#1 √ 75
Answer : 5√3 or ≈ 8.66025
#2 √ 48
Answer : 4√3 or ≈ 6.9282
#3 √ 128
Answer : 8√2 or ≈ 11.31371
#4 √ 300
Answer : 10√3 or ≈ 17.32051
~~
#5 √⅔
Answer :
#6 √16/4
Answer : 2
#7 √ 6/8
Answer :
#8 √27/9
Answer : √ 2 or ≈ 1.73205
~~
I hope that helps you out!!
Any more questions, please feel free to ask me and I will gladly help you out!!
~Zoey
Y=mx+b, m= slope, b= y intercept, plug in. Y=3x+5
Answer:
{5, 7, 9, 11}
Step-by-step explanation:
Roster form just means list the elements of the set in the brackets.
It says the elements are greater than or equal to five and less than or equal to 11. So we start at 5 and end at 11.
We list {5, 7, 9, 11} because these elements follow the rule that they must be odd natural numbers. Natural numbers are the numbers you can count starting from 0 like {0, 1, 2, 3, ...}.
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
<span>0.05 arc-second = 1 degree/72000 = (pi
radians)/(180*72000) = 2.424 x 10^(-7) radians</span>
<span>The distance is roughly: </span>
<span>R*(theta) = (600 light-years)*2.424 x 10^(-7) = 0.00014544 light-years = 1.275
light-hours = (3600 seconds)*(3 x 10^8 m/s)*(1.275) = 1.38 x 10^12 meters.</span>
