Answer:
(x, y) = (0, p)
Step-by-step explanation:
In the given equation, p is the distance from the vertex (0, 0) to the focus of the parabola. Any vertical "light beam" will bounce off the curve and pass through the focus at (0, p).
Answer:
bruh okie
Step-by-step explanation:
So first off, the answer to this is 24 now think of x as the real or original sum or amount, also y as the new sum or amount.
y * 1.5 is x, and y+12 is x. Now flip that to figure out what is y, also which is what we need, and you have x/1.5 = y as well as x-12 = y. Use the equal values method and make an equation x/1.5=x-12. solve this equation to get x, which is 36. to figure out the new amount, y, you need to minus 12, which can help you get the answer which is 24.
Answer:
214.5 seconds or less time will qualify individuals for such training
Step-by-step explanation:
given that the time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec.
i.e. X the time that it takes a randomly selected job applicant to perform a certain task is N(150,25)
The fastest 10% are the ones who are above the 90th percentile
We know 90th percentile for Z std normal score is 1.28
Corresponding X score would be

Round off to one decimal
214.5 seconds or less time will qualify individuals for such training
Answer:
.
Step-by-step explanation:
The equation of a circle of radius
centered at
is:
.
.
Differentiate implicitly with respect to
to find the slope of tangents to this circle.
![\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E%7B2%7D%20%2B%20y%5E%7B2%7D%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B25%5D)
.
Apply the power rule and the chain rule. Treat
as a function of
,
.
.
.
That is:
.
Solve this equation for
:
.
The slope of the tangent to this circle at point
will thus equal
.
Apply the slope-point of a line in a cartesian plane:
, where
is the gradient of this line, and
are the coordinates of a point on that line.
For the tangent line in this question:
,
.
The equation of this tangent line will thus be:
.
That simplifies to
.