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Vera_Pavlovna [14]
4 years ago
10

1+1 please help!!!!!!

Mathematics
2 answers:
Anna71 [15]4 years ago
6 0

Answer:

2

Step-by-step explanation:

I gotcha.

so.. you have to get your hands.

these are your fingers down below

| | | | /  \ | | | |

so you count 1...2..

then that’s your answer!!

Sedbober [7]4 years ago
3 0

1+1=2

If I have one apple and someone gives me another one I now have two apples.

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Jill lives miles away from her school, and Ben lives miles away from his school. Which statement is true?
dedylja [7]

Answer:

Ben and Jill travel exactly the same distance to the school

8 0
3 years ago
Read 2 more answers
Find set<br> A={1, 2, 6, 10}<br> B={3, 6, 9, 10, 11}<br> C = {1, 2, 4, 7, 11}
PilotLPTM [1.2K]

If <em>U</em> = {1, 2, 3, …, 12} is the universal set, and

<em>A</em> = {1, 2, 6, 10}

<em>B</em> = {3, 6, 9, 10, 11}

<em>C</em> = {1, 2, 4, 7, 11}

then

(1) <em>A</em> U <em>B</em> is the set containing all elements from <em>A</em> and <em>B</em>,

<em>A</em> U <em>B</em> = {1, 2, 3, 6, 9, 10, 11}

(2) <em>A</em> ∩ <em>B</em> is the set of elements that are contained in both <em>A</em> and <em>B</em>,

<em>A</em> ∩ <em>B</em> = {6, 10}

(3) Unfortunately, <em>A</em> ∩ <em>B</em> U <em>C</em> is somewhat ambiguous. It could mean (<em>A</em> ∩ <em>B</em>) U <em>C</em> or <em>A</em> ∩ (<em>B</em> U <em>C </em>). Then either

(<em>A</em> ∩ <em>B</em>) U <em>C</em> = {6, 10} U {1, 2, 4, 7, 11} = {1, 2, 4, 6, 7, 10, 11}

or

<em>A</em> ∩ (<em>B</em> U <em>C </em>) = {1, 2, 6, 10} ∩ {1, 2, 3, 4, 6, 7, 9, 10, 11} = {1, 2, 6, 10}

The first interpretation is probably the intended one, since that essentially reads the set operations from left to right.

(4) <em>A'</em> U <em>B</em> is the union of <em>A'</em> and <em>B</em>, where <em>A'</em> is the complement of <em>A</em>, or all elements in <em>U</em> that are not in <em>A</em>. We have

<em>A'</em> = <em>U</em> - <em>A</em> = {3, 4, 5, 7, 8, 9, 11, 12}

and so

<em>A'</em> U <em>B</em> = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(5) We have

<em>A</em> U <em>C</em> = {1, 2, 4, 6, 7, 10, 11}

so that

(<em>A</em> U <em>C </em>)<em>'</em> = <em>U</em> - (<em>A</em> U <em>C</em> ) = {3, 5, 8, 9, 12}

(6) We have

<em>B'</em> = <em>U</em> - <em>B</em> = {1, 2, 4, 5, 7, 8, 12}

and so

<em>A</em> ∩ <em>B'</em> = {1, 2}

(7) Using the complements found in (4) and (6), we have

<em>A'</em> U <em>B'</em> = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}

Alternatively, we can use the fact that

<em>A'</em> U <em>B'</em> = (<em>A</em> ∩ <em>B</em>)<em>'</em>

and since we know from (2) that <em>A</em> ∩ <em>B</em> = {6, 10}, we end up with the same result,

(<em>A</em> ∩ <em>B</em>)<em>'</em> = <em>U</em> - (<em>A</em> ∩ <em>B</em>) = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}

(8) We have

<em>A</em> U <em>B</em> U <em>C</em> = {1, 2, 3, 4, 6, 7, 9, 10, 11}

so that

(<em>A</em> U <em>B</em> U <em>C</em> )<em>'</em> = {5, 8, 12}

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3 years ago
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ivanzaharov [21]
4. By the fundamental theorem of calculus,

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g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x

which is clearly positive for x\in[-3,0), and in the second interval you have

-x+3>0\implies x

Together, this means g'(x)>0 for all x\in[-3,3).

5. When 0\le x\le6, f(x) reduces to -x+3, so you have

g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt
g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}
g(x)=6+3x-\dfrac12x^2
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4 years ago
Please help!! Someone answered this already as a joke, I seriously need help on this
iogann1982 [59]
Gotchu my dude

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3 years ago
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DerKrebs [107]

Answer:

22 degrees is the answer

Step-by-step explanation:

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3 years ago
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