Answer:
A) $23.41
Step-by-step explanation:
you would multiply 57.10 by 41%
Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :
B )
Acceleration a is given by :
Hence , this is the required solution .
Answer:the weight after 225 days is
22885 kilograms
Step-by-step explanation:
The initial weight of the blue whale calf at birth is 2725 kilograms. blue whale calf gains 90 kilograms of weight each day for the first 240 days after its birth. The weight increases in arithmetic progression. This means that the first term of the sequence, a is 2725, the common difference, d is 90.
The formula for the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
n is the number of terms of the sequence.
a is the first term
d is the common difference
We want to determine its weight, T225 after 225 days after it’s birth. It means that n = 225
Therefore
T225 = 2725 + (225 - 1)90
T225 = 2725 + 224×90 = 2725 + 20160
T225 = 22885
3-4x=-6x^2
+6x^2=+6x^2
_____________
6x^2+3-4x=0
6x^2-4x+3=0
(-4)^2-4(6)(3)
16-72
Final Answer: -56
