As soon as I read this, the words "law of cosines" popped
into my head. I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.
The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.
To find angle-B, the law of cosines says
b² = a² + c² - 2 a c cosine(B)
B = angle-B
b = the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9
(1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)
1.96 = (1) + (3.61) - (3.8) cos(B)
Add 3.8 cos(B) from each side:
1.96 + 3.8 cos(B) = 4.61
Subtract 1.96 from each side:
3.8 cos(B) = 2.65
Divide each side by 3.8 :
cos(B) = 0.69737 (rounded)
Whipping out the
trusty calculator:
B = the angle whose cosine is 0.69737
= 45.784° .
Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...
By gosh ! Choice 'B' is 45.8° ! yay !
I'll bet that's it !
Answer:
Step-by-step explanation:
ΔADB is a right angled triangle. So, use Pythagorean theorem.
Altitude² = hypotenuse² - base²
= (2√5)² - (3√2)²
= 2²* (√5)² - 3²*(√2)² = 4*5 - 9*2 = 20 - 18 = 2
Altitude = √2
Base of ΔABC = 3√2 + √2 = 4√2 cm
Area = 
Answer: the numbers are 9 and 2
Step-by-step explanation:
Let the numbers be a and b
a + b = 11.........equation 1
ab = 18.......equation 2
Multiply equation 1 by a
a ^2 + ab = 11a.......equation 3
Substitute the value of ab from equation 2 into equation 3
a ^2 + 18 = 11a
a^2 - 11a + 18 = 0
(a-9) x (a-2) = 0
If a = 9, then b =2
I hope this helps, please mark as brainliest answer.
Answer:
0.6 = 60% probability that he or she studies on a weeknight.
Step-by-step explanation:
We solve this question treating these events as Venn probabilities.
I am going to say that:
Probability A: Probability of a student studying on weeknights.
Probability B: Probability of a student studying on weekends.
Forty-two percent of students said they study on weeknights and weekends
This means that 
47% said they studied on weekends
This means that 
65% said they study either on weeknights or weekends.
This is 
If you were to pick one student at random, what is the probability that he or she studies on a weeknight?
This is P(A), and the equation used is:
Considering the values we have:
0.6 = 60% probability that he or she studies on a weeknight.
