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Reika [66]
3 years ago
7

Identify the center and radius of the circle with equation. ( please help me )

Mathematics
1 answer:
lutik1710 [3]3 years ago
6 0

Answer:

Center = (4, -5)

radius = 6

======================================================

Explanation:

The general template of any circle is

(x-h)^2 + (y-k)^2 = r^2

Rewrite the given equation into this form

(x-4)^2 + (y- (-5))^2 = 6^2

We can see that (h,k) = (4,-5) is the center and r = 6 is the radius.

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Ali's dog weighs 8 times as much as her cat. Together the two pets weigh 54 pounds. How much does Ali's dog weigh?
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). If a, ß are zeroes of the quadratic polynomial p(x)=kx²+4x+4 such
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Answer:

The values of k are 2/3 and -1

Step-by-step explanation:

Product of zeros = αβ= constant  / coefficient of x^2 =  4/k

Sum of zeros =α+β = - coefficient of x / coefficient of x^2= -4/k

Given

Consider a= α and b= β

(\alpha)^2 + (\beta)^2 = 24

(\alpha)^2 + (\beta)^2 can be written as (\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 if we add \pm 2 (\alpha)(\beta) in the above equation.

(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 -2(\alpha)(\beta)

(\alpha + \beta)^2 -2(\alpha)(\beta)

Putting values of αβ and α+β

(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1

The values of k are 2/3 and -1

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