Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:
Using the distance formula, we get
Similarly,
Now, the length of the rectangle is and the width of the rectangle is . So, the area of the rectangle is:
Therefore, the area of the rectangle is 20 square units.
Answer:60 male staff
Step-by-step explanation:
Ok, there are 96 female staff. The ratio of male to female is 3:4. We’re going to use a proportion which looks like this. 3/4 = x/96. X is a variable that takes the place of the male staff since we don’t know what it is. Now, the question is what do we do to get from 4 to 96. So 96/4 = 24, so 4 x 20 = 96. To keep this ratio equivalent whatever you do to the top you have to do to the bottom so 4 x 20 = 96 and 3 x 20 = 60. So there are 60 male staff.
Answer: x^2 + y^2 -10y = 0
Step-by-step explanation:
Cartesian coordinates, also called the Rectangular coordinates, isdefined in terms of x and y. So, for the problem θ has to be eliminated or converted using basic foundations that are described by the unit circle and the right triangle trigonometry.
r= 10sin(θ)
Remember that:
x= r × cos(θ)
y= r × sin(θ)
r^2= x^2 + y^2
Multiply both sides of the equation by r. This will give:
r × r = 10r × sin(θ)
r^2 = 10r × sin(θ)
x^2 + y^2= 10r × sin(θ)
Because y= r × sin(θ), we can make a substitution. This will be:
x^2 + y^2= 10y
x^2 + y^2 -10y = 0
The above equation is the Rectangular coordinate equivalent to the given equation.
Answer:
the student should score atleast 229 to be among the top 10%.
Step-by-step explanation:
in terms of the normal distribution, and if the table that you're using calculates the area of the normal distribution from the mean to a point x, only then what we are actually finding the value 'x' at which the z-score is at 40% (the rest 50% is already skipped by the table)
after finding the the value at this z-score, we can find the value of x at which the score is in the top 10% range.
we can find the z-score either using a normal distribution table or calculator. (but be sure what area is it calculating)
looking at the table the closest value we can find is, 0.4015 at z = 1.29 ((it is above 40% because we want to be in the top 10% range)
the student should score atleast 229 to be among the top 10%.
Substitute
, so that
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes
which has the characteristic equation
with roots at
. This means the characteristic solution for
is
and in terms of
, this is
From the given initial conditions, we find
so the particular solution to the IVP is
