Answer:
Approximate solution is 541.
Step-by-step explanation:
y' = 2xy, Δx = 0.4.
Make a table:
x y y' y' Δx + y
1 2 4 4*0.4 + 2 = 3.6 <---this is the new y value.
1.4 3.6 10.08 7.632
1.8 7.632 27.48 18.624
2.2 18.624 81.95 51.70
2.6 51.70 268.85 159.24
3 159.24 955.44 541.4
In order to solve or know the probability of having 2 girls
and 2 boys, assumed that a girl is as likely as a girl at each birth, pascal’s
triangle will be likely used. And we will be referring to the line 4 of pascal’s
triangle, which was 1 4 6 4 1. Then it
will look like this: 1 = 4 girls; 4 = 3 girls & 1 boy; 6 = 2 girls & 2
boys; 4 = 3 boys & 1 girl; 1 = 4 boys. And now for the solution in order to
get the probability of having 2 girls and 2 boys is to divided into the sum of 1+4+6+4+1.
Answer:
x²+1/x² = 51
Explanation:
Given x - 1/x = 7 ---(1)
We know the algebraic identity:
a²+b²-2ab = (a-b)²
Or
a²+b² = (a-b)²+2ab
Now,
x²+1/x²
= (x-1/x)²+2*x*(1/x)
= (x-1/x)²+2
:7²+2 [ from (1)] =
= 49+2
= 51
Therefore,
x²+1/x² = 51
Answer:
V = 486
, or approx. 1,526.814
Step-by-step explanation:
V = *r²*(h/3)
plug in the numbers: V = * (9)² * (18/3)
simplify: V = * 81 * 6
simplify: V = 486, or approx. 1,526.814
