You roll a six-sided number cube. what is the probability that you will roll a two?
1 answer:
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Answer:
54.0°
Step-by-step explanation:
We have that:
u=<cos 30°, 4sin30°>
This implies that:
u=<½√3, 2>
Also, v=<3cos 45°, 3sin 45°>
v=<3√2/2, 3√2/2>
u+v=<2.99,4.12)
The direction of u+v, is the angle, it makes with the positive x-axis.
This is given by;
To the nearest tenth, the direction is 54.0°
Part A
The pattern of squares is 1, 4, 9, ... which is the set of perfect squares
- 1 = 1^2
- 4 = 2^2
- 9 = 3^2
and so on
The 7th figure will have 49 squares because 7^2 = 49
<h3>Answer: 49</h3>==============================================================
Part B
Each pattern has one circle per corner (4 circles so far). In addition, there's one circle per unit side to form the perimeter.
- Pattern 1 has 4+4(1) = 8 circles
- Pattern 2 has 4+4(2) = 12 circles
- Pattern 3 has 4+4(3) = 16 circles
The nth term will have 4+4n circles. The first '4' is the number of circles at the corners. The 4n is the circles along the perimeter. If you wanted, 4+4n factors to 4(1+n).
Plug in n = 20 to find the 20th figure has 4+4n = 4+4(20) = 84 circles
<h3>Answer: 84</h3>==============================================================
Part C
- Pattern 1 has 1 square + 8 circles = 9 items total
- Pattern 2 has 4 squares + 12 circles = 16 items total
- Pattern 3 has 9 squares + 16 circles = 25 items total
This seems to suggest if the pattern number is odd, then we need an odd number of tiles (square + circular).
Let n be the pattern number. Pattern n needs n^2 square tiles and 4+4n = 4n+4 circular tiles. Overall, n^2+4n+4 tiles are needed.
It turns out that if n is odd, then n^2+4n+4 is always odd. The proof is shown below.
Side note: n^2+4n+4 factors to (n+2)^2
<h3>Answer: B) will always be odd</h3>