Answer:
A. 4(x - 4) + 2(3x² + 3x - 20)
C. (11x² + 7x - 55)-(5x² - 3x + 1)
F. (3x² + 5x - 28) + (3x² + 5x - 28)
Step-by-step explanation:
Given:
(3x-7)(2x+8)
= 6x² + 24x - 14x - 56
=6x² + 10x - 56
A. 4(x - 4) + 2(3x² + 3x - 20)
= 4x - 16 + 6x² + 6x - 40
= 6x² + 10x - 56
B. (3x² + 5x - 28) - (2x² + 4x + 28)
= 3x² + 5x - 28 - 2x² - 4x - 28
= x² + x - 56
C. (11x² + 7x - 55)-(5x² - 3x + 1)
= 11x² + 7x - 55 - 5x² + 3x - 1
= 6x² + 10x - 56
D. 4(x - 4) - 2(3x² + 3x - 20)
= 4x - 16 - 6x² - 6x + 40
= - 6x² - 2x + 24
E. (11x² + 7x - 55)-(5x² - 3x + 2)
= 11x² + 7x - 55 - 5x² + 3x - 2
= 11x² - 5x² + 7x + 3x - 55 - 2
= 6x² + 10x - 57
F. (3x² + 5x - 28) + (3x² + 5x - 28)
= 3x² + 5x - 28 + 3x² + 5x - 28
= 6x² + 10x - 56
Answer:
35,829,630 melodies
Step-by-step explanation:
There are 12 half-steps in an octave and therefore arrangements of 7 notes if there were no stipulations.
Using complimentary counting, subtract the inadmissible arrangements from to get the number of admissible arrangements.
can be any note, giving us 12 options. Whatever note we choose, must match it, yielding . For the remaining two white key notes, and , we have 11 options for each (they can be anything but the note we chose for the black keys).
There are three possible arrangements of white key groups and black key groups that are inadmissible:
White key notes can be different, so a distinct arrangement of them will be considered a distinct melody. With 11 notes to choose from per white key, the number of ways to inadmissibly arrange the white keys is .
Therefore, the number of admissible arrangements is:
We are given
The largest telescope in the world is powerful enough to identify a penny that is 5 miles away
it means that
distance =5 miles
we can change it into yards
we know that
1 mile=1760 yards
we can replace 1 mile
So, distance in yards is 8800 yards..........Answer
Answer:
The answer is zero
Step-by-step explanation:
2,6,15,31,......
6-2=4=2X2
15-6=9=3X3
31-15=16=4X4
So the next number should be 31+5X5=56
hope this helps :)