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bazaltina [42]
2 years ago
10

Pls help!!! math what the value?? --

Mathematics
2 answers:
bearhunter [10]2 years ago
4 0

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

\qquad \sf  \dashrightarrow \:  \sqrt{ - 25}

\qquad \sf  \dashrightarrow \:  \sqrt{  - 1 \times  25}

\qquad \sf  \dashrightarrow \:  \sqrt{   {i}^{2}  \times  25}

\qquad \sf  \dashrightarrow \: 5i

so, correct choice is D

elena-s [515]2 years ago
4 0
<h3>Given:</h3>

\sqrt{ - 25}

<h3>To find:</h3>

The equivalent value from the given options.

<h3>Solution:</h3>

\sqrt{ - 25}

=  \sqrt{ -  1\times 25}

=  \sqrt{ {i}^{2} \times 25 }

= 5i

<u>Therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>correct</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>option</u><u> </u><u>D</u>

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Answer:

The probability of the spinner landing on A is 0.1111.

Step-by-step explanation:

The possible outcomes of a board game is that it can lead on any three of the regions, A, B and C.

The probability of all the outcomes of an event sums up to be 1.

P(E) = P(E_{1})+P(E_{2})+...+P(E_{n})=1

According to the provided information:

P (B) = 3 P (A)

P (C) = 5 P (A)

Compute the probability of the spinner landing on A as follows:

P(A)+P(B)+P(C)=1\\P(A)+3P(A)+5P(A)=1\\9P(A)=1\\P(A)=\frac{1}{9}\\=0.111111\\\approx0.1111

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3 0
3 years ago
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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Step-by-step explanation:

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