$→{x}^{2} + 6x + {y}^{2} - 8y = 19 \\ ({x}^{2} + 2.3.x + {3}^{2} ) +({y}^{2} - 2.4.y + {4}^{2} ) = 19 + {3}^{2} + {4}^{2} \\ {(x + 3)}^{2} + {(y - 4)}^{2} = 19 + 9 + 16 \\ \boxed{ {(x + 3)}^{2} + {(y - 4)}^{2} = 44}✓$

3) (x+3)²+(y-4)²=44 is the right answer.

8 0

2 years ago

Read 2 more answers

ABCD Is a rectangle that represents a park.

Andru [333]

Answer:

Shortest distance from A to C = 102.9005 m.

Step-by-step explanation:

It is given that, ABCD is a rectangular park.

The length of the park is 80 m.

The breadth of the park is 50 m.

The diameter of the circle = 15 m.

We have to calculate the shortest distance from A to C across the park.

The distance AC = $\sqrt{50^{2}+80^{2} }$ = 94.339 m.

As one has to pass only through the lines shown, he cannot pass through the circle.

So, we have to subtract the diameter of 15 m from AC

=> 94.339 m - 15 m = 79.339 m.

One must pass through either half of the circumference of the circle.

Since, diameter of circle = 15 m, its radius(r) = $\frac{15}{2}$ = 7.5 m.

The circumference of the circle = 2×π×r = 47.123 m.

Half of the circumference = $\frac{47.123}{2}$ = 23.5615 m.

Distance from A to C passing through circumference = 79.339 m + 23.5615 m = 102.9005 m.

As we have to calculate the shortest distance from A to C, one cannot pass from A to C either through B or D,

since the distance ABC or ADC = 50+80 = 130 m.

Therefore, shortest distance from A to C = 102.9005 m.

3 0

3 years ago

Read 2 more answers

57; 27; 72; 36; 61; 81; 45 From the numbers above, choose a prime factor​

57; 27; 72; 36; 61; 81; 45 From the numbers above, choose a prime factor