Answer 1: A and C
Step-by-step explanation:
Use distribution law to see which answer match the equation at the top.
Answer 2: 16q + 4
10q and 6q are like terms
10q + 6q = 16q
+
5 - 1
Answer:
No
Step-by-step explanation:
The equation of a circle with center (a,b) and radius r is given as:
If a given point (x,y) does not lie on this circle, it will not satisfy its equation.
This means the distance from the point to the center is not equal to the radius.
It is either less or greater than the radius.
Hence you cannot write the equation of the circle.
An arithmetic sequence has a common difference.
143 - 130 = 13
156 - 143 = 13
169 - 156 = 13
The common difference is 13.
a1 = 130
a2 = 130 + 13
a3 = 130 + 2 * 13
a4 = 130 + 3 * 13
...
an = 130 + (n - 1) * 13
an = 130 + 13(n - 1)
an = 130 + 13n - 13
an = 117 + 13n
an = 13n + 117
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.
The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...
Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.