The correct question is
<span>Given cos theta=4/9 and csc theta < 0 find sin theta and tan theta
</span>
we know that
csc theta=1/sin theta
if csc theta < 0
then
sin theta < 0
we have that
<span>cos theta=4/9
we know that
sin</span>² theta+cos² theta=1
so
sin² theta=1-cos² theta-----> 1-(4/9)²----> 1-(16/81)----> 65/81
sin theta=-√(65/81)---->-√65/9
the answer Part a) is
sin theta=-√65/9
Part b) find tan theta
tan theta=sin theta/cos theta
tan theta=(-√65/9)/(4/9)-----> tan theta=-√65/4
the answer part b) is
tan theta=-√65/4
(a)
We can see that
this is a straight line
and this is the graph of velocity
we know that
acceleration is the derivative of velocity
so, slope of curve of velocity is acceleration
so, we will find slope of this line
We can select any two points
(0,4) and (5,7)
we can use slope formula
now, we can plug points
we know that slope of line is always constant irrespective of any value of t
so, acceleration will always be same irrespective of any value of t
so, we will get acceleration
............Answer
(b)
we can see that acceleration is constant
and we know that
derivative of constant is always 0
so, instantaneous rate of acceleration at t=10s is 0........Answer
5p-14=8p+19
8p-5p = -3p
-3p-14=18
14+18 = 33
33/-3=-11
We can create two points.
30 feet below the surface of the water 10 seconds after he entered the water
100 feet below the surface after 40 seconds
So we have (10, 30) and (40, 100)
Plug this values into the slope formula:
m = y2-y1/x2-x1
m = (100 - 30)/(40 - 10)
m = 70/30
m = 2 1/3
You sum the like terms
-6b+3b + 11+2
the answer is -3b+13
