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bekas [8.4K]
3 years ago
5

Lauren solved the equation $|x-5| = 2$. meanwhile jane solved an equation of the form $x^2+ bx + c = 0$ that had the same two so

lutions for $x$ as lauren's equation. what is the ordered pair $(b, c)$?

Mathematics
1 answer:
noname [10]3 years ago
4 0
You must first solve the equation with absolute value, which has two possible solutions that in this case are x = 7 and x = 3. The equation is solved by applying the definition of absolute value. Then, you must solve the second-degree polynomial by applying the resolver and matching both solutions to the absolute value solutions.The result will be (b, c) = (- 10,21). I attach the correct solution.

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The answer would be (4,2) I’m pretty sure I may be wrong tho
8 0
3 years ago
Graph y < x2 + 4x. Click on the graph until the correct graph appears.
jonny [76]

Answer:  The correct answer is:  

_________________________  

The given "graph" in the bottom right, lowest corner

Step-by-step explanation:

_________________________

Note:  When there is only one (1) equation give for a graph;

          and/or:  only one (1)  "inequality given";  

         we  look for the symbol.

If the symbol is "not" an "equals" symbol (i.e. not an:  =  symbol) ;

 we check for the type of "inequality" symbol.

If there is a:  "less than" (<) ;  or a "greater than" (>) symbol;  the graph of the "inequality" will have "dashed lines" (since there will be a "boundary").

If there is an "inequality" that is a: "less than or equal to" (≤) ;

                                                 or a: "greater than or equal to" (≥) ;

       →   then there will be not be a dashed line when graphed;

          but rather—a "solid line" ;  since "less than or equal to" ;

          or "greater than or equal to" —is similar to:

          "up to AND including"; or: "lesser/fewer than AND including".).

 _________________________  

Note:  We are given the "inequality" :

            →   " y <  x²  + 4x  "  .

_________________________________

Note that we have a "less than" symbol (< ) ; so the graph will have a:

   "solid line" [and not a "dotted line".].

_________________________________

Note that all of the graphs among our 4 answer choices have "dotted lines".

Not that all values (all x and y coordinated) within the "shaded portion" of the corresponding graph are considered part of the graph.  

As such, given any point within the shaded part,  the x and y coordinates must match the inequality (i.e. the given inequality must be true when one puts in the "x-coordinate" and "y-coordinate" into the "given inequality" :

→   " y <  x²  + 4x "  .

_________________________

Likewise, we can take any point within the "white, unshaded" portion of any of the graph, and take the "x-coordinate" and "y-coordinate" of that point, and the inequality: →   " y <  x²  + 4x  "  ;  will not hold true when the "x-coordinate" and "y-coordinate" values of that point— are substituted into the "inequality".  

_________________________

{Note:  Answer is continued on images attached.}.

Wishing you the best!

6 0
3 years ago
PLEASE HELP DUE TODAY
Masteriza [31]

Answer:

Five plus the difference of seventeen and eight

6 0
3 years ago
Read 2 more answers
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



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