Answer:
Step-by-step explanation:
A tiny hint. When you are looking for the opposite side of a triangle and the function you use is the tangent, the side opposite is always going to be larger than the side adjacent. That happens when theta is greater than 1 when the angle is larger than 45.
Phew! I didn't know the hint was going to be that long.
The trig function that does not involve the hypotenuse is the Tangent or Tan(x).
C
Tan(x) = opposite / Adjacent.
The opposite side is the side not connected to the reference angle (x)
The adjacent side makes up the angle and is not the hypotenuse.
Tan(46) = 1.036
Side adjacent = 8
So the answer is either A or C. I'm guessing it's C.
Tan(46) = x/8
1.036 = x / 8 Multiply by 8
8*1.036 = x
x = 8.28
Answer:
x = 9 Angle B = 40 degrees
Step-by-step explanation:
Angles A and B are congruent, so the equation are the same
5x - 5 = 3x + 13
-3x - 3x Subtract 3x from both sides
2x - 5 = 13
+ 5 +5 Add 5 to both sides
2x = 18 Divide both sides by 2
x = 9
Plug this into the equation for angle B
B = 3(9) + 13 Multiply
B = 27 + 13 Add
B = 40
If these answers are correct, please make me Brainliest!
Ok so basically, the number of student tickets is 3x, where x=the number of adult tickets sold. And we know that s(for student tickets)+x=480 total tickets sold. So if we replace s with 3x we have 3x+x=480, or 4x=480. We divide by 4 and get x=120, which is the amount of adult tickets sold.
2 liters = 20 deciliters
20 x 0.15 = 3.00
the answer is $3.00
Answer:
A) 0.1612
B) 0.8031
C) 0.1969
Step-by-step explanation:
For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
In this problem we have that:
Five tosses, so 
The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is 
(a) twice
This is 
(b) at most once
(c) at least two times.
Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So
We want 
So
