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borishaifa [10]
2 years ago
14

What are the zero(s) of the function f(x) = the quantity of 5 x squared minus 25 x, all over x ? (2 points)

Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
8 0

Answer:

Five

Step-by-step explanation:

5x square minus 25x=0

Divide by x gives 5x - 25=0

X=5

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why do we need imaginary numbers?explain how can we expand (a+ib)^5. finally provide the expanded solution of (a+ib)^5.(write a
zheka24 [161]

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

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3 years ago
Find an equation for the nth term of the arithmetic sequence.<br> 8,6,4,2,…
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Answer:

1st term= 8 2nd term=6 3rd term=4 4th term=2 5th term=0 6th term=-2 7th term=-4 8th term=-6 9th term=-8 10th term=-10

Step-by-step explanation:

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