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3 years ago

Six less than the product of five and a number is ten.

Mathematics

1 answer:

dlinn [17]3 years ago

6 0

The answer is 5 because it would like that’s

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Solve for k for the equation h+p=3(k-8)

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Answer:

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Step-by-step explanation:

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How do I solve this ?

timofeeve [1]

Answer:

$\large\boxed{\{4,\ 7,\ 10\}}$

Step-by-step explanation:

$X\ \cup\ Y\\\text{The union of a collection of sets is the set of all elements in the collection.}\\\\X\ \cap\ Y\\\text{The intersection}\ X\cap Y\ \text{of two sets X and Y is the set that contains}\\\text{all elements of X that also belong to Y, but no other elements.}\\\\A'\\\text{When A is a subset of a given set U, the absolute complement A}\\\text{is the set of elements in U, but not in A.}$

$U=\{3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10\}\\A=\{4,\ 6,\ 8\}\\B=\{3,\ 4,\ 7,\ 10\}\\C=\{3,\ 5,\ 9\}\\\\A\ \cup\ B=\{4,\ 6,\ 8\}\ \cup\ \{3,\ 4,\ 7,\ 10\}=\{3,\ 4,\ 6,\ 7,\ 8,\ 10\}\\C'=\{3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10\}-\{3,\ 5,\ 9\}=\{4,\ 6,\ 7,\ 8,\ 10\}\\B\ \cap\ C'=\{3,\ \b4,\ \b7,\ \b1\b0\}\ \cap\ \{\b4,\ 6,\ \b7,\ 8,\ \b1\b0\}=\{\b4,\ \b7,\ \b1\b0\}\\\\(A\ \cup\ B)\ \cap\ (B\ \cap\ C')=\{3,\ \b4,\ 6,\ \b7,\ 8,\ \b1\b0\}\ \cap\ \{\b4,\ \b7,\ \b1\b0\}=\{\b4,\ \b7,\ \b1\b0\}$

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4 years ago