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3 years ago

(x2 - 10x + 25) + (y2 - 16y +64) = 16 center (, ) radius =

1 answer:

4 0

The standard equation for a circle is

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

where the center is at point (h,k) and r is the radius.

To investigate this, check first if the quadratic equations are a perfect square. This is determined by dividing the middle term by 2 and taking its square. If it equals to the third term, it is a perfect square.

For (x^2 - 10x +25): (-10/2)^2 = 25 (perfect square)
For (y^2 - 16y + 64): (-16/2)^2 = 64 (perfect square)

Then, we simplify the equation by factoring:

(x - 5)^2 + (y -- 8)^2 = 16

Thus, the center is at point (5,8)
Also, the radius is square root of 16, which is 4.

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$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

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Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago