Time out = d/so = d/523
time back = d/sb = d/415
t = time out + time back
11 = d/523 + d/415 find common deneminator
11 = (415d + 523d)/(523*415)
11*523*415 = 415d + 523d
2387495 = 938d
d = 2545 miles
hope this help
Answer:
ΔPTS≅ΔRTA by AAS axiom of congruency
Step-by-step explanation:
Consider ΔPQA and ΔRQS
∠PQA=∠RQS (Vertically Opposite Angles)
∠QAP=∠QSR (Complementary of two equal angles, ∠RAT and∠PST)
Due to angle sum property of a triangle, we come to the conclusion that
∠APQ=∠SRQ
Consider ΔPTS and ΔRTA
TA=TS (Given)
∠RAT=∠PST(Given)
∠APQ=∠SRQ (Proved above)
Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.
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:
draw a perpendicular line from the directrix passing through the focus, this will be the line of symmetry.
The vertex(h, k) will be located on the line half way between the focus and directrix.
The distance from the focus to the vertex is called the focal length, call it a. The then equation is
(x - h)^2 = 4a(y - k)
the equation can be manipulated to
y = 1/4a(x - h)^2 + k
hope it helps
Answer:
Step-by-step explanation:
y=-1
because when y = - 1,
it is an equation of x axis.. hence it will parallel to x axis....
