Step-by-step explanation:
In right angled triangle ABC,
Taking alpha as reference angle,
By pythagoras theorem,
p=BC,h=AB,b=AC
Taking thita as reference angle,
p=AC,h=AB,b=BC
<em>Keep</em><em> </em><em>smiling </em><em>and</em><em> </em><em>hop</em><em>e</em><em> </em><em>u</em><em> </em><em>are</em><em> </em><em>satisfied </em><em>with</em><em> </em><em>my</em><em> </em><em>answer</em><em>.</em><em>Have</em><em> </em><em>a</em><em> </em><em>good</em><em> </em><em>day</em><em> </em><em>:</em><em>)</em>
4 equal sides, 4 right angles, 2 pairs of parallel and opposit sides I think
Step-by-step explanation:
change graphic into alohabet so it is easier to see
solve easy equation with only a type of unknown first..
once you found the value of the unknownm , substitute it in any equation so you can find the value of other unknown.
If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
