A student can take three subjects in 40 ways.
<u>SOLUTION:</u>
Given that, there are 4 different math courses, 5 different science courses, and 2 different history courses.
A student must take one of each, how many different ways can this be done?
Now, number ways to take math course = 4
Number of ways to take science course = 5
Number of ways to take history course = 2
So, now, total possible ways = product of possible ways for each course = 4 x 5 x 2 = 40 ways.
Hence, a student can take three subjects in 40 ways.
I is C because you use the pythag theorem
Here you go. Do you think that you could help me with my question
5a – 10b = 45
5a – 10(3) = 45
5a – 30 = 45
+30 +30
5a = 75/5
<span>a = 15</span>
36;
the square root and the square cancel out, or you can work it out and say
((36)^1/2)^2 = (6)^2= 36
X+0.0825x=216.5
1.0825x=216.5
x=216.5 / 1.0825
x=200
Answer: C.) x=200
