SOLUTION
In this question, we are meant to evaluate 4 x when x = -9.
4 x = 4 X -9 = -36
Answer:
Step-by-step explanation:
(-1,-2) , (3,10)
slope(m) = (10 - (-2) / (3 - (-1) = (10 + 2) / (3 + 1) = 12/4 = 3
y = mx + b
slope(m) = 3
u can use either of ur points....
(3,10)....x = 3 and y = 10
sub into the formula and find b, the y int
10 = 3(3) + b
10 = 9 + b
10 - 9 = b
1 = b
so ur equation is : y = 3x + 1
Answer:
(10, 3)
Step-by-step explanation:
Solve by Substitution
2x − 4y = 8 and 7x − 3y = 61
Solve for x in the first equation.
x = 4 + 2y 7x − 3y = 61
Replace all occurrences of x with 4 + 2y in each e quation.
Replace all occurrences of x in 7x − 3y = 61 with 4 + 2y. 7 (4 + 2y) − 3y = 61
x = 4 + 2y
Simplify 7 (4 + 2y) − 3y.
28 + 11y = 61
x = 4 + 2y
Solve for y in the first equation.
Move all terms not containing y to the right side of the equation.
11y = 33
x = 4 + 2y
Divide each term by 11 and simplify.
y = 3
x = 4 + 2y
Replace all occurrences of y with 3 in each equation.
Replace all occurrences of y in x = 4 + 2y with 3. x = 4 + 2 (3)
y = 3
Simplify 4 + 2 (3).
x = 10
y = 3
The solution to the system is the complete set of ordered pairs that are valid solutions.
(10, 3)
The result can be shown in multiple forms.
Point Form:
(10, 3)
Equation Form:
x = 10, y = 3
Answer:
56, 6.56, 0.23, -538 are the rational numbers
Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.
An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.
Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.
Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.
Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.
To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.
For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false
For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true
For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true
For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false
For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false
For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.
