Answer: The required equation of the circle is
Step-by-step explanation: We are given to find the equation of the circle with center at the point (7, -3) and radius of length 7 units.
We know that
the standard equation of a circle with center at the point (h, k) and radius of length r units is given by
For the given circle,
center, (h, k) = (7, -3) and radius, r = 7 units.
From equation (i), we get
Thus, the required equation of the circle is
|x| < -1
|x| = -1
|x| > -1
The solution sets is all real numbers in case of:
|x| > -1
We know that modulus is a function with the property such that:
if a<0 then |a|= -a
that is the modulus of a negative number is positive and if a≥0
then |a| =a
and modulus of a positive value is also positive.
i.e. modulus function always gives positive value.
Hence,
1)
|x|<-1
This is not possible as modulus function always gives a value ≥0 for all real numbers.
2)
|x|= -1
This is also not possible as modulus of any number can't be negative.
3)
|x| > -1
The modulus of any number will definitely be greater than or equal to zero.
Hence, the solution set contain all the real numbers.
a. 32
b. 8
c. -4
d. -8
Answer:
b. 8
Step-by-step explanation:
To solve, we need to plug in the x and y values since we already have the values given to us
1/2(4)(2^2)
We should do the exponents first, 2^2 is 4
Now we are left with 1/2(4)(4)
4 times 4 is 16
16 times 1/2 is the same as 16/2
16/2=8
b. 8