Answer:
sorry hindi ko alam hirapuwu
A. The (x,y) coordinates where the equations y =4x and y = 2x - 1 intersect on a graph is the solution to that system of equations. When solving for this system of equations by substitution of y, the equation 4x = 2x -1 is obtained. So the x-coordinate obtained is both a solution to the equation 4x = 2x - 1 and the intersection point of the system of equations y = 4x and y = 2x - 1.
B. Idk about this part... Says to make a tables to solve for 4x = 2x - 1.
If you make a table for Eq. #1. y = 4x and Eq. #2. y = 2x -1 maybe..
#1. y = 4x
x ...... y
-4 .. -16
-3 .. -12
-2 .. -8
-1 .. -4
0 ... 0
1 ... 4
2 .... 8
3 ... 12
4 ... 16
the solution to 4x = 2x - 1 ; the x-coordinate is not an integer value so Idk what the point of this is.....
4x = 2x - 1
4x - 2x = -1
2x = -1
x = -1/2
not an integer..
C. You can solve the equation 4x = 2x - 1 graphically by graphing y = 4x and y = 2x - 1 and identifying the x-coordinate at the point of intersection.
if we take 64 to be the 100%, how much is 6¼% off of it?
(a) If lies on both planes, then
and at the same time
(b) A plane with normal vector containing the point can be written in the form
Expanding the left side, we see that the components of correspond to the coefficients of . So the normal vector to is .
(c) Similarly, the normal to is .
(d) The cross product of any two vectors and is perpendicular to both of the vectors. So we have
(e) Solve the two plane equations for .
By substitution,
Let . Then and
Then the intersection can be parameterized by equations
for .
We can also set or first, then solve for the other variables in terms of the parameter , so this is by no means a unique parameterization.
We equate the derivative of the function to 0 to find the x-value at its minimum:
f'(x) = 4x + 28
0 = 4x + 28
x = -7
Now, we put this value into the original equation to find the minimum value:
f(-7) = -106
The function has a positive slope so it will increase. Thus, the range will be y > -106
The answer is B