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2 years ago

What are the center and the radius of the circle 4x2 + 4y2 - 8x + 16y -4 = 0?

Mathematics

1 answer:

Sedaia [141]2 years ago

8 0

Answer:

Center of circle: (1, -2)

Radius: √6

Step-by-step explanation:

First, we need to isolate the constant:

$4x^2 + 4y^2 - 8x + 16y = 4$

Then, we can divide the whole equation by 4:

$x^2+y^2-2x+4y=1$

Then, we can complete the square(reminder: (a+b)^2 = a^2 + 2ab + b^2)

$(x^2-2x)+(y^2+4y)=1\\(x^2-2x+1)+(y^2+4y)=1+1\\(x^2-2x+1)+(y^2+4y+4)=1+1+4\\(x-1)^2+(y+2)^2=6$

Now that we got the standard equation of the circle, we can find the center and radius by using the equation $(x-h)^2+(y-k)^2=r^2$ , where (h, k) is the coordinate of the center, and r is the radius.

Therefore, the center of the circle is (1, -2), and the radius of the circle is √6.

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Answer:

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Step-by-step explanation:

Option 1

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Exponential equation;

y=a(1-r)^t

where;

y=future value

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r=depreciation rate=2% per month

t=time interval=12×3=36 months

Replacing;

y=60000(1-2/100)^36

y=60000(0.98)^36=28,992.79

The value after 3 years=$28,992.79

Initial value-Final value=(60000-28992.79)=$31007.21

Percentage of initial value lost=((Final value-Initial Value)/(Initial Value))×100

(31007.21/60000)×100=51.68%

Option 2

New starfish for $40,000 with a depreciation of $200 per month for 3 years

Linear equation;

y=a-bt

where;

y=Future value

a=Initial value=$40,000

b=the depreciation amount per time interval=$200 per month for 3 years

t=time interval=(3×12)=36 months

Replacing;

y=40000-(200×36)

y=32,800

Final value=y=$32,800

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(7200/40000)×100=18%

Option 1(51.68%)>Option 2(18%) therefor Option 1 loses value at a faster rate than Option 2

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Step-by-step explanation:

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2 years ago

When 1,250 Superscript three-fourths is written in simplest radical form, which value remains under the radical?

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The value remains under the radical is 8 ⇒ last answer

Step-by-step explanation:

Let us revise how to write the exponent as a radical

$a^{\frac{m}{n}}$ can be written as $\sqrt[n]{a^{m}}$
To simplify the radical factorize the base "a" to its prime factors

Example:

$(54)^{\frac{2}{3}}=\sqrt[3]{(54)^{2}}$ ,
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$3^{6}$ can go out the radical because 6 is divisible by 3, then divide 6 by 3, so it will be 3² out the radical
$\sqrt[3]{(54)^{2}}=3^{2}\sqrt[3]{2^{2}}=9\sqrt[3]{4}$