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JulsSmile [24]
3 years ago
6

How do extraneous solutions arise from radical equations?

Mathematics
2 answers:
Blababa [14]3 years ago
5 0
In general, extraneous solutions arise<span> when we perform non-invertible operations on both sides of an </span>equation<span>. (That is, they sometimes </span>arise, but not always.) ... Solvingequations<span> involving square roots involves squaring both sides of an</span>equation<span>. I hope that this helps you out, Have a wonderful day!!</span>
eduard3 years ago
3 0
In general, extraneous solutions arise<span> when we perform non-invertible operations on both sides of an </span>equation
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Magic Video Games, Inc., sells an expensive video games
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Answer:

Magic Video Games, Inc., sells an expensive video games

package. Because the package is so expensive, the company wants to advertise an impressive guarantee

for the life expectancy of its computer control system.

The guarantee policy will refund the full purchase price if the computer fails during the guarantee period.

The research department has done tests that show that the mean life for the computer is 35 months, with standard deviation of 5 months. The computer

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Step-by-step explanation:

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3 years ago
Jenna buys 12 lb of meat. Some of the meat is chicken and costs $1.90/lb. The rest is beef, which costs $4.30/lb. She spends a t
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If the number of pounds of chicken is represented by c, then the number of the pounds of beef is 12 - c. The equation that best represent the given conditions above based on the total amount that Jenny spent is,
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3 years ago
Don's reading group has read 168 books this year his reading group has six members each member read the same amount of books how
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28 Books

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hope this helped

7 0
3 years ago
A drawing of a room has a scale of 1 4 inch = 1 ft. Which choice does NOT represent a proportional relationship, according to th
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5 0
4 years ago
Read 2 more answers
Prove that when a polynomial function p(x) is divided by a first degree polynomial ax + b the remainder is p(-b/a)
DochEvi [55]

Answer:

r = p(-b/a)

Step-by-step explanation:

Let p(x) = q(x)(ax + b) + r(x) where q(x) is the quotient when p(x) is divided by ax + b and r(x) is the remainder.

Since ax + b is a first degree polynomial, r(x) is one power less than ax + b is is just a constant, r.

So, p(x) = q(x)(ax + b) + r

Now, p(x) = r when q(x)(ax + b) = 0

since q(x) ≠ 0, ax + b = 0 ⇒ ax = -b ⇒ x = -b/a

⇒ p(x) = r when x = -b/a

So, r = p(-b/a)

So, the remainder when a polynomial function p(x) is divided by a first degree polynomial ax + b is p(-b/a)

8 0
3 years ago
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