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3 years ago

-6g + 3g + 12 > -18

Mathematics

1 answer:

suter [353]3 years ago

3 0

First of all, you can sum the two terms involving g on the left hand side:

$-6g + 3g + 12 > -18 \iff -3g + 12 > -18$

Subtract 12 from both sides:

$-3g > -30$

Divide both sides by -3: note that since we're dividing by a negative number, we must invert the inequality sign:

$g< 10$

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Xelga [282]

Answer:

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Step-by-step explanation:

To determine the nature of the solutions use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then 2 real solutions

• If b² - 4ac = 0 the 2 real and equal solutions

• If b² - 4ac < 0 then no real solutions

Given

x² + 9x + 20 = 0 ← in standard form

with a = 1, b = 9, c = 20, then

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= 9² - (4 × 1 × 20 ) = 81 - 80 = 1

Since b² - 4ac > 0 then there are 2 real solutions

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