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kramer
3 years ago
10

Find the area of the circle whose equation is x2+y2=6x-8y​

Mathematics
1 answer:
Semenov [28]3 years ago
6 0

Answer:

Given that the equation of a circle is :

\green{ \boxed{\boxed{\begin{array}{cc}  {x}^{2}  +  {y}^{2}  = 6x - 8y \\  =  >  {x}^{2}  +  {y}^{2}   - 6x + 8y = 0 \\  =  >  {x}^{2}  +  {y}^{2}  + 2 \times ( - 3) \times x + 2 \times 4 \times y = 0 \\  \\  \sf \: standard \: equation \: o f \: circle \: is :  \\   {x}^{2}  +  {x}^{2}  + 2gx + 2fy + c = 0 \\  \\  \sf \: by \: comparing \\  \\ g =  - 3 \\ f = 4 \\ c = 0 \\  \\  \sf \: radius \:  \: r =  \sqrt{ {g}^{2}  +  {f}^{2} - c }  \\  =  \sqrt{ {( - 3)}^{2} +  {4}^{2} - 0  }  \\  =  \sqrt{9 + 16}  \\   = \sqrt{25}  \\  = 5 \: unit \\  \\  \bf \: area \:  = \pi {r}^{2}  \\  = \pi \times  {5}^{2}  \\  =\pink{ 25\pi \:  { unit }^{2} }\end{array}}}}

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Damm [24]

Answer:

2.0 x 10^{1}

Step-by-step explanation:

1. length = 3.2 x 10^{5} m

width = 1.6 x 10^{4} m

Area of a rectangle = length x width

                                = (3.2 x 10^{5}) x (1.6 x 10^{4})

                                = 5.12 x 10^{9} m^{2}

2. length = 1.28 x 10^{7} m

   width = 8 x 10^{3} m

Area of a rectangle = length x width

                                 = (1.28 x 10^{7}) x (8 x 10^{3})

                                 = 1.024 x 10^{11} m^{2}

\frac{Area of rectangle 2}{Area of rectangle 1} = \frac{1.024*10^{11} }{5.12*10^{9} }

                       = 20

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Thus,

The area of rectangle 2 is 2.0 x 10^{1} times greater than the area of rectangle 1.

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A force of 50 pounds acts on an object at an angle of 45 degrees. A second force of 75 pounds acts on the object at an angle of
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Answer:

The magnitude of the resultant force is:

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The direction is:

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Step-by-step explanation:

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F_{tot-x}=F_{1}cos(45)+F_{2}cos(30)

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F_{tot-x}=100.3 \: pound

<u>Sum of y-component vector forces.</u>

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R=\sqrt{100.3^{2}+2.1^{2}}

R=100.3 \:pound

The direction is:

tan(\theta)=\frac{2.1}{100.3}

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