Let r = usual driving rate
let t = usual driving time
We need to figure out t
The distance she covers in her usual time at her usual rate is r*t
The distance she covers in her new time at her new rate is:
(1+t)*((2/3)r)
Set this equal to each other and solve for t.
rt = (2/3)r + (2/3)rt
(1/3)rt = (2/3)r
(1/3)t = (2/3)
t = 2
So her usual time is 2 hours. (There's probably a faster way to do this)
Answer:
I suppose it should be
6b² and (6b²)²
or
6b² and 36
Step-by-step explanation:
(a+b)² = a² + 2ab + b²
therefore
(6b²+2a)² = (6b²)² + 12ab² + 4a²
Answer:
a) f(-1/2) = -2 is NOT TRUE.
b) f(0) =3/2 is TRUE.
c) f(1) = -1 is NOT TRUE.
d) f(2) = 1 is NOT TRUE.
e) f(4) = 7/2 is TRUE.
Step-by-step explanation:
Here, the given function is 
Now, checking for each values for the given function:
a) Putting x = (-1/2):

and (5/4) ≠ -2
Hence, f(-1/2) = -2 is NOT TRUE.
b)Putting x = 0 :

Hence, f(0) =3/2 is TRUE.
c) Putting x = 1:

Hence, f(1) = -1 is NOT TRUE.
d)Putting x = 2:

and (5/2) ≠ 1
Hence, f(2) = 1 is NOT TRUE.
e)Putting x = 4:

Hence, f(4) = 7/2 is TRUE.
Answer: 2.1 Pull out like factors :
6s - 2r = -2 • (r - 3s)
Equation at the end of step
((0-(9•(6s-4r)))+3s)--14•(r-3s)
Pulling out like terms
4.1 Pull out like factors : 6s - 4r = -2 • (2r - 3s)
Equation at the end of step
((0--18•(2r-3s))+3s)--14•(r-3s)
Final result :
<u>50r - 93s</u>