The standard form of a quadratic equation is
, while the vertex form is:
, where (h, k) is the vertex of the parabola.
What we want is to write
as
First, we note that all the three terms have a factor of 3, so we factorize it and write:
.
Second, we notice that
are the terms produced by
, without the 9. So we can write:
, and substituting in
we have:
![\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11]](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29%3D3%5B%28x-3%29%5E2-9-2%5D%3D3%5B%28x-3%29%5E2-11%5D)
.
Finally, distributing 3 over the two terms in the brackets we have:
![y=3[x-3]^2-33](https://tex.z-dn.net/?f=y%3D3%5Bx-3%5D%5E2-33)
.
Answer:
Answer:
a)g: 3x + 4y = 10 b) a:x+y = 5 c) c: 3x + 4y = 10
h: 6x + 8y = 5 b:2x + 3y = 8 d: 6x + 8y = 5
Step-by-step explanation:
a) Has no solution
g: 3x + 4y = 10
h: 6x + 8y = 5
Above Equations gives you parallel lines refer attachment
b) has exactly one solution
a:x+y = 5
b:2x + 3y = 8
Above Equations gives you intersecting lines refer attachment
c) has infinitely many solutions
c: 3x + 4y = 10
d: 6x + 8y = 5
Above Equations gives you collinear lines refer attachment
i) if we add x + 2y = 1 to equation x + y = 5 to make an inconsistent system.
ii) if we add x + 2y = 3 to equation x + y = 5 to create infinitely system.
iii) if we add x + 4y = 1 to equation x + y = 5 to create infinitely system.
iv) if we add to x + y =5 equation x + y = 5 to change the unique solution you had to a different unique solution
Substitute 
, so that 
. Then the ODE is equivalent to
which is separable as
Split the left side into partial fractions,
so that integrating both sides is trivial and we get
Given the initial condition 
, we find
so that the ODE has the particular solution,
Answer:
a) 75/3=25/1
Step-by-step explanation:
Answer:
a = 2 5/8
Step-by-step explanation:
14a - (2a + 9) = 3 (12a - 18)
14a - 2a + 9 = 36a - 54
12a + 9 = 36a - 54
12a - 36a = -54 - 9
-24a = -63
a = -63/-24
a = 21/8
a = 2 5/8
