The easiest way to solve this problem is by using the Pythagorean theorem :D
Pythagorean theorem: [-b +- sqrt(b^2 - 4ac)]/2a
5 = a
-1 = b
6 = c
therefore, by plugging these values in!
[-(-1) +- sqrt((-1)^2 - 4(5)(6))]/2(5)
[1 +- sqrt(1 - 120)]/10
Oh so we're using imaginary numbers, so in this case you'll need to know that i = sqrt(-1), so keep this in mind (I can see why this was so difficult)
[1 +- sqrt(119)*i]10
answers:
( 1 + i*sqrt(119) ) / 10
( 1 - i*sqrt(119) ) / 10
The box plot shows the range and quartiles of the data. It does not show numbers or the mean. The only appropriate choice is
... The lower quartile for the data
Step-by-step explanation:
This is known as the triple tangent identity. Start with the fact that the three angles add up to 0.
(x − y) + (z − x) + (y − z) = 0
Subtract two terms to the other side and take the tangent:
x − y = -((z − x) + (y − z))
tan(x − y) = tan(-((z − x) + (y − z)))
Use reflection property:
tan(x − y) = -tan((z − x) + (y − z))
Now use angle sum identity:
tan(x − y) = -[tan(z − x) + tan(y − z)] / [1 − tan(z − x) tan(y − z)]
tan(x − y) = [tan(z − x) + tan(y − z)] / [tan(z − x) tan(y − z) − 1]
tan(x − y) [tan(z − x) tan(y − z) − 1] = tan(z − x) + tan(y − z)
tan(x − y) tan(z − x) tan(y − z) − tan(x − y) = tan(z − x) + tan(y − z)
tan(x − y) tan(z − x) tan(y − z) = tan(x − y) + tan(z − x) + tan(y − z)
Answer:
Step-by-step explanation:
Using the section formula
= = = =