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Karo-lina-s [1.5K]
3 years ago
10

Help I WILL MARK IF RIGHT

Mathematics
2 answers:
Roman55 [17]3 years ago
7 0
0.95 is the answer brainliest please
melisa1 [442]3 years ago
6 0

Answer:

0.95

Step-by-step explanation:

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Find the equation of the line which passes through the point (-4,12) and is perpendicular to the given line. Express your answer
Vika [28.1K]

Answer:

y = 1/7x + 88/7

Step-by-step explanation:

y = mx +b

since it is perpendicular you have to get the negative reciprocal of the slope which is 1/7

y = 1/7x + b then you would plug in your points

12 = 1/7(-4) + b and then solve for b

12 = -4/7 + b

+ 4/7

88/7 = b

y = 1/7x + 88/7

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3 years ago
Find the missing number of each unit rate.
UNO [17]
15/5 = 3/1 24/6 = 4/1
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A total of 1482 earthquakes occurred in a 7-day period at the end of April
topjm [15]

Answer:

211.7 Per Day 8.8 Per Hour

Step-by-step explanation:

First you do 1482/7 which gives you an average for each day (211.7)

Then you take 211.7  and divide that by 24 because there are 24 hours in a day and doing so would give you 8.8

6 0
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Write these numbers in order starting with the smallest 2.303,2.3,2.33,2.03
zavuch27 [327]
2.03, 2.3, 2.303, 2.33

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3 years ago
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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
3 years ago
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