Answer:
21 yrs old
Step-by-step explanation:
m= michael age
b=Brandon age
m=b+12
m-17=4(b-17)= 4b-68 --»»m = 4b -51
//put b+12 instead m// b+12=4b-51
3b=63 --»»b=21
Brandon age is 21
Michael age is 33
Answer:
B: (4x - 7)(16x^2 + 28x + 49)
Step-by-step explanation:
64x^3-343 can be rewritten as 4^3x^3 - 7^3 or (4x)^3 - 7^3. This is the difference of two cubes. The appropriate formula for factoring such is
a^3 - b^3 = (a - b)(a^2 + ab + b^2). Therefore,
our (4x)^3 - 7^3 = (4x - 7)(16x^2 + 28x + 49). This is Answer B.
Answer:
The estimate of In(1.4) is the first five non-zero terms.
Step-by-step explanation:
From the given information:
We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)
So, by the application of Maclurin Series which can be expressed as:

Let examine f(x) = In(1+x), then find its derivatives;
f(x) = In(1+x)

f'(0) 
f ' ' (x) 
f ' ' (x) 
f ' ' '(x) 
f ' ' '(x) 
f ' ' ' '(x) 
f ' ' ' '(x) 
f ' ' ' ' ' (x) 
f ' ' ' ' ' (x) 
Now, the next process is to substitute the above values back into equation (1)



To estimate the value of In(1.4), let's replace x with 0.4


Therefore, from the above calculations, we will realize that the value of
as well as
which are less than 0.001
Hence, the estimate of In(1.4) to the term is
is said to be enough to justify our claim.
∴
The estimate of In(1.4) is the first five non-zero terms.
Answer:
7/40
Step-by-step explanation:
There are 3 girls+9 boys = 12 students in the 7th grade
P (girl in 7th grade) = girls/ total
= 3/12 = 1/4
There are 7 girls+3 boys = 10 students in the 8th grade
P (girl in 8th grade) = girls/ total
= 7/10
P(7th grade girl, 8th grade girl) = 1/4 * 7/10 = 7/40