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3 years ago

8

Negation of all owls fly?

2 answers:

Reptile [31]3 years ago

8 0

Looking at this in terms of sets, let's call O the set of all owls, and F the set of all things that can fly. What this original statement is saying every animal that's a member of the set of all owls is also a member of the set of all things that can fly, or in other words, O⊂F (O is a subset of F). Negating this tells us that, while there's <em>at least one</em> element of O that also belongs to F, O is not contained entirely in F (O⊆F, in notation), so a good negation or our original statement might be:

<em>Not all owls can fly.</em>

Ratling [72]3 years ago

6 0

...wow the other person was fantastic.. I’m not even gonna try competing...

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Take a factor out of the square root:

Allisa [31]

Answer:

Question A)

$=\sqrt{6}x$

Question B)

$=3a\sqrt{a}$

Question C)

$=5\sqrt{2}b^2$

Step-by-step explanation:

A)

We are given:

$\sqrt{6x^2}\, \text{ where } x\geq 0$

We can rewrite the expression:

$=\sqrt{6}\cdot \sqrt{x^2}$

The square root and square will cancel each other out. Thus:

$=\sqrt{6}x$

B)

We are given:

$\sqrt{9a^3}$

Rewrite:

$=\sqrt{9}\cdot \sqrt{a^3}$

Note that the square root of 9 is simply 3. We can also factor the second part:

$=3\cdot \sqrt{a^2\cdot a}$

Rewriting:

$=3\cdot\sqrt{a^2}\cdot\sqrt{a}$

Simplify:

$=3a\sqrt{a}$

C)

We are given:

$\sqrt{50b^4}$

Rewrite. Note that 50 = 25(2):

$=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}$

Simplify. We can rewrite the factor as:

$=5\cdot \sqrt{2}\cdot \sqrt{(b^2)^2}$

The square and square root will cancel out. Thus:

$=5\sqrt{2}b^2$

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3 years ago

Lucy purchased 5 equally-priced shirts for $60. Frank bought 4 shirts. Each of his shirts were $2 more than the price of a shirt

4vir4ik [10]

Answer:

$56

Step-by-step explanation:

Lucy's shirts were $12 each since 60/5 is 12. Frank's shirts were 2 more than lucys so his were $14. 14 times 4 is 56.

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3 years ago

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Nimfa-mama [501]

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Answer:

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Step-by-step explanation:

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