Answer:
725.51
Step-by-step explanation:
Check the picture below.
![~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&5\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&9\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-4.9t^2+5t+9\implies \stackrel{h(t)}{0}=-4.9t^2+5t+9 \implies 4.9t^2-5t-9=0](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Binitial%20velocity%20in%20meters%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-4.9t%5E2%2Bv_ot%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%265%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%269%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20h%28t%29%3D-4.9t%5E2%2B5t%2B9%5Cimplies%20%5Cstackrel%7Bh%28t%29%7D%7B0%7D%3D-4.9t%5E2%2B5t%2B9%20%5Cimplies%204.9t%5E2-5t-9%3D0)
since we don't get any neat integer values from it hmmm let's plug that in the quadratic formula to see what we get as h(t) = 0.
![~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{4.9}t^2\stackrel{\stackrel{b}{\downarrow }}{-5}t\stackrel{\stackrel{c}{\downarrow }}{-9}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-5) \pm \sqrt { (-5)^2 -4(4.9)(-9)}}{2(4.9)}\implies t=\cfrac{5\pm\sqrt{25+176.4}}{9.8} \\\\\\ t=\cfrac{5\pm \sqrt{201.4}}{9.8}\implies t\approx \begin{cases} 1.96~~\checkmark\\ -0.94 \end{cases}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bquadratic%20formula%7D%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B4.9%7Dt%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B-5%7Dt%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B-9%7D%3D0%20%5Cqquad%20%5Cqquad%20t%3D%20%5Ccfrac%7B%20-%20b%20%5Cpm%20%5Csqrt%20%7B%20b%5E2%20-4%20a%20c%7D%7D%7B2%20a%7D%20%5C%5C%5C%5C%5C%5C%20t%3D%20%5Ccfrac%7B%20-%20%28-5%29%20%5Cpm%20%5Csqrt%20%7B%20%28-5%29%5E2%20-4%284.9%29%28-9%29%7D%7D%7B2%284.9%29%7D%5Cimplies%20t%3D%5Ccfrac%7B5%5Cpm%5Csqrt%7B25%2B176.4%7D%7D%7B9.8%7D%20%5C%5C%5C%5C%5C%5C%20t%3D%5Ccfrac%7B5%5Cpm%20%5Csqrt%7B201.4%7D%7D%7B9.8%7D%5Cimplies%20t%5Capprox%20%5Cbegin%7Bcases%7D%201.96~~%5Ccheckmark%5C%5C%20-0.94%20%5Cend%7Bcases%7D)
notice, we do not use the negative value for "t", since the seconds cannot be less than 0.
Answer:
67.38
Step-by-step explanation:
SinX=opp/hyp
Angles then arcsin
You have the following expression:
![(1+cosx)(1-\cos x)=\sin ^2x](https://tex.z-dn.net/?f=%281%2Bcosx%29%281-%5Ccos%20x%29%3D%5Csin%20%5E2x)
To verify the identity, you have that the left side of the equation becomes:
(1 + cosx)(1 - cosx) = (1 - cos^2 x)
because of the product of an expression by its conjugate result in a difference of squares. Or simplfy by expanding the factors and simplying.
Next, you have:
1 - cos^2 x = sin^2 x
because the Pythagorean identity.
Then, the identity is verified.
The answer is -7/5 which is answer d