well, looking at the picture of this vertically opening parabola, it has a vertex at 0,0 and it passes through 2,1 hmm ok
![~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = a(x-0)^2+0\qquad \stackrel{\textit{we also know that}}{x=2\qquad y = 1}\qquad \implies 1=a(2-0)^2+0 \\\\\\ 1=4a\implies \cfrac{1}{4}=a~\hspace{10em} \boxed{y=\cfrac{1}{4}x^2}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20y%20%3D%20a%28x-0%29%5E2%2B0%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bwe%20also%20know%20that%7D%7D%7Bx%3D2%5Cqquad%20y%20%3D%201%7D%5Cqquad%20%5Cimplies%201%3Da%282-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%201%3D4a%5Cimplies%20%5Ccfrac%7B1%7D%7B4%7D%3Da~%5Chspace%7B10em%7D%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B4%7Dx%5E2%7D)
Answer:
8 , 5 , 5.785
Step-by-step explanation:
y^2 - x^3
first just plug in the numbers and you get
(4)^2 - (2)^3
Multiply
16-8 and this equals 8
x1 + x^3 - y^2
same pattern plug in the numbers
(3)^(1) + (3)^3 - (5)^2
multiply
3 + 27 - 25
P
E
M
D
Addition
Subtraction so do 3+27 first and you get
30 - 25 = 5
x^4 / x^3 plug it in
(6)^4 / (6)^3
multiply
1269 / 216
divide this and you get
5.875
After simplifying the real and imaginary parts you'd get -8i
Answer:
a
Step-by-step explanation:
Given
u = (x₁, y₁, z₁ ) and v = (x₂, y₂, z₂ ) then the dot product is
u • v = (x₁ x₂ ) + (y₁ y₂ ) + (z₁ z₂ )
Then
v • w
= (5 × 9) + (- 4 × - 2) + (4 × 7)
= 45 + 8 + 28
= 81 → a
