Lets say thoseodd integers are x,x+2
From the given equation
X+Y=20-->i
X-Y=14--->ii
add eq i and eq ii
2x=34
x=17
putting the value of x=17 in eq i
17+y=20
then y = 3 so the right answer is
17 and 3
QUESTION 1
the solution is (-1,-5)
To verify the solution algebraically, we need to multiply the first equation with -1 and then sum them up
-y=5
y=3x-2
0=3x+3
3x=-3
x=-3/3
x=-1
Then, we use this value to find the other one
y=3x-2
y=3(-1)-2
y=-3-2
y=-5
Therefore, the solution is (-1,-5), which is the same showed graphically.
QUESTION 2.
Using the same process as we did in the QUESTION 1. The image attached shows both lines and the interception point which is the solution. So, the solution is (6,0).
To verify the solution algebraically, we must multiply the second equation by -6
x+6y=6
-6y=-2x+12
x=-2x+18
x+2x=18
x=18/3
x=6
Then, we use this value to find the other one
x+6y=6
6+6y=6
6y=6-6
y=0/6
y=6
Therefore, the solution is (6,0),
Answer:
Parent function: y = sin(x)
Transformations:
1) horizontal translation of pi units towards left
2) stretch parallel to y-axis with factor "2'
3) vertical translations of 1 unit downwards
Range: [-3 , 1]
Step-by-step explanation:
Parent function: y = sin(x)
Transformations:
1) horizontal translation of pi units towards left
2) stretch parallel to y-axis with factor "2'
3) vertical translations of 1 unit downwards
Sketch the graph using 5 points listed below:
1) (0, -1)
2) (pi/2, -3)
3) (pi, -1)
4) (3pi/2, 1)
5) (2pi, -1)
Range: [-3 , 1]
Answer:
z = 3y + x
Step-by-step explanation:
z - x / 3 = y
multiply both sizes by 3
z - x = 3y
add x
z = 3y + x
