Question

1.Докажите, что значение дроби равно нулю: 2. Расстояние между двумя пунктами по реке равно 2 км. Лодка совершила путь в оба конца за 1 час 30 минут. Найдите скорость лодки в стоячей воде, если скорость течения реки равна 1 км/ч. . 4. Докажите, что середины сторон произвольного четырехугольника являются вершинами параллелограмма. 5. Отрезок - хорда окружности с центром в точке . Угол между равен . Найдите величину угла между прямой и касательной к окружности, проходящей через точку . решите пожалуйста

Answer 1

1)Пусть скорость лодки в стоячей воде х км/ч, тогда по течению она движется со скоростью х+1 км/ч, а против течения х-1 км/ч. В оба конца 2км она пройдет за. 

Answer 2

1. Докажите, что значение дроби равно нулю: To prove that the value of the fraction is equal to zero, we need to show that the numerator of the fraction is equal to zero. Let's consider the given fraction:

fraction = 0/n

where n is any non-zero number. If we can show that the numerator, which is 0, is equal to zero, then the value of the fraction will be zero.

We can see that the numerator of the fraction is already zero. Therefore, the value of the fraction is indeed zero.

2. Расстояние между двумя пунктами по реке равно 2 км. Лодка совершила путь в оба конца за 1 час 30 минут. Найдите скорость лодки в стоячей воде, если скорость течения реки равна 1 км/ч. To find the speed of the boat in still water, we can use the concept of relative velocity. Let's assume the speed of the boat in still water is v km/h.

When the boat is moving upstream (against the current), its effective speed is reduced by the speed of the current. So, the boat's speed relative to the riverbank is (v - 1) km/h.

Similarly, when the boat is moving downstream (with the current), its effective speed is increased by the speed of the current. So, the boat's speed relative to the riverbank is (v + 1) km/h.

We know that the distance between the two points is 2 km, and the boat completes the round trip in 1 hour 30 minutes, which is equivalent to 1.5 hours.

Using the formula: speed = distance / time, we can set up the following equation for the upstream journey:

2 = (v - 1) * 1.5

Simplifying the equation, we have:

3 = 1.5v - 1.5

Adding 1.5 to both sides:

4.5 = 1.5v

Dividing both sides by 1.5:

v = 3

Therefore, the speed of the boat in still water is 3 km/h.

4. Докажите, что середины сторон произвольного четырехугольника являются вершинами параллелограмма. To prove that the midpoints of the sides of an arbitrary quadrilateral form the vertices of a parallelogram, we can use the concept of parallel lines.

Let ABCD be an arbitrary quadrilateral, and let E, F, G, and H be the midpoints of sides AB, BC, CD, and DA, respectively.

We need to show that EF is parallel to HG and EH is parallel to FG.

First, consider triangle ABC. Since E is the midpoint of AB, we can apply the midline theorem, which states that the line connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. Therefore, EF is parallel to AC and EF = 1/2 * AC.

Similarly, in triangle CDA, we can apply the midline theorem to show that HG is parallel to AC and HG = 1/2 * AC.

Since EF and HG are both parallel to AC and have the same length (EF = HG), they must be parallel to each other.

Next, consider triangle BCD. By applying the midline theorem, we can show that EH is parallel to BD and EH = 1/2 * BD.

Likewise, in triangle DAB, we can apply the midline theorem to show that FG is parallel to BD and FG = 1/2 * BD.

Since EH and FG are both parallel to BD and have the same length (EH = FG), they must be parallel to each other.

Therefore, we have shown that EF is parallel to HG and EH is parallel to FG, proving that the midpoints of the sides of the quadrilateral form the vertices of a parallelogram.

5. Отрезок - хорда окружности с центром в точке O. Угол между равен 60 градусов. Найдите величину угла между прямой и касательной к окружности, проходящей через точку B.

To find the angle between the line and the tangent to the circle passing through point B, we need to use the properties of tangents and chords.

Let's consider a circle with center O and a chord AB. The angle between the chord and the tangent at point B is equal to half the angle subtended by the chord at the center of the circle.

Given that the angle between is 60 degrees, we can conclude that the angle subtended by the chord AB at the center O is 2 * 60 = 120 degrees.

Since the angle between the chord AB and the tangent at point B is equal to half the angle subtended by the chord at the center, the angle between the line and the tangent is 1/2 * 120 = 60 degrees.

Therefore, the angle between the line and the tangent to the circle passing through point B is 60 degrees.