Answer:
a) Rumble: $1
Pedal: $0.6
b) more than 5 songs
c) No, should be integers only
d) No, has to be integers greater than or equal to 2
Step-by-step explanation:
Part a
Rate per song is the slope.
Rumble:
Slope is 0.6
Rate is $0.6 per song
Pedal:
Slope: (4-2)/(2-0) = 1
Rate is $1 per song
Part b
Pedal would be a better deal if it's cost is lower than Rumble's
Rumble's cost: C = 2 + S
Pedal's cost: C = 4 + 0.6S
4 + 0.6S < 2 + S
0.4S > 2
S > 5
c) No, because no. of songs has to be positive integers only.
For example graph shows S = 2.5, but we know that's not possible
d) No, since the cost is 2 + S,
cost can't be in decimals either.
Cost too should be only integers greater than equal to 2
Answer:
or 
Step-by-step explanation:
We use casework on when
and when
.
For the first case,
, we add 9 to both sides to get
.
Dividing both sides by 3 gives

For the second case,
, we add 9 to both sides to get
.
Dividing both sides by 3 gives
.
Checking both cases, we plug in
and
.
For the first case, we have
, which satisfies the equation.
For the second case, we have
, which also satisfies the equation.
This gives us two solutions to the equation;
and
.
its 14n+21. 7×2n is 14n, and 3×7 is 21
Answer:
45 or multiples of 45 around the centre of octagon
Step-by-step explanation:
Given that ABCDEFGH is a regular octagon. i.e. it has 8 sides.
Since regular ocagon, all interior angles would be equal.
Sum of all interior angles of octagon = 2(8)-4 right angles
= 12 right angles
Hence each angle = 12(90)/8 = 135 degrees
Thus the octagon when rotated will take the same shape if vertices interchange also due to the property that all sides and angles are equal
Since each angle is 135 imagine an octagon with one vertex at origin O, and adjacent vertex B on x axis. OB has to be coincident with BC the next side or the previous side to get it mapped onto itself
The centre will be at the middle with each side subtending an angle of 45 degrees.
Hence if rotation is done around the centre with 45 degrees we will get octagon mapped onto itself.
45, 90, 135 thus multiples of 45