A) 144 in
is your answer. good luck
Answer:
B. (1,5) and (5.25, 3.94)
Step-by-step explanation:
The answer is where the 2 equations intersect.
We need to solve the following system of equations:
y = -x^2 + 6x
4y = 21 - x
From the second equation:
x = 21 - 4y
Plug this into the first equation:
y = -(21 - 4y)^2 + 6(21 - 4y)
y = -(441 - 168y + 16y^2)+ 126 - 24y
y = -441 + 168y - 16y^2 + 126 - 24y
16y^2 + y - 168y + 24y + 441 - 126 = 0
16y^2 - 143y + 315 = 0
y = [-(-143) +/- sqrt ((-143)^2 - 4 * 16 * 315)]/ (2*16)
y = 5, 3.938
When y = 5:
x = 21 - 4(5) = 1
When y = 3.938
x = 21 - 4(3.938) = 5.25.
Ok so any number tat makes the denomenator 0 or makes the inside of a square root negative is restricted
we only have a denomenaor so
100v=0
v=0
therefor 0 is the excluded value since 0/0 doesn't make sense
No.
The distance is |-5 -3| = 8 units. (Y-coordinates are the same, so the distance is measured entirely in the x-direction. Distance is non-negative.)
a is the beggining number and n is the amount of times needed to do this. if is 17 you would need to subtract 7 from the originol number 12 (a) that many times.
