Hello,
y=2xe^x
y'=2(e^x+xe^x)=2(x+1)e^x
y''=2(e^x+(x+1)e^x)=2(x+2)e^x
x |-infinite -2 0 +infinite
e^x | ++++++++++++++++++
x+2 |------------0 +++++++++++
y'' | -----------0 +++++++++++
y''<0 if x<-2
<span>The interval on which the graph is concave down is (-infinite -2[</span>
Answer: a) y=-29x-20 b) y=3/5x-18
Step-by-step explanation:
a) m = (-49-38)/(1 - (-2)) = -87/3 = -29
y=-29x+b --> Substitute (1,-49) as a point
(-49)=-29(1)+b
-49=-29+b
-b=-29+49
b=-20
Therefore, the equation of the line is y=-29x-20
b) m = (-18-(-9)/(0-15)) = -9/-15 = 3/5
The y-intercept is (0,-18) as it's where the line intersects the y-axis and x equals 0
Therefore, the equation of the line is y=3/5x-18
Number one: Just plot down on the number like the numbers you have. Just took the test for this.
Answer:
C
Step-by-step explanation:
The n th term formula of a geometric sequence is
= a 
where a is the first term and r the common ratio
Using the second and fourth term, then
ar = 6 → (1)
ar³ = 54 → (2)
Divide (2) by (1)
= 
= 9
r² = 9 ⇒ r = 
= 3 → C
Answer:
and the graph would fall at a slower rate to the right
Step-by-step explanation:
we have
Observing the graph
The initial value Ao of the truck is equal to the y-coordinate of the y-intercept of the graph
so
If the depreciation rate was 15% per year, the new formula would be
----> the graph would fall at a slower rate to the right
using a graphing tool
compare the graphs
