Answer:
Step-by-step explanation:
The inequality will be split into two
It is know that, if a<b<c
Then a<b and b<c
-8<2x-4<4
Apply that to this
Then,
-8<2x-4. Equation 1
Also,
2x-4<4 equation 2
Solving equation 1
-8<2x-4
Add 4 to both side of the equation
-8+4<2x-4+4
-4<2x
Divide both sides by 2
-4/2<2x/2
-2<x
Note, if a is less than b, then, b is greater than a, e.g. 4 is less than 10, this implies 10 is greater than 4
Therefore,
-2<x
Then, x greater than -2
Equation 2
2x-4<4
Add 4 to both side of the inequalities
2x-4+4<4+4
2x<8
Divide both side by 2
Then,
2x/2<8/2
x<4
Therefore x is between -2 and +4.
Check attachment for graphical solution
Step-by-step explanation:
Bc is also 20 answer is 400
R = 8 tan(x) sec(x)
In order to solve this, we are going to use the following:
r = sqrt(x^2 + y^2) //but in this case, we don't need this.
tan(x) = Y / X
X = r cos(x)
Y = r sin(x)
<span>(r cos x)^2 = 8(r sin x)</span>
x^2 = 8y
A good place to start is to set
to y. That would mean we are looking for
to be an integer. Clearly,
, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since
is a radical, it only outputs values from
, which means y is on the closed interval:
.
With that, we don't really have to consider y anymore, since we know the interval that
is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval
, which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.