Answer:
Dont tell me you're trolling. Anyway, its 16
The easiest way is to try the point (-4,1), that is, x=-4, y=1,
to see which equation works.
b works.
The usual way to do it is to find the equation of the circle
standard form of a circle is (x-h)²+(y-k)²=r², (h,k) are the coordinates of the center, r is the radius.
in this case, the center is (-2,1), so (x+2)²+(y-1)²=r²
the given point (-4,1) is for you to find r: (-4+2)²+(1-1)²=r², r=2
so the equation is (x+2)²+(y-1)²=2²
expand it: x²+4x+4+y²-2y+1=4
x²+y²+4x-2y+1=0, which is answer b.
The graph of the line y = -6 is a horizontal line which passes through the y axis at -6. On the other hand, the graph of the equation, x = 1/6 is a vertical line that passes through the x-axis at 1/6. The intersection of these lines is at a point that is expressed as (1/6, -6).
9514 1404 393
Answer:
-3 ≤ x ≤ 19/3
Step-by-step explanation:
This inequality can be resolved to a compound inequality:
-7 ≤ (3x -5)/2 ≤ 7
Multiply all parts by 2.
-14 ≤ 3x -5 ≤ 14
Add 5 to all parts.
-9 ≤ 3x ≤ 19
Divide all parts by 3.
-3 ≤ x ≤ 19/3
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<em>Additional comment</em>
If you subtract 7 from both sides of the given inequality, it becomes ...
|(3x -5)/2| -7 ≤ 0
Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.
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For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...
|f(x)| < (some number) . . . . or ≤
If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:
|f(x)| > g(x) ⇒ f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0
Any solutions to these inequalities must respect their domains.