He equation of a parabola is x = -4(y-1)^2. What is the equation of the directrix?
<span>You may write the equation as </span>
<span>(y-1)^2 = (1) (x+4) </span>
<span>(y-k)^2 = 4p(x-h), where (h,k) is the vertex </span>
<span>4p=1 </span>
<span>p=1/4 </span>
<span>k=1 </span>
<span>h=-4 </span>
<span>The directrix is a vertical line x= h-p </span>
<span>x = -4-1/4 </span>
<span>x=-17/4 </span>
<span>------------------------------- </span>
<span>What is the focal length of the parabola with equation y - 4 = 1/8x^2 </span>
<span>(x-0)^2 = 8(y-4) </span>
<span>The vertex is (0,4) </span>
<span>4p=8 </span>
<span>p=2 (focal length) -- distance between vertex and the focus </span>
<span>------------------------------- </span>
<span>(y-0)^2 = (4/3) (x-7) </span>
<span>vertex = (7,0) </span>
<span>4p=4/3 </span>
<span>p=1/3 </span>
<span>focus : (h+p,k) </span>
<span>(7+1/3, 0)</span>
Answer:
x = 16.5
Step-by-step explanation:
Use pythagorean theorem
a² + b² = c²
Plug in terms
7.1² + x² = 18²
Square each term
50.41 + x² = 324
Subtract a from c
x² = 273.59
Find square root
16.5
15/3 = 35/x
Cross multiply.
15x=105
105/15=x
x=7
Abe have to take 7 weeks of school this quarter.
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function
Part a
If we do the dimensional analysis for v we got:
Part b
For the acceleration we can use the result obtained from part a and we got:
Part c
From definition if we do the integral of the velocity respect to t we got the position:
And the dimensional analysis for the position is:
Part d
The integral for the acceleration respect to the time is the velocity:
And the dimensional analysis for the position is:
Part e
If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:
