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
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I think it is right
Answer:
y = (5/2)x + 6
Step-by-step explanation:
The given line has a slope (coefficient of x) of -2/5. The perpendicular line will have a slope that is the negative reciprocal of this:
-1/(-2/5) = 5/2
Since you are given a point you want the line to go through, the point-slope form of the equation of the line is useful. That form for slope m and point (h, k) is ...
y = m(x -h) +k
For your slope m=5/2 and point (h, k) = (-2, 1), the equation of the line is ...
y = 5/2(x +2) +1
y = 5/2x +6 . . . . . eliminate parentheses
Answer:
Hello some parts of your question is missing below is the missing part
c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces
Answer: A) 0.0099
B) 0.6796
C) 0.13956
Step-by-step explanation:
weight of Navel oranges evenly distributed
mean ( u ) = 8 ounces
std ( б )= 1.5
navel oranges = X
A ) percentage of oranges weighing more than 11.5 ounces
P( x > 11.5 ) = 
= P ( Z > 2.33 ) = 0.0099
= 0.9%
B) percentage of oranges weighing less than 8.7 ounces
P( x < 8.7 ) = 
= P ( Z < 0.4667 ) = 0.6796
= 67.96%
C ) probability of orange selected weighing between 6.2 and 7 ounces?
P ( 6.2 < X < 7 ) = 
= P ( -1.2 < Z < -0.66 )
= Ф ( -0.66 ) - Ф(-1.2) = 0.13956