It’s 120.......................
Answer:
Width=x
Length =x+5
Perimeter =38
2(x+x+5)=38
=>4x+10=38
=>x=28/4
Step-by-step explanation:
1,3 this the answer ejdbdkdbndndjdbd
Answer:
See below
Step-by-step explanation:
π, pronounced as <em>pi</em> is the ratio between 2πr which is circumference to 2r which is diameter.
In word, π is circumference/diameter which has exact value of 3.14159... and it is also an irrational number too.
π is often used to be defined as 3, 3.14 or 22/7 [Keep in mind, these numbers are only approximation.] in some fields. In trigonometry, π is mostly used to refer as <u>π radians</u> or <u>π rad</u> which is equivalent to 180°.
Why do we need it? Because it’s useful in a field that requires rotation. If you notice, figure with circular base will always have π with it, the volume is an example.
In calculus, you need π to do solid of revolution method which is:
![\displaystyle \large{V =\pi \int\limits^a_b {[f(x)]^2} \ dx }\\\\\displaystyle \large{V = \pi \int\limits^a_b {[g(y)]^2} \ dy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7BV%20%3D%5Cpi%20%5Cint%5Climits%5Ea_b%20%7B%5Bf%28x%29%5D%5E2%7D%20%5C%20dx%20%7D%5C%5C%5C%5C%5Cdisplaystyle%20%5Clarge%7BV%20%3D%20%5Cpi%20%5Cint%5Climits%5Ea_b%20%7B%5Bg%28y%29%5D%5E2%7D%20%5C%20dy%7D)
In trigonometry, you need it for measurement/angle.
So anything that’s related to revolution, rotation or related to circle must require π at least.
In general life, you do not use it but this is mostly applied in engineering field and anything related to science that involves with circle or revolve.
Let me know if you have any questions!
Answer:
.
Step-by-step explanation:
Let the
-coordinate of
be
. For
to be on the graph of the function
, the
-coordinate of
would need to be
. Therefore, the coordinate of
would be
.
The Euclidean Distance between
and
is:
.
The goal is to find the a
that minimizes this distance. However,
is non-negative for all real
. Hence, the
that minimizes the square of this expression,
, would also minimize
.
Differentiate
with respect to
:
.
.
Set the first derivative,
, to
and solve for
:
.
.
Notice that the second derivative is greater than
for this
. Hence,
would indeed minimize
. This
value would also minimize
, the distance between
and
.
Therefore, the point
would be closest to
when the
-coordinate of
is
.