If 1/5 are chocolate chip, then 4/5 are not chocolate chip because 1 - 1/5 = 4/5.
Half of 4/5 are peanut butter. Half of 4/5 is 2/5, so 2/5 of the cookies are peanut butter.
Answer:
add, subtract, multiply and divide complex numbers much as we would expect. We add and subtract
complex numbers by adding their real and imaginary parts:-
(a + bi)+(c + di)=(a + c)+(b + d)i,
(a + bi) − (c + di)=(a − c)+(b − d)i.
We can multiply complex numbers by expanding the brackets in the usual fashion and using i
2 = −1,
(a + bi) (c + di) = ac + bci + adi + bdi2 = (ac − bd)+(ad + bc)i,
and to divide complex numbers we note firstly that (c + di) (c − di) = c2 + d2 is real. So
a + bi
c + di = a + bi
c + di ×
c − di
c − di =
µac + bd
c2 + d2
¶
+
µbc − ad
c2 + d2
¶
i.
The number c−di which we just used, as relating to c+di, has a spec
Answer:
D. Continue cutting smaller and smaller slices.
Step-by-step explanation:
While, in theory, this could be a step, it would be a pointless one. You would have to eventually rearrange the pieces, but now there would be more of them.
Hope this helps!
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
