What does a quadrilateral look like, i'll look it up but its so fun on this
1 answer:
You might be interested in
f(x) > 0 - I and II quadrant
f(x) < 0 - III and IV quadrant
Look at the picture.
A. F(x) < 0 on the interval x < 0. TRUE
B. F (x) > 0 on the interval x <0. FALSE
C. F (x) < 0 on the interval 0 < x < 1. TRUE
D. F (x) > 0 on the interval 0 < x < 1. FALSE
E. F (x) < 0 on the interval 1 < x < 3. FALSE
F. F (x) > 0 on the interval 1 < x < 3. TRUE
G. F (x) < 0 on the interval x > 3. TRUE
H. F (x) > 0 on the interval x > 3. FALSE
Answers:
a = -6/37
b = -1/37
============================================================
Explanation:
Let's start things off by computing the derivatives we'll need
Apply substitution to get
I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.
The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero
a-6b = 0
a = 6b
At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)
-6a -b = 1
-6(6b) - b = 1 .... plug in a = 6b
-36b - b = 1
-37b = 1
b = -1/37
Use this to find 'a'
a = 6b
a = 6(-1/37)
a = -6/37