Answer:
a. We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution.
b. (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
Step-by-step explanation:
a. Why do we need imaginary numbers?
We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution. For example, the equation of the form x² + 2x + 1 = 0 has the solution (x - 1)(x + 1) = 0 , x = 1 twice. The equation x² + 1 = 0 has the solution x² = -1 ⇒ x = √-1. Since we cannot find the square-root of a negative number, the identity i = √-1 was developed to be the solution to the problem of solving quadratic equations which have the square-root of a negative number.
b. Expand (a + ib)⁵
(a + ib)⁵ = (a + ib)(a + ib)⁴ = (a + ib)(a + ib)²(a + ib)²
(a + ib)² = (a + ib)(a + ib) = a² + 2iab + (ib)² = a² + 2iab - b²
(a + ib)²(a + ib)² = (a² + 2iab - b²)(a² + 2iab - b²)
= a⁴ + 2ia³b - a²b² + 2ia³b + (2iab)² - 2iab³ - a²b² - 2iab³ + b⁴
= a⁴ + 2ia³b - a²b² + 2ia³b - 4a²b² - 2iab³ - a²b² - 2iab³ + b⁴
collecting like terms, we have
= a⁴ + 2ia³b + 2ia³b - a²b² - 4a²b² - a²b² - 2iab³ - 2iab³ + b⁴
= a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴
(a + ib)(a + ib)⁴ = (a + ib)(a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴)
= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b + 4i²a³b² - 6ia²b³ - 4i²ab⁴ + ib⁵
= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b - 4a³b² - 6ia²b³ + 4ab⁴ + ib⁵
collecting like terms, we have
= a⁵ + 4ia⁴b + ia⁴b - 6a³b² - 4a³b² - 4ia²b³ - 6ia²b³ + ab⁴ + 4ab⁴ + ib⁵
= a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
So, (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
Answer:
C. I can’t really enter square roots
Step-by-step explanation:
a^2/3 = a^2*a^1/3
Which is the cube root of a squared
1.
"The spending limit on John’s credit card is given by the function f(x)=15,000+1.5x"
means that if the monthly income of John is $ 5,000 ,he can spend at most
f(5,000)=15,000+1.5*5,000=15,000+ 7,500=22, 500 (dollars)
Or for example
if Johns monthly income is $8,000, then he can spend at most
f(8,000)=15,000+1.5*8,000=15,000+ 12,000=27,000 (dollars)
2.
Now, assume that the maximum amount that John can spend is y.
Then, y=15,000+1.5x
we can express x, the monthly income, in terms of y by isolating x:
y=15,000+1.5x
1.5x = y-15,000
X=y-15,000/1.5
thus, in functional notation, x, the monthly income, is a function , say g, of variable y, the max amount:
X=g(y) y-15000/1.5
since we generally use the letter x for the variable of a function, we write g again as:
G (x) x-15000/1.5
tells us that if the maximum amount that John can spend is 50,000 $, then his monthly income is 23,333 $.
3.
If John's limit is $60,000, his monthly income is
G(600,000)=60,000-15,000/ 1.5=45,000/1.5 =30,000
dollars.
Answer: $ 30,000
Remark: g is called the inverse function of f, since it undoes what f does.
instead of g(x), we could use the notation
times the length of one of the legs. In this case 14. So the hypotenuse of the lower triangle is also the short side of the upper triangle. Using Pythagoras
