Answer:
3 rings
Step-by-step explanation:
10 / 2 = 5 necklaces
(5/5) x 2 = 2 bracelets
10 - 5 - 2 = 3<u> rings</u>
Answer:
Simplifying
4cos(10x) + 2 = 2
Remove parenthesis around (10x)
4cos * 10x + 2 = 2
Reorder the terms for easier multiplication:
4 * 10cos * x + 2 = 2
Multiply 4 * 10
40cos * x + 2 = 2
Multiply cos * x
40cosx + 2 = 2
Reorder the terms:
2 + 40cosx = 2
Add '-2' to each side of the equation.
2 + -2 + 40cosx = 2 + -2
Combine like terms: 2 + -2 = 0
0 + 40cosx = 2 + -2
40cosx = 2 + -2
Combine like terms: 2 + -2 = 0
40cosx = 0
Solving
40cosx = 0
Solving for variable 'c'.
Move all terms containing c to the left, all other terms to the right.
Divide each side by '40'.
cosx = 0.0
Simplifying
cosx = 0.0
The solution to this equation could not be determined.
Step-by-step explanation:
Answer:
![a = 9\\b = 48\\c = -1](https://tex.z-dn.net/?f=a%20%3D%209%5C%5Cb%20%3D%2048%5C%5Cc%20%3D%20-1)
Step-by-step explanation:
We know that:
In a deck of 52 cards there are 4 aces.
Therefore the probability of obtaining an ace is:
P (x) = 4/52
The probability of not getting an ace is:
P ('x) = 1-4 / 52
P ('x) = 48/52
In this problem the number of aces obtained when extracting cards from the deck is a discrete random variable.
For a discrete random variable V, the expected value is defined as:
![E(V) = VP(V)](https://tex.z-dn.net/?f=E%28V%29%20%3D%20VP%28V%29)
Where V is the value that the random variable can take and P (V) is the probability that it takes that value.
We have the following equation for the expected value:
![E(V) = \frac{4}{52}(a) + \frac{b}{52}(c)](https://tex.z-dn.net/?f=E%28V%29%20%3D%20%5Cfrac%7B4%7D%7B52%7D%28a%29%20%2B%20%5Cfrac%7Bb%7D%7B52%7D%28c%29)
In this problem the variable V can take the value V = 9 if an ace of the deck is obtained, with probability of 4/52, and can take the value V = -1 if an ace of the deck is not obtained, with a probability of 48 / 52
Therefore, expected value for V, the number of points obtained in the game is:
![E(V) = \frac{4}{52}(9) + \frac{48}{52}(-1)](https://tex.z-dn.net/?f=E%28V%29%20%3D%20%5Cfrac%7B4%7D%7B52%7D%289%29%20%2B%20%5Cfrac%7B48%7D%7B52%7D%28-1%29)
So:
![a = 9\\b = 48\\c = -1](https://tex.z-dn.net/?f=a%20%3D%209%5C%5Cb%20%3D%2048%5C%5Cc%20%3D%20-1)
Answer:
10. 9
In the first question we can solve using the Pythagorean theorem. It states that the hypotenuse in a right triangle or the longest side of the triangle, squared is equal to the other 2 sides, squared. Its expressed as so: C^2 = A^2 + b^2 where c is the hypotenuse and a and b are the other sides oft eh triangle.
Therefore,
15^2 = 12^2+x^2
225=144+x^2
81=x^2
x=9
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11. x=12
In this figure, there are 2 right triangles on the sides of the square. If we can find the lengths of the base of both triangles then we can find x using the Pythagorean theorem. the total base is 21 and the square is 11. 11+a+b=21. We can assume that both triangles are congruent and therefore we can solve this equation:
11+a+b=21
a+b=10
5+5=10
b=5
a=5
The bases of the two triangles are 5 and the hypotenuse is 13. Now we can solve using the Pythagorean theorem:
c^2 = a^2+b^2
13^2=5^2+x^2
169=25+x^2
144=x^2
144= 12 x 12
x=12
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12.
The diagnol of a triangle is the hypotenuse and since the length is the square root of 3, we have all the information we need.
c^2 = a^2+b^2
2^2 = (square root of 3)^2+b^2
4=3+b^2
1=b^2
b=1
Step-by-step explanation: