Answer:
The last equation x2 - 2x -4 = 0
has solution (x - 1)^2 - 5 = 0, x = 1 + root(5) or x = 1 - root(5)
Step-by-step explanation:
If a quadratic function has roots 1 and 5
f(x) = (x -1)(x- 5)
f(x) = x^2 - 6x + 5
Unless you meant. -4 and 6 ?
g(x) = (x + 4)(x - 6)
g(x) = x^2 -2x -24
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Or did you mean x = 1 and x =4 ?...
x^2 + 2x + 4 = 0 : complete square x^2 + 2x + 1 + 3 = 0, (x+1)^2 + 3 = 0
x^2 - 2x + 4 = 0 : complete square: (x -1)^2 + 3 = 0
0x^2 + 2x - 4 = 0, 2x - 4 = 0, x = 2
x^2 - 2x - 4 = 0 becomes: x^2 - 2x + 1 - 1 -4 = 0 ; (x - 1)^2 - 5 = 0
I believe 10 bald eagles plz correct me if I'm wrong
Answer:
Increase by 72.5 percent
Step-by-step explanation:
Just dived 435 by 600 and you get your answer
1.0g/cm3 means that the mass of one cm3 is 1.0g
The easiest method to use is the rule of three, and let x be the mass of 10.0 cm3 of water
1g -- > 1.0 cm3
x --> 10.0 cm3
x= (10*1)/1
x=10.0 g
So the mass of 10.0 cm3 of water is 10.0g
Hope this Helps! :)
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.