Answer:
(1252+252)(52−1)
=(1252+252)(52+−1)
=(1252)(52)+(1252)(−1)+(252)(52)+(252)(−1)
=390625−15625+15625−625
=390000
divide by 39 is
=10000
6a - (b - (3a - (2b + c + 4a - (a + 2b - c))))
6a - (b - (3a - (2b + c + 4a - a - 2b + c)))
6a - (b - (3a - (2b - 2b + 4a - a + c + c)))
6a - (b - (3a - (3a + 2c)))
6a - (b - (3a - 3a - 3c))
6a - (b - 3a + 3a + 3c)
6a - (b + 3c)
6a - b - 3c
x³ + x² - 25x - 25
x²(x) + x²(1) - 25(x) - 25(1)
x²(x + 1) - 25(x + 1)
(x² - 25)(x + 1)
(x² - 5x + 5x - 25)(x + 1)
(x(x) - x(5) + 5(x) - 5(5))(x + 1)
(x(x - 5) + 5(x - 5))(x + 1)
(x + 5)(x - 5)(x + 1)
36x² + 60x + 25
36x² + 30x + 30x + 25
6x(6x) + 6x(5) + 5(6x) + 5(5)
6x(6x + 5) + 5(6x + 5)
(6x + 5)(6x + 5)
(6x + 5)²
Answer:
I think it would be b
Step-by-step explanation:
He doesn't want to spend more than 50 and he needs 8 for lunch so the 8 dollars and the book can't go over 50.
Its either a or b im just not sure if its plus 8 or minus 8. sorry
Answer:

Step-by-step explanation:
The standard equation of a horizontal hyperbola with center (h,k) is

The given hyperbola has vertices at (–10, 6) and (4, 6).
The length of its major axis is
.



The center is the midpoint of the vertices (–10, 6) and (4, 6).
The center is 
We need to use the relation
to find
.
The c-value is the distance from the center (-3,6) to one of the foci (6,6)





We substitute these values into the standard equation of the hyperbola to obtain:


Answer:
vertex: (2,-18)
Step-by-step explanation:
y = ax² + bx + c
(-1,0) : a - b + c = 0 ...(1)
(5,0) : 25a +5b +c = 0 ...(2)
(0,-10): 0a + 0b + c = -10 c=-10
(1) x5: 5a - 5b + 5c = 0 ...(3)
(2)+(3): 30a + 6c = 0 30a = -6c = 60 a = 2
(1): 2 - b -10 = 0 b = -8
Equation: y = 2x² - 8x -10 = 2 (x² -4x +4) - 18 = 2(x-2)² -18
equation: y = a(x-h)²+k (h,k): vertex
vertex: (2,-18)