Ms. Cassidy instructed Miguel to change one sign of the graph of y < 2x – 4 so that point (2, 3) can be included in the solution set.
To check which of the given options might Miguel write we check the inequality that holds true for the point (2,3).Substituting x=2 ,y=3 we have:
1) y < 2x – 1
3<2(2)-1
3<3 Not True.
2)y ≤ 2x – 4
3≤ 2(2) -4
3≤ 0 .Not true.
3) y > 2x – 4
3> 2(2)-4
3> 0 True.
4) y < 2x + 4
3<2(2)+4
3<8 True
5) .y < 3.5x – 4
3< 3.5(2)-4
3<3 Not true
6) y < 4x – 4
3<4(2)-4
3<4 True.
Options 3 ,4 ,6 holds true for the point (2,3)
Answer: 2/3
Step-by-step explanation:
Answer:
Neither linear nor exponential
Step-by-step explanation:
To check for a linear relationship. Find slope.
slope= (-1 - (-2)) / ( 5 - 2) = 1/3
check other points
slope = (1 - (-1) )/ (8 - 5) = 2/3
check more
slope = (4 - 1) / (11 - 8) = 3/ 3 = 1
Nope.
try assuming an exponential:
y = c * (a^x)
-2 = c* (a^2); -2/c = a^2
-1 = c *(a ^5); -1/c = a^5
1 = c * (a^8), 1/c = a^8
(-2/c)^4 = a^8 = 1/c
16/(c^4) = 1/c
c^3 = 16, then a = root (-2/ cube-root(16) )
The change from negative to postive would not work for y = c(a^x)
so...
assume y = a^x + k
-2 = a^2 + k
-1 = a^5 + k
... I would say neither..
The formula is
A=p e^rt
A future value?
P present value 2600
R interest rate 0.085
T time 5years
E constant
A=2,600×e^(0.085×5)
A=3,976.94
Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|
Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5
Required answer is f(x) = one fourth (x - 3)^2 + 5
