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Gnesinka [82]
3 years ago
10

((x-h)^2)+((y-k)^2)​

Mathematics
1 answer:
Schach [20]3 years ago
5 0

Answer:

Step-by-step explanation:

= (x^2 + h^2 - 2xh) + (y^2 + k^2 - 2yk)     As, (a-b)^2 = a^2 + b^2 - 2ab

= x^2 + h^2 - 2xh + y^2 + k^2 - 2yk

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HERES MY QUESTION???!!!!!!!!!!!!!!!
Ivan

Answer:   x=55           y=145

x=55

35+90=125

125-180=55

y=145

180-35=145

I hope this is good enough:

5 0
3 years ago
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HELP MEEEEEEEEE PLEASEEEEEEEEEEEEEE!!!!!!!!!!!!!!!!!!
Nadya [2.5K]

Answer: 58, 122

Step-by-step explanation:

x+2x+6=180\\\\3x+6=180\\\\x+2=60\\\\x=58\\\\\therefore 180-x=122

4 0
1 year ago
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Do the ratios 10/8 and 16/20 form a proportion<br><br> Yes or no?
ziro4ka [17]

Answer:

No. Reduce 10/8 by dividing by 2. 10/8= 5, 8/2= 4. 10/8= 5/4. Reduce 16/20 by dividing by 4. 16/4= 4 , 20/4= 5. 16/20= 4/5.

6 0
3 years ago
Graph the system of inequalities. y is greater than 3x-4 4y-x is less than or equal to 8
Ugo [173]

Answer:

no;no because it is going to be negative 12.

Step-by-step explanation:

first you do 3x-4=-12 and so if you get -12 it is not even close to 8.

4 0
3 years ago
What's the next number? 0 , 1/3 , 1/2 , 3/5 , 2/3​
madam [21]

Answer:

The next number of the series 0, 1/3, 1/2, 3/5, and 2/3 is 5/7

Step-by-step explanation:

The given numbers are;

0, 1/3, 1/2, 3/5, and 2/3

The number sequence is formed adding \dfrac{1}{\left (\dfrac{n^2 + n}{2} \right ) } to each (n - 1)th term to get the nth term number in the sequence, with the first term equal to 0, as follows;

For the 2nd term, the (n - 1)th term is 0, and n = 2, gives;

The

0 +\dfrac{1}{\left (\dfrac{2^2 + 2}{2} \right ) } = 0 + \dfrac{1}{3} = \dfrac{1}{3}

For the 3rd term, the (n - 1)th term is 1/3, and n = 3, gives;

\dfrac{1}{3} +\dfrac{1}{\left (\dfrac{3^2 + 3}{2} \right ) } = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}

For the 4th term, the (n - 1)th term is 1/2, and n = 4, gives;

\dfrac{1}{2} +\dfrac{1}{\left (\dfrac{4^2 + 4}{2} \right ) } = \dfrac{1}{2} + \dfrac{1}{10} = \dfrac{3}{5}

For the 5th term, the (n - 1)th term is 3/5, and n = 5, gives;

\dfrac{3}{5} +\dfrac{1}{\left (\dfrac{5^2 + 5}{2} \right ) } = \dfrac{3}{5} + \dfrac{1}{15} = \dfrac{2}{3}

For the next or 6th term, the (n - 1)th term is 2/3, and n = 6, gives;

\dfrac{2}{3} +\dfrac{1}{\left (\dfrac{6^2 + 6}{2} \right ) } = \dfrac{2}{3} + \dfrac{1}{21} =  \dfrac{15}{21} = \dfrac{5}{7}

The next number of the series 0, 1/3, 1/2, 3/5, and 2/3 = 5/7.

6 0
3 years ago
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