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2 years ago

He statement below describes a situation in which opposite quantities combine to make zero:

Mathematics

1 answer:

kozerog [31]2 years ago

3 0

False I’m pretty sure

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What is the common difference of the following sequence? 4.3,5,7,9,11___<br>5.16,19,22,25__

myrzilka [38]

Answer:

Step-by-step explanation:

4). Given sequence → 3, 5, 7, 9, 11........

Common difference of an arithmetic sequence is defined by,

Common difference = Successive term - previous term

= 2nd term - 1st term

= 5 - 3

= 2

5). Given sequence → 16, 19, 22, 25........

Common difference of an arithmetic sequence = Second term - first term

= 19 - 16

= 3

8 0

3 years ago

What can be used to show the difference in variation for two populations that have equivalent measures of center?

irakobra [83]

1. C

2. D

3. A

4. A

5. C

6. A

7. Do yourself

4 0

3 years ago

Use the function below to find f(4). F(x) = 3x

Sonja [21]

Since F(X) = 3^X,

F(4) = 3^4 = 81

5 0

3 years ago

Read 2 more answers

Billy is hiking in Colorado. He walks eastward four miles, then turns $60$ degrees northward and walks six miles. How far is he

nikklg [1K]

He is 2√19 miles far way from starting point.

Suppose Billy starts at point A, turns at point B, and ends at point D, as shown below.

If Billy turns 60◦ northward and walks six miles, then we can draw a 30 − 60 − 90 triangle whose hypotenuse is 6 miles

It follows that Billy traveled 6/2 = 3 miles eastward during these 6 miles, and that he traveled 3√3 miles northward during these 6 miles. In total, Billy traveled 4 + 3 = 7 miles eastward and 3√ 3 miles northward.

By the Pythagorean Theorem, the distance from his starting point is q:

(7)2 + (3√3)2

=√49 + 27

= √76

= 2√19 .

Learn more about Pythagorean Theorem on:

brainly.com/question/20545047

#SPJ4

5 0

2 years ago

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT(x, y,

Gnesinka [82]

Answer:

a) ∀x∃y ¬∀zT(x, y, z)

∀x∃y ∃z ¬T(x, y, z)

b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]

∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

∀x∀y [¬P(x, y) ^ ¬Q(x, y)]

c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))

∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))

d) ∀x¬∃y (P(x, y) → Q(x, y))

∀x∀y ¬(P(x, y) → Q(x, y))

∀x∀y (¬P(x, y) ^ Q(x, y))

3 0

3 years ago

Read 2 more answers