Answer:
Yes
Step-by-step explanation:
1 : The athlete's hands push the medicine ball forward. The medicine ball pushes the athlete's hands backward.
2: Friction
3: The first pair of action-reaction force pairs is: foot A pushes ball B to the right; and ball B pushes foot A to the left. The second pair of action-reaction force pairs is: foot C pushes ball B to the left; and ball B pushes foot C to the right
There is no difficulty in this problem until you construct the figures. How can we do it is shown in the attached picture. After drawing PRST, from the point P, we can draw PMKD and later we can complete PMCT as a result. From this picture, we can see that the side of PMCT is also a. Then, the area of this square is
Answer:
x = 4 ±√26
Step-by-step explanation:
Please write this as x^2 - 8x = 10, or x^2 - 8x - 10 = 0. " ^ " indicates exponentiation.
Let's complete the square:
x^2 - 8 x can be rewritten as
x^2 - 8x + 16 - 16, and so
x^2 - 8x = 10 becomes x^2 - 8x + 16 - 16 - 10 = 0, or
(x - 4)^2 = 26
Taking the square root of both sides, we get:
x - 4 = ±√26, or
x = 4 ±√26
Answer:
I think s(2s) = 205
Step-by-step explanation:
Answer:
Las longitudes solicitadas en yardas son:
- <u>Trayecto A = 109.361 yardas.</u>
- <u>Trayecto B = 20.231785 yardas.</u>
Step-by-step explanation:
Para hacer la conversión de unidades que requieres en el ejercicio, debes saber que:
Con ese factor de conversión tú puedes hacer reglas de tres para calcular las medidas que requieres. En el caso del trayecto A:
Si:
- 1 metro = 1.09361 yardas
- 100 metros = X
Entonces:
Cancelamos metros y obtenemos:
- x = 100 * 1.09361 yardas
- <u>x = 109.361 yardas</u>
En este caso, <u>el trayecto A en yardas corresponde a 109.361 yardas</u>. El mismo procedimiento puede aplicarse para el trayecto B:
Si:
- 1 metro = 1.09361 yardas
- 18.50 metros = X
Entonces:
Cuando se cancelan los metros se obtiene:
- x = 18.50 * 1.09361 yardas
- <u>x = 20.231785 yardas</u>
Así, <u>el trayecto B en yardas corresponde a 20.231785 yardas</u>.