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kiruha [24]
3 years ago
8

(30x^4 - 6x^3- 25x^2- 5x - 6) /(5x - 1)

Mathematics
1 answer:
denis-greek [22]3 years ago
7 0
150x^5-60x^4-119x^3+5x-6/5x-1
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Use the drop-down menus to describe the key aspects of the function f(x) = -x2 – 2x – 1.
sdas [7]

Answer:

1st : Maximum Value

2nd: When x<-1

3rd : x>-1

4th: All real Numbers

5th : All numbers less than or equal to 0

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4 years ago
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How much does a customer save for buying a sound system marked $1,200 at a discount of 15%​
kramer

Answer:

Step-by-step explanation:

Mp=$1200

Let sp be x

SP=MP-discount% of MP

x=1200-15/100 * 1200

x=120000-18000/100

x=102000/100

x=1020

the amount of money person saves=MP-SP

=$1200-$1020

=$180

therefore he saves $180

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3 years ago
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mixas84 [53]
Answer is B:1 because she needs 32 lenses and she already has 24. And in each packages 8 lenses come, so she needs one more package
7 0
3 years ago
A robot can complete 5 tasks in 3/4
Bezzdna [24]

Answer:

[A] 9 minutes to complete one task

[B] 6 tasks completed, with 2/3rds of another task completed.

Step-by-step explanation:

5 tasks in 45 mins is equal to one task in 9 minutes

60 minutes in an hour, 9 minutes per task = 6 tasks completed, with 2/3rds of another task completed.

8 0
4 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
4 years ago
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