Answer:
17.907082 unit
Step-by-step explanation:
According to the Question,
Given, A circle with centre F, ∠EFG=54 and EF=19 .
length of arc EG = Radius(EF) × ∠EFG(in Radian)
- We Know, 1 degree = 0.0174533 Radian
- 54 degree = 0.942478 Radian
length of arc EG = 19 x 0.942478 ⇔ 17.907082 unit
(For Diagram please find in attachment)
Answer: 13%
Step-by-step explanation:
We know that the formula to find the simple interest :
, where P is the principal amount , r is rate (in decimal )and t is the time period (in years).
Given : Albert borrowed $19,100 for 4 years. The simple interest is $9932.00. Find the rate.
i.e. P = $19,100 and t= 4 years and I = $9932.00
Now, Substitute all the values in the formula , we get
![9932=(19100)r(4)\\\\\Rightarrow\ r=\dfrac{9932}{19100\times4}\\\\\Rightarrow\ r=0.13\ \ \ \text{[On simplifying]}](https://tex.z-dn.net/?f=9932%3D%2819100%29r%284%29%5C%5C%5C%5C%5CRightarrow%5C%20r%3D%5Cdfrac%7B9932%7D%7B19100%5Ctimes4%7D%5C%5C%5C%5C%5CRightarrow%5C%20r%3D0.13%5C%20%5C%20%5C%20%5Ctext%7B%5BOn%20simplifying%5D%7D)
In percent, 
Hence, the rate of interest = 13%
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It is 2 that what i got and i think i got it write if not delete
Step-by-step explanation:
La posici´on de una part´ıcula que se mueve unidimensionalmente esta definida por la ecuaci´on:
x(t) = 2t
3 − 15t
2 + 24t + 4 donde 0x
0 y
0
t
0
se expresan en metros y segundos respectivamente. Determine:
a. ¿Cu´ando la velocidad es cero?
b. La posici´on y la distancia total recorrida cuando la aceleraci´on es cero.
Soluci´on:
a. Recordemos que:
v(t) =
dx
dt =
d
dt(2t
3 − 15t
2 + 24t + 4) = 6t
2 − 30t + 24
Sea t
0
el tiempo en que la velocidad se anula, entonces v(t
0
) = 0.
De este modo:
0 = v(t
0
) = 6(t
0
)
2 − 30(t
0
) + 24 = 6[(t
0
)
2 − 5(t
0
) + 4] = 6[(t
0
) − 4][(t
0
) − 1]
As´ı tenemos que:
t
0
1 = 4, t
0
2 = 1
De este modo, tenemos que la velocidad se anula al primer segundo y a los cuatro segundos.
b. Recordemos que:
a(t) =
dv
dt =
d
dt(6t
2 − 30t + 24) = 12t − 30
Ahora sea t
0
el instante en que la aceleraci´on se anula, entonces a(t
0
) = 0
Ahora:
0 = a(t
0
) = 12t
0 − 30
As´ı tenemos que: t
0 =
30
12 =
5
2
Por lo tanto, la posici´on en este instante es:
x(t
0
) = x
5
2
= 2
5
2
3 − 15
5
2
2 + 24
5
2
+ 4 = 125
4 − 3
125
4 + 60 + 4 = −2
125
4 + 64 = −
125
2 +
128
2 =
3
2
De este modo, la posici´on de la part´ıcula cuando la aceleraci´on es cero es de 3
2 metros.
Adem´as la distancia total recorrida esta dada por:
distancia = |x(t
0
) − x(0)| = |
3
2 − 4| =
5
2
Finalmente la distancia total recorrida es: 5
