Answer:
a.2s b.43.6m
Step-by-step explanation:
h(t)=−4.9t2+19.6t+24
24m = inittial possition
-4.9 t2= -1/2 g t2
19.6t= Vo t
as the initial velocity is positive, the ball is thrown up.
the maximum height of the ball is reached when the velocity is 0
V= Vo-gt=19.6m/s-9.8m/s2 t=0
t=19.6/9.8=2s
<u>it takes 2seconds for the ball to reach its maximum height</u>
h(t)=−4.9t2+19.6t+24
h(2)=-19.6m+39.2m+24=43.6m
<u>the maximum height of the ball is 43.6m</u>
The sum of two polynomials (polynomial plus polynomial) will always result in another polynomial
<span>9 + 0.07 + 0.005 this is expanded form i hope this is what u wanted as an answer</span>
Answer:
Step-by-step explanation:
The required t score would be determined from the t distribution table. In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 50 - 1 = 49
Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01
α/2 = 0.01/2 = 0.005
the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995
Looking at the t distribution table,
t score = 2.678
Let
![\{\varphi_i~|~i\in\mathbb N,1\le i\le k\}](https://tex.z-dn.net/?f=%5C%7B%5Cvarphi_i~%7C~i%5Cin%5Cmathbb%20N%2C1%5Cle%20i%5Cle%20k%5C%7D)
be a set of orthogonal vectors. By definition of orthogonality, any pairwise dot product between distinct vectors must be zero, i.e.
![\varphi_i\cdot\varphi_j=\begin{cases}\|\varphi_i\|^2&\text{if }i=j\\0&\text{if }i\neq j\end{cases}](https://tex.z-dn.net/?f=%5Cvarphi_i%5Ccdot%5Cvarphi_j%3D%5Cbegin%7Bcases%7D%5C%7C%5Cvarphi_i%5C%7C%5E2%26%5Ctext%7Bif%20%7Di%3Dj%5C%5C0%26%5Ctext%7Bif%20%7Di%5Cneq%20j%5Cend%7Bcases%7D)
Suppose there is some linear combination of the
![\varphi_i](https://tex.z-dn.net/?f=%5Cvarphi_i)
such that it's equivalent to the zero vector. In other words, assume they are linearly dependent and that there exist
![c_i\in\mathbb R](https://tex.z-dn.net/?f=c_i%5Cin%5Cmathbb%20R)
(not all zero) such that
![\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_1+\cdots+c_k\varphi_k=\mathbf 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Ekc_i%5Cvarphi_i%3Dc_1%5Cvarphi_1%2B%5Ccdots%2Bc_k%5Cvarphi_k%3D%5Cmathbf%200)
(This is our hypothesis)
Take the dot product of both sides with any vector from the set:
![v_j\cdot\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_j\cdot\varphi_1+\cdots+c_k\varphi_j\varphi_k=\varphi_j\cdot\mathbf 0](https://tex.z-dn.net/?f=v_j%5Ccdot%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Ekc_i%5Cvarphi_i%3Dc_1%5Cvarphi_j%5Ccdot%5Cvarphi_1%2B%5Ccdots%2Bc_k%5Cvarphi_j%5Cvarphi_k%3D%5Cvarphi_j%5Ccdot%5Cmathbf%200)
By orthogonality of the vectors, this reduces to
![c_j\|\varphi_j\|^2=0](https://tex.z-dn.net/?f=c_j%5C%7C%5Cvarphi_j%5C%7C%5E2%3D0)
Since none of the
![\varphi_i](https://tex.z-dn.net/?f=%5Cvarphi_i)
are zero vectors (presumably), this means
![c_j=0](https://tex.z-dn.net/?f=c_j%3D0)
. This is true for all
![j](https://tex.z-dn.net/?f=j)
, which means only
![c_i=0](https://tex.z-dn.net/?f=c_i%3D0)
will allow such a linear combination to be equivalent to the zero vector, which contradicts the hypothesis and hence the set of vectors must be linearly independent.