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slega [8]
2 years ago
11

Negative three times a number increased by seven is less than negative eleven

Mathematics
1 answer:
Damm [24]2 years ago
6 0
Express the question as an inequality and solve

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Which graph is that one of the inequality shown below?
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Answer:

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Step-by-step explanation:

3 0
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What makes a square a square <br> why isn't a rectangle?
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A square is a geometrical figure that has 4 congruent sides and angles. This means all side lengths and angles are the same.

A rectangle is a geometrical figure that has 4 similar sides but are not the same length. However, a rectangle does have congruent angles. A rectangle has 2 sides that are double the other 2 sides, in which both of each (length & width) have the same measurement.

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3 0
3 years ago
two chefs are trying to break world record for the longest spaghetti. chef a measures his spaghetti strand to be 503 2/3 ft chef
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Lenght spaghetti chef a:
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Because 503,(6)>503,66
=> lenght spaghetti chef a > lenght spaghetti chef b
So, the winner must be chef a, but, if 503.66 rounds to 503.7, also 503(6) rounds to 503.7, and in this case
lenght spaghetti chef a= lenght spaghetti chef b
So, <span>chef a should protest the judges' decision, because his </span>lenght spaghetti is ≥ lenght spaghetti chef b.





4 0
3 years ago
Read 2 more answers
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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