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3 years ago

6

Which is the best description of an extraneous solution?

1 answer:

Degger [83]3 years ago

8 0

Answer:

A solution that results in a false statement when substituted back into the original equation.

Step-by-step explanation:

An extraneous solution is one that arises in a solution of equations, but which on closer inspection is not a solution to the original equation. Therefore, they imply an error in the development of the solution of the equation, so they cannot be taken as a valid solution since, when replacing the result with the missing variable, the equation does not obtain the desired result.

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Enter the numerator and the denominator of the fraction in simplest form that is equal to 0.7 repeating

valentinak56 [21]

Answer:

$\frac{7}{9\\}$

Step-by-step explanation:

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1 year ago

Find the equivalent ratio for 32:__=12:24

Igoryamba

32 : 64 = 12 : 24

or

$\displaystyle\\
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3 years ago

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a bottle of ketchup contains 40 oz of ketchup. How many 1/2 oz servings can you get out of one bottle?

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Answer:

you can get 80 1/2 oz servings.

Step-by-step explanation:

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2 years ago

Kristin drew a triangle with 2 congruent sides and 1 obtuse angle. Which term best describes the triangle mark all that appl. A

yarga [219]

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3 years ago

Check all that apply. If csc =13/5 , then:

nirvana33 [79]

Answer:

The answer are the options B and C

$sin(x)=\frac{5}{13}$

$tan(x)=\frac{5}{12}$

Step-by-step explanation:

we know that

$csc(x)=\frac{1}{sin(x)}$

In this problem we have

$csc(x)=\frac{13}{5}$

substitute

$\frac{13}{5}=\frac{1}{sin(x)}$

solve for sin(x)

$sin(x)=\frac{5}{13}$

$cos^{2}(x)+sin^{2}(x)=1$

substitute the value of sin(x) and solve for cos(x)

$cos^{2}(x)+(\frac{5}{13})^{2}=1$

$cos^{2}(x)=1-(\frac{5}{13})^{2}$

$cos^{2}(x)=(\frac{13^{2}-5^{2}}{13^{2}})$

$cos^{2}(x)=(\frac{144}{13^{2}})$

$cos^{2}(x)=(\frac{12^{2}}{13^{2}})$

$cos(x)=(\frac{12}{13})$

The function tangent is equal to

$tan(x)=\frac{sin(x)}{cos(x)}$

substitute the values

$tan(x)=\frac{(5/13)}{(12/13)}=5/12$

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3 years ago

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