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(10 Peala)TE TAnkar Iaakl x enly n ned [Ee pruntlt} & PUX > I70):and #adLd dUL() PiX < 1021P(1ed < X < [70)P017i< I <1)[atn fullig atnd 1s, Ex...

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(10 Peala)TE TAnkar Iaakl x enly n ned [Ee pruntlt} & PUX > I70):and #adLd dUL() PiX < 1021P(1ed < X < [70)P017i< I <1)[atn fullig atnd 1s, Exl Frocebillty & geitine Hami 0

(10 Peala)TE TAnkar Iaakl x enly n ned [Ee pruntlt} & PUX > I70): and #adLd dUL () PiX < 1021 P(1ed < X < [70) P017i< I <1) [atn fullig atnd 1s, Exl Frocebillty & geitine Hami 0



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Solve the inequality by using the graph of the function. [Hint: The graphs were drawn in Problems $7-10 \text { of Section } 5.3 .]$ Solve $R(x)<0,$ where $R(x)=\frac{x}{(x-1)(x+2)}$

As we all know that the thermal stability depends upon caramel. The stability depends upon the iconic character. The iconic get it did of the oxide of the oxide. So here the order will be weird. I know after that Cdo after that HD you because from Falling Skills Law, the electronic negativity of dead and 8 1.65 CD 1.69 and HDH two. Therefore, therefore later the electron negative of the matter more will be the eye. Any character. So here, in this problem of senses, correct?

To solve this compound in equality. I'll solve each inequality separately and then find the union of the two solution sets. So for the 1st 1 will start by adding three to both sides. So five Z is greater than 10 and then dividing both sides by five. So Z is greater than two, and for the 2nd 1 I'll start by adding six to both sides. So for Z is less than negative for and then dividing both sides by four Z is less than negative one, and we need the word or in between. So we have A Z is greater than two or Z is less than negative one. And for the graph, we'll make a number line and make some marks on it, and we need to show negative one. And we need to show, too, and maybe some extra numbers if we want to. So Z is greater than two is an open circle onto, and an arrow to the right Z is less than negative. One is an open circle on negative one and an arrow to the left

The function given to us is our affix is equal. Toe to X plus four, divided by X minus one now Benito graphically determine very evenly function. Are effects be greater than or equal to zero? That is very little I above the X axis. So let's plot the graph and find out the solution. So from the graph, what we can see is that the function lies above the X axis for X. Such that X is less than or equal to minus two or X is greater than zero, which means that it will be in the intervals off minus infinity to minus two Union zero to positive infinity.

So let's go ahead and distribute the numbers outside. The prince sees into the princes so five times ours five are and then 5 10 streets of plus 15 the minus four times ours minus for our in the minus four tons of minus two is a plus eight. Seven times er seven are and seven times a minus four is a minus 28 and then we have the minus. Seven are in the end there. Let's combine like terms on the left side. Five R minus four are is one are and 15 plus an eight is a plus 23 seven ar minus seven r zero. Are they cancel? And so he said, the negative 28. So let's subtract 23 off both sides to get rid of the 23 get our by itself. So our is greater than this is gonna be 31 41 51 negative 51 r is greater than


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