Well you would multiply the number of boxes (24) by how much each one is ($87) to get you answer ($2,088) :)
One root belonged to an old tree next to the road, and the other is part of a small tree.
Answer:

Step-by-step explanation:
First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.
There are 4 ace in the deck
There are 16 cards of 10 points in the deck
To make this calculation we use the formula of combinations

Where n is the total number of letters and r are chosen from them
The number of ways to choose 1 As is:

The number of ways to choose a 10-point letter is:

Therefore, the number of ways to choose an Ace and a 10-point card is:

Now the number of ways to choose any 2 cards from a deck of 52 cards is:


Therefore, the probability of obtaining an "blackjack" is:





4sin²(x) = 5 - 4cos(x)
4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
2 - 2cos(2x) = 5 - 4cos(x)
- 2 - 2
-2cos(2x) = 3 - 4cos(x)
-2[2cos²(x) - 1] = 3 - 4cos(x)
-4cos²(x) + 2 = 3 - 4cos(x)
- 2 - 2
-4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
4cos²(x) - 4cos(x) + 1 = 0
[2cos(x) - 1]² = 0
2cos(x) - 1 = 0
+ 1 + 1
2cos(x) = 1
2 2
cos(x) = ¹/₂
cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
x = 60, 300
x = π/3, 5π/3
[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π