Answer:
14
Step-by-step explanation:
The least common multiple is a number that both numbers can multiply something to get to, and the smallest possible common one.
Let's try listing out the multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
And the multiples of 7:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70
Notice that both have the number 14 in common, and it is the first number that they do so. There are lots of other multiples that they have in common too, but 14 is the least.
the answer in the attached figure
Answer
Find out the number of hours when the cost of parking at both garages will be the same.
To prove
As given
There are two parking garages in beacon falls .
As given
Let us assume that the y is representing the cost of parking at both garages will be the same.
The total number of hours is represented by the x.
First case
Garage a charges $7.00 to park for the first 2 hours ,and each additional hour costs $3.00 .
As garage charges $7.00 for the first 2 hours so the remaning hours are (x -2)
Than the equation becomes
y = 3.00 (x -2) + 7.00
written in the simple form
y = 3x - 6 +7
y = 3x + 1
Second case
Garage b charges $3.25 per hour to park.
than the equation becomes
y = 3.25x
Compare both the equations
3x +1 = 3.25x
3.25x -3x = 1
.25x = 1

x = 4hours
Therefore in the 4 hours the cost of parking at both garages will be the same.
If you check the picture below
the rectangular solid is a 6x6x6, so all sides are equal thus is a cube
is painted on all faces, so the only part that's painted is the faces, ok
then you split it like in the picture, you have all 6 faces painted there,
now, let's say each of those 6 faces are cut in 2cm
how many of those 2cm pieces will each have? so get that amount for each and sum them up, that's how many of the cubes have one face painted
only the cubes that are part of the face, which was painted, are the ones that have a painted face themselves
keep in mind that each face is a 6x6 rectangle
notice the 2nd picture below
Answer:
sec²(x) - sec(x) + tan²(x) = (sec(x) - 1)(2sec(x) + 1)
Step-by-step explanation:
sec²(x) - sec(x) + tan²(x) =
= sec²(x) - sec(x) + [sec²(x) - 1]
= sec²(x) - sec(x) + [(sec(x) + 1)(sec(x) - 1)]
= sec(x)[sec(x) - 1] + [(sec(x) + 1)(sec(x) - 1)]
= (sec(x) - 1)(sec(x) + sec(x) + 1)
= (sec(x) - 1)(2sec(x) + 1)