MHB Did I Simplify These Precalculus Problems Correctly?

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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.
 
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