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vichka [17]
2 years ago
8

2.) Of a classroom filled with 20 students, 2 will be selected to stay after school and correct

Mathematics
1 answer:
harina [27]2 years ago
8 0

Answer:

190

Step-by-step explanation:

20C2 = (20*19)/(1*2) = 190

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Simplify (5x−2+3x^2)+(4x+3)
Art [367]

Answer:

3x^2 + 9x + 1

Or

3x ( x + 3 ) + 1

Step-by-step explanation:

(5x - 2 +3x^2 ) + (4x + 3 )

To make this a little bit more easier to read, you can remove the parentheses:

5x - 2 + 3x^2 + 4x + 3

Now, write in a way so that the like terms are next to each other:

3x^2 + 5x + 4x - 2 + 3

Now simplify the 'x' terms to get:

3x^2 + 9x - 2 + 3

Now, simplify the integers (the ones with now variables with them) to get:

3x^2 + 9x + 1

If you want, you can factor out the 3x for two of the terms to get :

3x ( x + 3 ) + 1

Therefore, your simplest form can either be 3x^2 + 9x + 1 OR 3x (x + 3 ) + 1

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3 years ago
An equipment rental company charges a flat rate of $25, plus $13 per day for insurance. Kyle has $121. Write an inequality to re
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121 \geqslant38d
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38. 38


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4 0
4 years ago
2x -2 = -200 <br><br><br> whats the value of x
Elden [556K]

Answer:

Step-by-step explanation:

2x - 2 = - 200

Bringing like terms on one side

2x = - 200 + 2 = - 198

2x = - 198

x = - 198/2 = - 99

4 0
3 years ago
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katen-ka-za [31]

Answer:

0

Step-by-step explanation:

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∫ u² du

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0

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3 years ago
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Check the picture below.

\bf \textit{using the pythagorean theorem}&#10;\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\&#10;-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}&#10;\\\\\\&#10;csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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3 years ago
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