MHB Did I Simplify These Precalculus Problems Correctly?

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The discussion revolves around simplifying two precalculus problems. The first problem, 2x^-3 + 9x^-7, was simplified correctly to 2x^4 + 9/x^7. However, the second problem, ((5x^4+y^-8)/(10x^5y^3))^-3, was incorrectly solved, with another user providing a step-by-step correction. The correct approach involves flipping the fraction and applying exponent rules accurately. The thread emphasizes the importance of careful manipulation of exponents and fractions in precalculus.
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Just joined this website for math help like right now :)

Background info: I was sick for 2 years before going to university so my level in math is horrible; though I will try my best in all questions which I would ask.1. Simplify 2x^-3 + 9x^-7

Here's what I did

2/x^3 + 9/x^7
find common denominator
2/x^3 multiplied top and bottom by x^4

2x^4/x^7+9/x^7= 2x^4+9/x^7

Did I get the correct answer?

2. Simplify ((5x^4+y^-8)/(10x^5y^3))^-3

I first multiplied everything by ^-3

so (5^-3 x^-12+ y^-24)/ (10^-3 x^-15 y^-9)

(1/5^3 x^-12 + y^-24 ) / (1/10^3 x^-15 y^-9)

1/125 multiplied by 1000/1 (flipped this one)

I get 1000/125 = 8 so 8 on top

rest I can use the property of subtracting variables with exponents when it is multiplication...anyways fast forward and I get.

(8x^3)/(y^15)
 
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tenthfire said:
Just joined this website for math help like right now :)

Background info: I was sick for 2 years before going to university so my level in math is horrible; though I will try my best in all questions which I would ask.1. Simplify 2x^-3 + 9x^-7

Here's what I did

2/x^3 + 9/x^7
find common denominator
2/x^3 multiplied top and bottom by x^4

2x^4/x^7+9/x^7= 2x^4+9/x^7

Did I get the correct answer? ... Yes

2. Simplify ((5x^4+y^-8)/(10x^5y^3))^-3

...

(8x^3)/(y^15)

Good morning,

your 2nd result is wrong.

$$\left( \frac{5x^4+\frac1{y^8}}{10x^5 \cdot y^3} \right)^{-3}$$

1. step:

$$\left( \frac{5x^4+\frac1{y^8}}{10x^5 \cdot y^3} \right)^{-3} = \left( \frac{10x^5 \cdot y^3} {5x^4+\frac1{y^8}}\right)^{3}$$

2. step:

$$\left( \frac{10x^5 \cdot y^3} {5x^4+\frac1{y^8}}\right)^{3}= \left( \frac{10x^5 \cdot y^3} {\frac{5x^4 y^8+1}{y^8}}\right)^{3}$$

3. I'll leave the rest for you.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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