hello :<span>
<span>an equation of the circle Center at the
A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a =0 and b = 0 (Center at the origin)
r = OP....p(-8,3)
r² = (OP)²
r² = (-8-0)² +(3-0)² = 64+9=73
an equation of the circle that satisfies the stated conditions.
Center at the origin, passing through P(-8, 3) is : x² +y² = 73</span></span>
Answer:
- none
- none
- x ≥ 4
Step-by-step explanation:
The restrictions placed on the independent variable in a function are those necessary to ensure that the function is defined for all allowed values of that variable.
In the graphs of problems 1) and 2), we see that the functions are defined for all values of x, so there are no restrictions.
__
3. For the function ...

the value under the radical cannot be negative. The square root function is not defined for negative values, so the restriction is ...
x -4 ≥ 0
x ≥ 4 . . . . . . . add 4 to both sides of the inequality
1 hundred can also be 10 tens so when you add 10 tens and 4 tens you 14 tens and that is the same thing as 14 tens
Step-by-step explanation:
is it correct ?? think that d answer is helpful to u
Answer:
Step-by-step explanation:
x²+2 ×x×5+5²=-20x+2×x×5-3
(x+5)²=-20+10x-3
(x+5)²=-10x-3