I do y get this but it might be 4
Answer:
x^87
Step-by-step explanation:
Multiplying x^22 and x^7 results in x^29.
Then we have:
(x^29)^3 = x^87
Recall that (a^b)^c = a^(bc) and that a^b*a^c = a^(b + c)
For this parabola we have:
f ( 0 ) = 8
and : f ( 1 ) = 24
In the first equation ( A) :
f ( 0 ) = - 16 * ( 0 - 1 )² + 24 = - 16 * 1 + 24 = 8 ( correct )
f ( 1 ) = - 16 * ( 1 - 1 )² + 24 = 24 ( correct )
For B:
f ( 0 ) = - 16 * ( 0 + 1 )² + 24 = - 16 + 24 = 8 ( correct )
f ( 1 ) = - 16 * ( 1 + 1 )² + 24 = - 16 * 4 + 24 = - 64 + 24 = 40 ( false )
For C:
f ( 0 ) = - 16 * ( 0 - 1 )² - 24 = - 16 - 24 = - 40 ( false )
f ( 1 ) = - 16 * ( 1 - 1 )² - 24 = - 24 ( false )
For D:
f ( 0 ) = - 16 * ( 0 + 1 )² - 24 = - 16 - 24 = - 40 ( false )
f ( 1 ) = - 16 * ( 1 - 1 )² - 24 = - 24 ( false )
Answer:
A ) f ( t ) = - 16 * ( t - 1 )² + 24
Answer: Simple, it got Translated according to the rule (x, y) → (x + 2, y + 8) and reflected across the y-axis
Step-by-step explanation:
Hey there!!
How do we solve this problem :
We will use the combinations formula to solve this :
c ( n , r ) where n = 11 and r = 2
c ( n , r ) = n ! / r ! ( n - r ) !
... 11 ! / 2 ! ( 11 - 2 ) !
... 11! / 2! × 9!
... 11! / 2 × 9!
... 11×10×9×8×7×6×5×4×3×2 / 2×9×8×7×6×5×4×3×2
... 11×10 / 2
... 11 × 5
... 55 combinations.
Hence, the required answer = 55 , option ( d )
Hope my answer helps!
