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Brums [2.3K]
3 years ago
8

1. Suppose a and b are real numbers where a > 0 and b < 0.

Mathematics
1 answer:
Nat2105 [25]3 years ago
3 0

a) –a + b is negative

b) a – b is positive

c) b-a is negative

Step-by-step explanation:

1. Suppose a and b are real numbers where a > 0 and b < 0.

a. Is –a + b positive or negative. Explain how you know.

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

–a + b

= -(6)+(-5) = -6-5 = -11

So, –a + b is negative

b. Is a – b positive or negative? Explain how you know.

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

a - b

=6-(-5) = 6+5 = 11

So, a – b is positive

c. Is b – a positive or negative. Explain how you know

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

b-a

=(-5)-(6)

= -5-6

= -11

So, b-a is negative

Keywords: Solving Integers:

Learn more about solving integers at:

  • brainly.com/question/5496711
  • brainly.com/question/11549940
  • brainly.com/question/11334714

#learnwithBrainly

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Answer:

Step-by-step explanation:

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The discriminant is b^2 - 4ac, or 49 - 4(1)(3), or 49-12, or 37.

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