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Q4 ence $ Qesl nlv V8 f 5 € U 4h e { Yhe nIne x neley covcras o< Dive{oe > , uence Seq 2"+6 A n 7 3'...

Question

Q4 ence $ Qesl nlv V8 f 5 € U 4h e { Yhe nIne x neley covcras o< Dive{oe > , uence Seq 2"+6 A n 7 3'

Q4 ence $ Qesl nlv V8 f 5 € U 4h e { Yhe nIne x neley covcras o< Dive{oe > , uence Seq 2"+6 A n 7 3'



Answers

A 7.nSO $_{4}$ solurion is boilcd wirh an NallCO $_{3}$ solution ro producc (a) $7 . n] 1 C O_{3} .7 . n 0$ (b) $7 n(011)_{2}$ (c) $7 \mathrm{nCO}_{3}$ (d) $\angle \mathrm{L} C O_{3} \cdot \mathrm{Na}_{2} \mathrm{SO}_{4}$

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.

Question. Six states that f of X equals X to the third, minus one all to the third for part A. They would like you to find f prime of zero. So first taking that first derivative F prime of X is equal to three times X to the third minus one squared Ah, and then using the chain role multiplied by three x squared. Simplified. That's nine x squared times X to the third minus one squared Ah, and then taking f prime of zero. That would just be zero for Part B. They would like f double prime of zero. So taking the derivative again after Prime of X. Ah, we need to use the product rule here. So that comes out to nine X squared times two times x to the third minus one multiplied by three X squared plus next to the third minus one squared times 18 x. From there, we can start multiplying out to just simplify this so 54 x to the fourth times x to the third, minus one plus 18 x times x to the six minus two x to the third plus one again multiply everything out. So 54 x to the seventh, minus 54 x to the fourth plus 18 x to the seventh, minus 36 x to the fourth plus 18 x Then you have 72 x to the seventh, minus 90 x to the fourth plus 18 x and using that after all, prime of zero is just zero part C. They like, um, the third derivative. So after the third of X is equal to just take a derivative of this expression, we simplified. That's 504 x to the sixth, minus 3. 60 x to the third plus 18. Now third derivative of X or of zero, is now equal to 18. Taking the fourth derivative of X for part D, that would be five. I'm sorry. 3024 x to the fifth minus 1080 x to the fourth. Plugging in zero to our fourth derivative is just zero, then for part E when, after the end of zero. So the end derivative for n greater than nine, they would like you to find what that is. Essentially, what that would be is you're taking six more derivative. So after the fifth of X would have the highest power of X to the fourth F to the fourth back to the six of X be extra the third. Then, after the seventh of X would be X squared after the eighth of X would be X after the ninth of X would be when you get to a constant and then for after the 10 of X would just be zero, and anything after that would be zero. Therefore, after the end of zero for and greater than nine not equal to just greater, um is going to be just zero. And those are all your answers for Question six.

Okay to review. Let's first start cancelling out what we can here in the denominator. I have a five to the third and that cancels with the five to the third in the numerator in the numerator have a three squared and I have a three cubed. So that will cancel and leave three in the denominator in the numerator, I have four to the fifth, which will cancel with the Ford of the six and leave a four in the denominator, so we get an answer of one over 12.

So we have two types off things in our snack mix. And the two things are we have sunflower seeds and there are £3 off them. And the cost per house is because for pound is s. Which means that the value which is equal to a C would people took three. Yes, and then we have ah reasons And there are £4 off the reasons and the cost per pop pop is our So the total value is for our now in the mixture we have three plus four is equal to £7 the value history s plus for our and then we do not know the sea. But we do know the equation is a C is equal to V so we have seven c is equal to three s plus four are in order to get value for C. All we need to do is divide both sides by seven and that gives us the cost per pound off The mixture is three s plus four are over seven and that is option for


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