because y^3 is raised to the power of two, we will have to multiply the exponents rather than adding them.
By distributing the power of 2, we will get y^2(6).
Because now the exponents are being multiplied, we can just add them to get y^8. The other y has a power of 1, so we'll just add the power of that y as well to get y^9.
summary:
multiply exponents if they are being raised to a power.
add exponents if they are being multiplied, and only add them if they have the same base (in this case, both the bases of the exponents are y, so we can add them)
Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.
An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.
Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C:
Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
</span>
Yes, HL would be correct.
This states that if two hypotenuses are congruent (they have the same one)
and if one pair of legs are congruent,
the figure is congruent.
I hope this helps!
~cupcake
7 - 4 ( d - 3) = 23
7 - 4d + 12 = 23
Subtract 12 from both sides,
7 - 4d = 11
Subtract 7 to both sides
- 4d = 4
Divide -4 to both sides
d = -1
