Answer:
The graph now opens in the opposite direction. The y intercept is shifted up 4.
Step-by-step explanation:
F(x) = 2x^2-8
F(X)= -2x^2-4
We know that the graph is reflected over the y-intercept becuase of the negative sign in the second equation ( F(X)= -2x^2-4)
We know that the intercept is shifted up four, because the number decreased from 8 to 4, and the graph is reflected.
Attacthed is both of the equations graphed to help you visualize this shift!
Answer:
The length of BE is 27 units ⇒ 3rd answer
Step-by-step explanation:
In circle A:
∠BAE ≅ ∠DAE
Line segments A B, A E, and A D are radii
Lines are drawn from point B to point E and from point E to point D to form secants B E and E D
The length of B E is 3 x minus 24 and the length of E D is x + 10
We need to find the length of BE
∵ AB and AD are radii in circle A
∴ AB ≅ AD
In Δs EAB and EAD
∵ ∠BAE ≅ ∠DAE ⇒ given
∵ AB = AD ⇒ proved
∵ EA = EA ⇒ common side in the two triangles
- Two triangles have two corresponding sides equal and the
including angles between them are equal, then the two
triangles are congruent by SAS postulate of congruence
∴ Δ EAB ≅ Δ EAD ⇒ SAS postulate of congruence
By using the result of congruence
∴ EB ≅ ED
∵ EB = 3 x - 24
∵ ED = x + 10
- Equate the two expressions to find x
∴ 3 x - 24 = x + 10
- Add 24 to both sides
∴ 3 x = x + 34
- Subtract x from both sides
∴ 2 x = 34
- Divide both sides by 2
∴ x = 17
Substitute the value of x in the expression of the length of BE to find its length
∵ BE = 3 x - 24
∵ x = 17
∴ BE = 3(17) - 24
∴ BE = 51 - 24
∴ BE = 27
The length of BE is 27 units
Answer:
its b
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
I used a calculator and the answer is 6
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
