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Reptile [31]
2 years ago
8

Find an equation of the line having the given slope and containing the given point.

Mathematics
1 answer:
Nadusha1986 [10]2 years ago
4 0
1.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that reacts with water to produce a lilac flame

2.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element used as an inert atmosphere

3.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that has a valency of 3

4.Write a balanced chemical equation for the reaction between potassium and water. (Non-anonymous question). .
(1 Point)

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5.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element with a fixed valency of 2 that not is not in group 2

HELP
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Zinaida [17]

Answer:

The 3rd one 50 - 12p = 26

Step-by-step explanation:

7 0
4 years ago
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Find the value of this expression if x = -9.<br> x2 + 3<br> x +6
iren [92.7K]

Answer:

Step-by-step explanation:

x² + 3 = (-9)² + 3          {  (-9)² = -9 *-9 = 81}

     = 81 + 3

       = 84

x + 6 = -9 + 6 = -3

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Berry waked the dog for 20 minutes.
Sergeu [11.5K]
Chris walked the dog for 80 minutes or 1 hour and 20 minutes
5 0
3 years ago
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three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

7 0
3 years ago
1 point<br> Captionless Image
algol [13]

Answer:

uhhh ok

Step-by-step explanation:

6 0
3 years ago
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