Answer:
it goes by 2 two times then subtracts 3 then again
Step-by-step explanation:
10,12,9,11,8,10,7
<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>
Note that the domain of h is <span>[0,∞]</span>.
By differentiating,
<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>
by factoring out <span>34</span>,
<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>
by finding the common denominator,
<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>
<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>
Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.
Hence, the critical numbers are <span>t=0,1</span>.
I hope that this was helpful.
Answer:
sin θ . tan θ
Step-by-step explanation:
Note : -
sec ( - θ ) = sec θ
Formula / Identity : -
sec θ = 1 / cos θ
sec ( - θ ) - cos θ
= [ 1 / cos θ ] - cos θ
{ LCM = cos θ }
= [ 1 / cos θ ] - [ cos²θ / cos θ ]
= [ 1 - cos²θ ] / cos θ
{ 1 - cos²θ = sin²θ }
= sin²θ / cos θ
{ sin²θ = sin θ . sin θ }
= sin θ . sin θ / cos θ
{ sin θ / cos θ = tan θ }
= sin θ . tan θ
Hence, simplified.
