The parents could have 3 boys and 3 girls in 400 ways
<h3>How to determine the number of ways?</h3>
The given parameters are
Children, n = 6
Boys, r = 3
Girls, r =3
The number of combination is:
So, we have:
Apply the combination formula
This gives
Ways = 400
Hence, the parents could have 3 boys and 3 girls in 400 ways
Read more about combination at:
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Chain rule:
if
y=y(u) and u=u(x)
The dy/dx=(dv/du)(du/dx)
In our case
y=arcsin(u)
u=sin(x)
dy/du=1/√(1-u²) = 1/√(1-sin²x)
du/dx=cos x
dy/dx=cos x /√(1-sin²x)
Answer: dy/dx=cos x /√(1-sin²x)
Answer:
d 2
Step-by-step explanation:
x/7 +13/14 = 17/14
double the first one so all the denominators are the same
2x/14+13/14=17/14
then subtract the second fraction from the thrid
2x/14=4/14
then we can divide by two so we are back to 7 as the demoninator
x/7=2/7
The Answer is C: (24-15) x 4-3 x 2=30
Answer:
D
Step-by-step explanation:
4(x-7)+8=4x-4*7+8=4x-28+8
