X+y+5=0
y=-x-5
If a solution exists y=y so we can say
x^2-9x+10=-x-5 add x+5 to both sides
x^2-8x+15=0 now factor
x^2-3x-5x+15=0
x(x-3)-5(x-3)
(x-5)(x-3) so x=3 and 5, using y=-x-5
y(3)=-8 and y(5)=-10
So the two solutions are:
(3,-8) and (5,-10)
The possibilities are:
bbbb
bbbg
bbgg
bggg
This shows that there is a 1 in 4 chance of having three boys and one girl
So the answer could be 1/4 or a 25% chance
the anwser would be -2 because -2*3*3-2*8 equals -2
L=Lim tan(x)^2/x x->0
Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.
d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)
d(x)/dx = 1
=>
L=2tan(x)sec^2(x)/1 x->0
= (2(0)/1^2)/1
=0/1
=0
Another way using series,
We know that tan(x) = x+x^3/3+2x^5/15+.....
then tan^2(x), using binomial expansion gives
x^2+2*x^4/3+.... (we only need two terms)
and again apply l'Hôpital's rule, we have
L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1
=0 as x->0
Triangle angles add to 180 degrees. Thus, x - 17, x - 22, and 3x + 19 add to 180.
Add like terms to get 5x - 20 = 180
5x = 200
x = 40
