Given the function g(x) = x^3 − (8+1)x^2-4(9+2)x+(5x2-1) determine where the function is quickly(slowly) increasing (decreasing)
1 answer:
Answer:
hence g(x) is increasing at point x= 3+14i,3-14i
Step-by-step explanation:
given, g(x)=x^3-(8+1)x^2-4(9+2)x+5x2-1
g(x)=x^3-9x^2-44x+10x-1
g(x)=x^3-9x^2-34x-1
=3x^2-18x-34
for increasing and decreasing function,
3x^2-18x-34=0
x=3+14i 0
x=3-14i 0
hence g(x) is increasing at point x= 3+14i,3-14i answer
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Answer:
The relationship between the circumference of a circle and its diameter represent a direct variation and the constant of proportionality is equal to the constant
Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form
where K is the constant of proportionality
In this problem we know that
The circumference of a circle is equal to
therefore
the relationship between the circumference of a circle and its diameter is a direct variation and the constant of proportionality is equal to the constant
Answer:
C
Step-by-step explanation:
To make it easy let's start by organizing our information :
- AC=12 AND BD=8
- ABCD is a rhombus
- K and L are the midpoints of sides AD and CD
- we notice that the rhombus ABCD is divided into four right triangles
What do you think of when you hear a right triangle ?
- The pythagorian theorem !
AC and BD are khown so let's focus on them .
If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?
Simply because they are the diagonals of a rhombus .
ow let's apply the pythagorian theorem :
- (AC/2)² + (BD/2)² = BC²
- 6²+4²=52
- BC²= 52⇒
=BC
Now we khow that : AB=BC=CD=AD=
This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle
- KL is parallel to AC
- k is the midpoint of AD and L of DC
I smell something . yes! Thales theorem
- KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
- KL/12=
- KL/12=1/2⇒ KL=6
Now we have the length of the base kl
Now the big boss the height :
- notice that you khow the length of KL
- BD crosses kl from its midpoint and DL =
What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D
- DL²=(KL/2)²+D²
- 52/4= 9+ D²
- D² = 52/4-9 +4 SO D=2
now the height of the trigle is H= BD-D= 8-2=6
NOw the area of the triangle is :
- A=(KL*H)/2 ⇒ A= (6*6)/2=18
THE ANSWER IS 18 SQ.UN