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Fin| the devintle thp fundionc 9= lo( s4 = Jxlo X 4 0 * khx + Cs ~) 47 838e*...

Question

Fin| the devintle thp fundionc 9= lo( s4 = Jxlo X 4 0 * khx + Cs ~) 47 838e*

Fin| the devintle thp fundionc 9= lo( s4 = Jxlo X 4 0 * khx + Cs ~) 47 838e*



Answers

$$ \begin{array}{l} \text {
} \frac{6}{a^{2}+14 a+45}, \frac{5 a}{a^{2}-81} \\ \text { LCD }(a+9)(a+5)(a-9) \end{array} $$

Question asked us to divide and the first thing I'm gonna do is rewrite this in terms of decreasing exponential order will notice inter numerator that air highest degree. Each of the six is not our first term. So the correct order would be 58 to the six plus 49 eight to the fourth minus 14. A cube minus 15 a squared all over seven a cube. And now we can rewrite this into four separate fractions. So 58 to the six over minus seven. A cube plus 49 8 to the fourth over, minus seven. A cubed minus 14. A cube over minus seven. A cube minus 15. A squared over minus seven a cute. And we just have to remember that when we do division of exponents of variables were exponents we have to use to track their exponents. So for this first term, in terms of exponents we have six minus three which gives us three and we can't simplify 57 civil leave that this five sevens and we have three as our exponents for are variable. We're the second term 49 divided by seven gives us minus seven and for our exponents, We have four minus three, which gives us one suggest minus seven A. Or this third term we have minus 14 divided by minus seven, which gives us a positive too. And for exponents, we have three minus three, which simplifies to zero. So that tells us our A's cancel out. And for this last term, we can't simplify 15/7, but a negative divided by a negative gives us a positive. And for our exponents, we have two minus three, which gives us minus one. And when we have a negative exponents weaken, just rewrite it in the denominator. So this here is our final answer.

And yes and every day of so correct. Absolutely be energy because larger size that can have high yesterday. So we ended these directives in that case plus and cl minus minus have larger radiator because of negative charge. So it will be incorrect in that plus four M n plus seven is incredible because th place for in that lie in the group North period. Not or not our size. Electronic. So in the ISO electronic, please check the charge, but these are not isolate tronic. So titanium have larger size than the mm past seven. It is also being destroyed.

Okay, so if we add our like terms going down, we have nine a to the fourth by this 88 to the fourth. So that will give us a to the fourth that you have 4.3 a que tu minus 2.8 a cubed and that'll give us plus 2.5 a cute. And then we have negative seven A squared plus four ace were So that'll give us negative. Three a squared. And then we have negative 1.2 a minus 3.9 a. So that will give us 5.1 a. And then lastly, we have six plus five just out on our constant. That'll be 11. So our answer here is gonna be h of the fourth plus 2.5 a huge minus three squared minus 5.18 plus 11.

Although we're now dealing with decimals, the process is still the same. Or adding these two polynomial is together because we just have to find, like terms. So to start 4.9 a cubed does not have any term to be combined with because there are no other A cube terms in E equation. So this will be all by itself. Then we have to find our another a squared term, and that will be our 2.1 a squared, so we'll have positive 3.2 a squared plus 2.1 a squared. Then we will have our just regular a term negative 5.1 and positive 3.7 and then, lastly, our Austin term has no other term to be combined with, and that will just be all by itself. Now let's let to do is combine, so we'll have 4.9. A cute 3.2 plus 2.1 is going to be 5.3 because two plus one is three and three plus two is five a squared. Then we have negative 5.1 plus 3.7, and with the help of a calculator, weaken, get negative. 1.4 a. We just have to add that constant term of 4.6 that that is the answer to our I wish.


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