The correct answer would be D. It represents a nonlinear function because it's points are not on a straight line.
This is correct because a linear function would be a straight line, and because when graphed the table doesn't create a line, it would be a nonlinear function.
Answer:
(x)^2 (y)^2
---------- + --------- = 1
4 3
Step-by-step explanation:
The standard equation for an ellipse is
(x-h)^2 (y-k)^2
---------- + --------- = 1
a^2 b^2
The center is at (h,k)
The vertices are at (h±a, k)
The foci are at (h±c,k )
Where c is sqrt(a^2 - b^2)
It is centered at the origin so h,k are zero
(x)^2 (y)^2
---------- + --------- = 1
a^2 b^2
The center is at (0,0)
The vertices are at (0±a, 0)
The foci are at (0±c,0 )
The vertices are (±2,0) so a =2
The foci is 1
c = sqrt(a^2 - b^2)
1 = sqrt(2^2 - b^2)
Square each side
1 = 4-b^2
Subtract 4 from each side
1-4 = -b^2
-3 = -b^2
3= b^2
Take the square root
b=sqrt(3)
(x)^2 (y)^2
---------- + --------- = 1
4 3
Answer:
Jack's age is 36
Wade's age is 16
Step-by-step explanation:
Let jack's age be "j"
and
Wade's age be "w"
First, we have:
6 yrs AGO, Jack was 3 times as Wade, so we can write:
j-6 = 3(w-6)
Then, we have:
4 yrs from NOW, Wade will be HALF as Jack, so we can write:
w+4 = 0.5(j+4)
Lets simplify the first equation:
j-6 = 3(w-6)
j - 6 = 3w - 18
Solving for j:
j = 3w - 18 + 6
j = 3w - 12
Now, simplifying the second equation:
w+4 = 0.5(j+4)
w+4 = 0.5j+2
w = 0.5j - 2
Now we replace this equation with j's expression we found from first equation:
w = 0.5j - 2
w = 0.5(3w - 12) - 2
w = 1.5w - 6 - 2
0.5w = 8
w = 16
Now, finding j:
j = 3w - 12
j = 3(16) - 12
j = 36
So,
Jack's age is 36
Wade's age is 16
Answer:
7.8
3.3 x 10^3
2.1 x 10^-3
3.7
Step-by-step explanation:
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The answer is 14 do you need anymore help
