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3 years ago

Which compound inequality could this graph be the solution of?

Mathematics

1 answer:

vodomira [7]3 years ago

4 0

-4 ≤ x ≤ 6

x ≥ -4

3x ≥ -12

3x + 12 ≥ 0

3x + 14 ≥ 2

6 ≥ x

6 - x ≥ 0

1 - x ≥ -5

A. 3x + 14 ≥ 2 and 1 - x ≥ -5

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Read 2 more answers

Solve the proportion

mario62 [17]

Answer:

$\boxed{\sf x = 5}$

Step-by-step explanation:

$\sf Solve \: for \: x \: over \: the \: real \: numbers: \\ \sf \implies \frac{2}{x - 3} = \frac{5}{x} \\ \\ \sf Take \: the \: reciprocal \: of \: both \: sides: \\ \sf \implies \frac{x - 3}{2} = \frac{x}{5} \\ \\ \sf Expand \: out \: terms \: of \: the \: left \: hand \: side: \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} = \frac{x}{5} \\ \\ \sf Subtract \: \frac{x}{5} - \frac{3}{2} \: from \: both \: sides: \\ \sf \implies \frac{x}{2} - \frac{3}{2} - ( \frac{x}{5} - \frac{3}{2} ) = \frac{x}{5} - ( \frac{x}{5} - \frac{3}{2} ) \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} - \frac{x}{5} + \frac{3}{2} = \frac{x}{5} - \frac{x}{5} + \frac{3}{2} \\ \\ \sf \frac{x}{5} - \frac{x}{5} = 0 : \\ \sf \implies \frac{x}{2} - \frac{x}{5} - \frac{3}{2} + \frac{3}{2} = \frac{3}{2} \\ \\ \sf \frac{3}{2} - \frac{3}{2} = 0: \\ \sf \implies \frac{x}{2} - \frac{x}{5} = \frac{3}{2} \\ \\ \sf \frac{x}{2} - \frac{x}{5} = \frac{5x - 2x}{10} = \frac{3x}{10} : \\ \sf \implies \frac{3x}{10} = \frac{3}{2} \\ \\ \sf Multiply \: both \: sides \: by \: \frac{10}{3} : \\ \sf \implies \frac{3x}{10} \times \frac{10}{3} = \frac{3}{2 } \times \frac{10}{3} \\ \\ \sf \frac{3x}{10} \times \frac{10}{3} = \cancel{\frac{3}{10} } \times( x) \times \cancel{ \frac{10}{3} } = x : \\ \sf \implies x = \frac{3}{2} \times \frac{10}{3} \\ \\ \sf \frac{3}{2} \times \frac{10}{3} = \cancel{ \frac{3}{2} } \times \cancel{ \frac{3}{2} } \times 5 : \\ \sf \implies x = 5$

8 0

3 years ago