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olya-2409 [2.1K]
3 years ago
7

Pls help me pls i will give brainlist

Mathematics
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

1. Pour the mixture through a filter to separate the sand; 2. Evaporate the liquid to separate the salt.

Step-by-step explanation:

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In a bank account a paid out expense is called a debit and a deposit is called a credit. Would you use positive or negative inte
Allisa [31]

Answer:

positive integer

Step-by-step explanation:

In a bank account,

A paid out expense is called as debit

A deposit is called as credit

To find: sign used to represent credits

Solution:

Integers denote a set of natural numbers, their negative counterpart, and 0.

As credit denotes the amount deposited, the positive integer is used to denote credits.

5 0
4 years ago
Determine the center and radius of the following circle equation:
UkoKoshka [18]

Answer:

The equation of the circle  <em>(x + 5 )² + ( y + 10 )² = (4)²</em>

<em>Center of the circle ( h, k) = ( -5 , -10)</em>

<em>Radius of the circle     ' r' = 4</em>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given circle equation is x² + y² +10 x + 20 y +109 =0

                                     x² +10 x +  y² + 20 y +109 =0

  x² +2 (5) (x)+(5)² -(5)²+ y² +2(10) y + (10)²-(10)² +109 =0

<em>By using formula </em>

<em>(a+b)² = a² + 2 a b + b²</em>

<em>(x + 5 )² + ( y + 10 )² - 25 - 100 + 109 = 0</em>

<em>(x + 5 )² + ( y + 10 )² - 16 = 0</em>

<em>(x + 5 )² + ( y + 10 )² = 16</em>

(x + 5 )² + ( y + 10 )² = (4)²

The standard equation of the circle  ( x - h )² + ( y -k)² = r²

<em> Center of the circle ( h, k) = ( -5 , -10)</em>

<em>Radius of the circle     ' r' = 4</em>

<u><em>Conclusion</em></u><em>:-</em>

The equation of the circle  <em>(x + 5 )² + ( y + 10 )² = (4)²</em>

<em>Center of the circle ( h, k) = ( -5 , -10)</em>

<em>Radius of the circle     ' r' = 4</em>

<em></em>

3 0
3 years ago
Use the diagram of ◾ABC to find m&lt;D<br>M&lt;D=​
Yuliya22 [10]

Answer:

In quadrilateral ABCD we have

AC = AD

and AB being the bisector of ∠A.

Now, in ΔABC and ΔABD,

AC = AD

[Given]

AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

∴ Using SAS criteria, we have

ΔABC ≌ ΔABD.

∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

∴ BC = BD.

7 0
3 years ago
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