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mariarad [96]
3 years ago
13

Hey Bestie! 50 pts Pls help! Real answers only pls <3

Mathematics
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

1. \: 3i

2. option D

3. option C

4. option D

5. option C

6. option B

7. option C

8. option D

9. option C

10. option C

Step-by-step explanation:

<h2>1. \:  \sqrt{ - 9}</h2>

\sqrt{ - 1(9)}

\sqrt{ - 1}  \times  \sqrt{  9}

i \times  \sqrt{9}

i \times  \sqrt{ {3}^{2} }

i \times 3

3i

<h2>2. \:  \sqrt{ - 8}</h2>

\sqrt{ - 1(8)}

\sqrt{ - 1}  \times  \sqrt{8}

i \times  \sqrt{8}

i \times  \sqrt{ {2}^{2} \times 2 }

2i \sqrt{2}

<h2>3. \:  \sqrt{ - 80}</h2>

\sqrt{ - 1}  \times  \sqrt{80}

i \times  \sqrt{80}

4i \:  \sqrt{5}

<h2>4. \:  \sqrt{ - 75}</h2>

\sqrt{ - 1}  \times   \sqrt{75}

i \times  \sqrt{75}

i \times  \sqrt{ {5}^{2} \times 3 }

5i \sqrt{3}

<h2>5. \:  \sqrt{ - 72}</h2>

\sqrt{ - 1}  \times  \sqrt{72}

i \times  \sqrt{72}

i \times ( {6}^{2}  \times 2)

6i \sqrt{2}

<h2>6.  \sqrt{ - 20}</h2>

\sqrt{ - 1}  \times  \sqrt{20}

i \times  \sqrt{20}

i \times  \sqrt{ {2}^{2}  \times 5}

2i \sqrt{5}

<h2>7. \:  \sqrt{ - 27}</h2>

\sqrt{ - 1}  \times  \sqrt{27}

i \times  \sqrt{27}

i \times  \sqrt{ {3}^{2}  \times 3}

3i \sqrt{3}

<h2>8. \:  \sqrt{ - 12}</h2>

\sqrt{ - 1 \times 12}

i \times  \sqrt{12}

i \times  \sqrt{4(3)}

2i \sqrt{3}

<h2>9. \:  \sqrt{ - 125}</h2>

\sqrt{ - 1}  \times  \sqrt{125}

i \times  \sqrt{ {5}^{2} \times 5 }

5i \sqrt{5}

<h2>10. \:  \sqrt{ - 180}</h2>

\sqrt{ - 1}  \times  \sqrt{180}

i \times  \sqrt{ {6}^{2} \times 5 }

6i \sqrt{5}

<h3>Hope it is helpful...</h3>
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An oak tree cast a shadow of 14 feet. A 30 foot flagpole casts a shadow of 60 how tall the Oak tree?
Alina [70]

Answer:

7 feet

Step-by-step explanation:

Oak tree = x

14 : x  as  60 : 30

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14 : x  as 2:1

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14 : x as 14 : 7

Since the first number in each ratio is the same, we know that that x and 7 are equal

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Hope this helps, feel free to let me know if you would like a further explanation :)

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3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o
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Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

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(2) Answer

Solution: There are two questions in this problem.

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on the previous page.  The part of this problem most groups had the most issue with was the second. Everyone should  be comfortable with carefully proving a function is surjective by the time we get to the midterm.

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4 years ago
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Step-by-step explanation:

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Fgbdrgbsdrhbrhrsewhsrthbrdehbeddb.
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Answer:

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Step-by-step explanation:

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