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2 years ago

Write the additive inverse of rational number -11/5

Mathematics

2 answers:

sertanlavr [38]2 years ago

6 0

Answer:

If " a " represent any rational number then " a + ( - a ) = 0" or " ( - a ) + a = 0 ". Thus " a " and " - a " are additive inverse of each other. Hence " 11 / 5 " is the additive inverse of " - 11 / 5 " .

Step-by-step explanation:

Makovka662 [10]2 years ago

4 0

Answer:

Additive inverse of -11/5 => + 11/5

hope that helps uh..☺

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Suppose that an individual has a body fat percentage of 19.7% and weighs 157 pounds. How many pounds of his weight is made up of

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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

Solve for a 2.5+5a=3a+10.5

Oksi-84 [34.3K]

Answer:

a = 4

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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lianna [129]

Your interest formula is given to you.

Interest in a year = principal (the amount invested) * rate (the interest rate) * period (the time you're measuring) 

Interest = 55,000 * 2% * 1 year = 55,000 * 0.02 * 1 = $1,100 

How much would you need to have made for your spending power to keep with inflation? Your interest rate would have needed to match the inflation rate, otherwise, prices are going up faster than you're saving. 

Required interest = 55,000 * 3.24% * 1 year = 55,000 * 0.0324 * 1 = $1,782 

How much buying power did you lose? The difference between your required interest and your actual interest. 
Buying power lost = 1,782 - 1,100 = $682. You lost this much in buying power.

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3 years ago

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The mean rent of a 3-bedroom apartment in Orlando is $1300. You randomly select 10 apartments around town. The rents are normall

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Answer:

96.49% probability that the mean rent is more than $1100

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .