Answer:
x =5 ±sqrt( 31)
Step-by-step explanation:
x² + 10x = 6
b = 10
Take b/2 and square it
(10/2) = 5 5^2 = 25
Add 25 to each side
x² + 10x+25 = 6+25
The left side will be (x+b/2)^2
(x+5)^2 = 31
Take the square root of each side
sqrt((x+5)^2) = ±sqrt( 31)
x+5 = ±sqrt( 31)
Subtract 5 from each side
x+5-5 = -5 ±sqrt( 31)
x =5 ±sqrt( 31)
Answer: Option A) x=3π/4, x=5π/4
The asymptotes of the function tan (z) are the values of z that are the odd multiples of <span>π/2:
</span>z=(2n+1)π/2, witn n= ..., -3, -2, -1, 0, 1, 2, 3, ...
In this case z=<span>2x − π, then:
</span>2x − π=(2n+1)π/2
Solving for x: Adding π both sides of the equation:
2x − π+ π=(2n+1)π/2+ π
Adding the terms on the right side of the equation:
2x=[(2n+1)π+2π]/2
Common factor π:
2x=[(2n+1)+2]π/2
2x=(2n+1+2)π/2
2x=(2n+3)π/2
Multiplying both sides of the equation by 1/2:
(1/2)(2x)=(1/2)[(2n+3)π/2]
(2/2)x=(2n+3)π/[(2)(2)]
x=(2n+3)π/4
For n=-1:
x=[2(-1)+3]π/4
x=(-2+3)π/4
x=π/4
x=π/4<π/2 No
For n=0:
x=[2(0)+3]π/4
x=(0+3)π/4
x=3π/4
π/2<x=3π/4<3π/2 Ok
For n=1:
x=[2(1)+3]π/4
x=(2+3)π/4
x=5π/4
π/2<x=5π/4<3π/2 Ok
For n=2:
x=[2(2)+3]π/4
x=(4+3)π/4
x=7π/4
x=7π/4>3π/2 No
Answer: x=3π/4, x=5π/4
You didn't include the graphs. However, I can describe the graph for you, so you can select the correct graph.
First, the car slows. So look for a line segment that is pointing down diagonally. Then, it maintains the speed. Look for a line that is horizontal. Finally, it slows down again to a stop. Look for a line segment that slants downward and touches the x axis.