In this you have to do 6(total) /(Divided) 2/3. And when you do 6/(2/3) you also have to do KCF(keep, change, flip) and that brings you to 18/2 after multiplying across. So the answer would be 9 people.
300 - 297 = 3
294 - 291 = 3
288 - 285 = 3
So, that is a sequence of sums of number 3: 3 + 3 + 3 + .... + 3
How may times will the number 3 be added?
Note that 300 is reduced in 6 units each time => 300 / 6 = 50 => 3 will be added 50 times.
=> 50 * 3 = 150
Answer: 150
Answer:
The required y-value is -121.
Step-by-step explanation:
There are several ways in which we could approach this problem. From the two binomial factors (x + 8) and (x - 14), we know that the two roots are x = -8 and x = 14. The axis of symmetry is the vertical line x = 3. This value is halfway between x = -8 and x = 14.
Substituting x = 3 into f(x)= -(x+8)(x-14) results in f(3) = -(11)(-11) = -121.
The vertex is at (3, -121). The required y-value is -121.
Answer: No, Titus is incorrect.
Step-by-step explanation: Two shapes are <u>congruent</u> when they have the same size and shape, but one is created of rotation, reversion or translation of the other.
So, hexagons FEDCBA and F'E'D'C'B'A' are congruent because they have the same size and shape, however they are reversed and translated from each other, i.e.:
Comparing the coordinates of both hexagons:
F (-6,6) → F' (6, -4)
A (-10,6) → A' (10,-4)
E (-4,4) → E' (4, -6)
B (-12,4) → B' (12, -6)
D (-6,2) → D' (6, -8)
C (-10,2) → C' (10, -8)
We notice that the transformation necessary to transform FEDCBA into F'E'D'C'B'A' is
- multiply x-coordinate by (-1);
- subtract y-coordinate by 10;
Therefore, it is (x,y) → ( -x, y-10).
So, Titus is incorrect about the transformations that prove the hexagons are congruent.
