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blagie [28]
2 years ago
6

Write the prime factorization of 36. Use Exponents when appropriate and order the factors from least to greatest​

Mathematics
1 answer:
Setler [38]2 years ago
7 0

Answer:

2 x 2 x 3 x 3

Step-by-step explanation:

36

6 x 6

3 x 2 x 3 x 2

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Suppose that the die is rolled twice, rather than once. find the probability of rolling an even number on at least one die. find
puteri [66]
We know P = 1/2

X~B(2, 0.5)
n = 1

2C1 from pascal triangle = 2

P(X=0.5)= 2C1 x 0.5 x 0.5

= 0.5
= 50%
7 0
3 years ago
What is a complete graph.
Kay [80]

Answer:

A complete graph is 100 or over

Step-by-step explanation:

hope this helps G

3 0
2 years ago
Tavon has a gift card for $165 that loses $4 for each 30-day period it is not used. He has another gift card for $145 that loses
sweet [91]

Answer:

Equation 1 = Equation 2

165 - 4x = 145 - 3.50x

Step-by-step explanation:

Let x represent the number of 30-day periods.

Tavon has a gift card for $165 that loses $4 for each 30-day period it is not used.

Therefore the equation =

$165 -$4 × x

= 165 - 4x.......... Equation 1

He has another gift card for $145 that loses $3.50 for each 30-day period it is not used.

$145 -$3.50 × x

= 145 - 3.50x........... Equation 2

Hence, an equation for the number of 30-day periods until the value of the gift cards will be equal is obtained by equating Equation 1 and Equation 2 together

So, we have

Equation 1 = Equation 2

165 - 4x = 145 - 3.50x

We simplify further:

165 - 145 = -3.50 + 4.0x

20 = 0.5x

x = 20/0.5

x = 40

Therefore, number of each 30-day periods until the value of the gift cards will be equal is 40

3 0
3 years ago
Kayla won 37 bouncy ball playing hoops at her schools game night then she gave two to each of her friends she only has 7 left ho
Pavel [41]

Answer:

Kayla has 15 friends

Step-by-step explanation:

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

5 0
3 years ago
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