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
<u>We are given:</u>
An even number 'n', multiplied by the next consecutive even number is 168
<u>Solving for n:</u>
From the given statement, we can say that:
n(n+2) = 168 [<em>n multiplied by the next even number 'n+2'</em>]
n² + 2n = 168
n² + 2n - 168 = 0 [<em>subtracting 168 from both sides</em>]
We can see that we now have a quadratic equation, solving using splitting the middle term
n² + 14n - 12n - 168 = 0
n(n + 14) -12(n + 14) = 0 <em>[factoring out common terms</em>]
(n-12)(n+14) = 0
Here, we can divide both sides by either (n-12) OR (n+14)
Checking the result in both the cases:
(n + 14) = 0/(n-12) (n-12) = 0/(n+14)
n + 14 = 0 n - 12 = 0
n = -14 n = 12
Both these values are even and since we are not told if the number 'n' is positive or negative, both 12 and -14 are the possible values of n
Answer:
first graph shows the function and it's inverse
Subtract the second equation from the first equation.
x = -1
Plug -1 back in to either equation and solve for y.
3(-1) + y =-10
-3 + y = -10
Add 3 to both sides
y = -7
CHECK
3(-1) -7 = -10
2(-1) -7 = -9
(-1, -7)
