Answer: Choice (D): x<=-2 or x>=8
Explanation:
An absolute value |x| is a function that has a "tricky" definition: it equals x is x >=0, but changes abruptly to -x for values of x<0.
This two-case scenario has to be respected and built into a solution of any equation or inequality involving absolute values.
So we start treating the inequality for two cases:
(1) for the case when x-3 >=0, fo which the absolute value is the same what is inside the vertical brackets:
(2) for the other case of x-3<0, in which case we need that extra minus sign:
So putting both cases together, the solution to the inqueality is an x falling to either of the two intervals: x>=8 or x<=-2, which corresponds to choice (D).
The three undefined terms are point, line, and plane. Thus, figure D represents an undefined term as it's a line
The answer is: " 12 " .
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→ " x = 12 " .
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To solve:
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3x - 1 = x + 23 ; Solve for "x" ;
Subtract "x" from each side of the equation; & add "1" to each side of the equation;
3x - 1 - x + 1 = x + 23 - x + 1 ;
to get:
2x = 24 ;
Divide each side of the equation by "2" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
2x/2 = 24/2 ;
to get:
x = 12 .
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Let us check our answer, by plugging in "12" for "x" ; on EACH side of the original equation; to see if the equation holds true; {that is, to see if each side of the original equation is equal, when "x = 12" } ; as follows:
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3x - 1 = x + 23 ;
Substitute "12" for "x" on each side ;
3(12) - 1 =? (12) + 23 ??
36 - 1 =? 35 ??
35 =? 35?? Yes!
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Answer:
Step-by-step explanation:
p(x) has degree 3
(x - 4) has degree 1
So p(x) can be rewritten as a polynomial of degree 2 multiplying (x-4)
So
To find a, b and c, we have to divide p(x) by x - 4.
So
Finding a:
Dividing the first term of p(x) by x - 4.
So a = 3.
Now multiplying 3x² by, x - 4, we have:
Subtracting p(x) from this:
Finding b:
Dividing the first term, after the subtraction, by x - 4.
So b = -8.
Multiplying -8x by x - 4, we have:
Then
Finding c:
So c = 5.
Just to verify if the remainder is 0.
Ok
Then:
OQ is equal to QP so the first option is correct
