Question

СРОЧНО!ПОМОГИТЕ! найдите два решения данных неравенств 4х - у² ≥ 1 Координаты каких точек являются решением неравенства у≦ -х2+81 А(4:-2) В(9:0) С(-10:1) Д(11:-11) На координатной плоскости изобразите штриховкой решения неравенства х2+у2≥4 На координатной плоскости изобразите штриховкой решение неравенства х2+у2≦2,25 Можно решение с фото,пожалуйста,очень нужно!

Answer 1

Решение во вложениииииииииииииииииииииииии

Answer 2

1
y²≤4x-1
Ответ (2;2)  (3;3)
2
у≦ -х2+81
А(4:-2)          -2≤-16+85⇒-2≤69 является
В(9:0)            0≤-81+81⇒0≤0       является
С(-10:1)          1≤-100+81⇒-1≤-19 не является
Д(11:-11)      -11≤-121+81⇒-11≤-40  не является
3
х2+у2≥4
Решением будет внешняя область круга радиусом 2
4
х2+у2≦2,25
Решением будет внутренняя область круга радиусом 1,5

Answer 3

Solution to the inequality 4x - y^2 ≥ 1:

To find the solutions to the inequality 4x - y^2 ≥ 1, we can start by rearranging the inequality to isolate y^2:

4x - y^2 ≥ 1 -y^2 ≥ 1 - 4x y^2 ≤ 4x - 1

Now, we can solve for y by taking the square root of both sides of the inequality. However, since we have a "≥" sign, we need to consider both the positive and negative square roots:

y ≤ √(4x - 1) y ≥ -√(4x - 1)

Therefore, the solutions to the inequality 4x - y^2 ≥ 1 are given by the range of values for x and y that satisfy the above inequalities.

Coordinates that satisfy the inequality y ≤ -x^2 + 81:

To find the coordinates that satisfy the inequality y ≤ -x^2 + 81, we can substitute the given x and y values into the inequality and check if the inequality holds true.

The given coordinates are: A(4, -2) B(9, 0) C(-10, 1) D(11, -11)

Substituting the x and y values of each coordinate into the inequality, we have:

For point A(4, -2): -2 ≤ -(4^2) + 81 -2 ≤ -16 + 81 -2 ≤ 65 (True)

For point B(9, 0): 0 ≤ -(9^2) + 81 0 ≤ -81 + 81 0 ≤ 0 (True)

For point C(-10, 1): 1 ≤ -(-10^2) + 81 1 ≤ -100 + 81 1 ≤ -19 (False)

For point D(11, -11): -11 ≤ -(11^2) + 81 -11 ≤ -121 + 81 -11 ≤ -40 (True)

Therefore, the coordinates that satisfy the inequality y ≤ -x^2 + 81 are A(4, -2), B(9, 0), and D(11, -11).

Graphing the solution to the inequality x^2 + y^2 ≥ 4:

To graph the solution to the inequality x^2 + y^2 ≥ 4, we can plot the boundary curve and shade the region that satisfies the inequality.

The boundary curve is given by x^2 + y^2 = 4, which represents a circle with a radius of 2 centered at the origin (0, 0). To graph this, we can plot points on the circle and draw a smooth curve passing through them.

Next, we need to shade the region that satisfies the inequality x^2 + y^2 ≥ 4. Since the inequality is "≥", we need to shade the region outside the circle.

Graphing the solution to the inequality x^2 + y^2 ≤ 2.25:

To graph the solution to the inequality x^2 + y^2 ≤ 2.25, we can follow a similar process as above.

The boundary curve is given by x^2 + y^2 = 2.25, which represents a circle with a radius of 1.5 centered at the origin (0, 0). Plot points on the circle and draw a smooth curve passing through them.

Next, shade the region that satisfies the inequality x^2 + y^2 ≤ 2.25. Since the inequality is "≤", we need to shade the region inside the circle.

Unfortunately, I am unable to provide a photo or visual representation here. However, you can easily graph these solutions on a coordinate plane using graphing software or by hand.