratio:

Tan(b°) = opposite/adjacent.

Tan(b°) = 6/8.

Tan(b°) = 3/4.

Find the diagram for the new triangle included (2).

Diagram 3 provides a depiction of both triangles.

As shown, the angle remains unchanged when a triangle is scaled by a factor.

Thus, we conclude that tan(b°) = three fourths (C).

Please clarify the question as I don't quite understand it.

Answer:

Step-by-step breakdown:

Given

Required

Simplify

The first task is to find the LCM of the given expression

Proceed with arithmetic operations on the numerator

Divide both numerator and denominator by 2

This expression cannot be further simplified;

Consequently, =