Y’’(x)= 6x + 1
y’(x)= 3x^2 + x + 2
y(x)= x^3 + 1/2x^2 + 2x + 5
Answer:
B. 4
Step-by-step explanation:
Determine the constant of variation for the direct variation given.
(0, 0), (3, 12), (9, 36)
A. 3
B. 4
C.12
Direct variation is given by:
y = kx
Where,
k = constant of variation
(3, 12)
x = 3; y = 12
y = kx
12 = k*3
12 = 3k
k = 12 / 3
k = 4
(9, 36)
x = 9; y = 36
y = kx
36 = k * 9
36 = 9k
k = 36 / 9
= 4
k = 4
Constant of the variation = 4
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
So there is an identity we'll need to use to solve this:
cos(x+y) = cosxcosy - sinxsiny
replace the numerator with the right hand side of that identity and we get:
(cosxcosy - sinxsiny)/cosxsiny
Separate the numerator into 2 fractions and we get:
cosxcosycosxsiny- sinxsiny/cosxsiny
the cosx's cancel on the left fraction, the siny's cancel on the right fraction and we're left with:
cosy/siny - sinx/cosx
which simplifies to:
coty - tanx
