Irrational numbers are imaginary. true or false

Mathematics

2 answers:

maxonik [38]2 years ago

7 0

Answer: true true true

Vedmedyk [2.9K]2 years ago

3 0

Imaginary numbers are not rational which may lead you to believe that they must be irrational. Though logical, you would still be incorrect because “irrational” also applies only to real numbers.

Step-by-step explanation:

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Simplify: 3(4x – 2) + 9 + 2

kirill [66]

Answer:

-6x+5y+8

Step-by-step explanation: subtract 9xfrom 3x

Subtract 2 from 10

-6x+5y+10-2

-6x+5y+8

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3 years ago

According to the scores on the last math test, 80%, or 20, of the students in the class received an A. Find the number of studen

Sladkaya [172]

80% is the same as 4/5, so 4/5 of the class= 20. That means that if there was a picture, 4 parts would be shaded, and 1 would not. 1/4 of 20 is 5, so 1 part= 5 people. 5 x 5 = 25. 25 is the answer. :)

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2 years ago

Read 2 more answers

Let f and g be the functions defined by f(x) = 10^ (x+2 / 3) and g(x) = log (x3 / 100) for all positive real numbers,

Vaselesa [24]

Answer:

F(x) and g(x) are not inverse functions.

Step-by-step explanation:

In order to calculate the inverse function of a function, we have to isolate X and after that, we change the variables.

As our function f(x) is a exponentian function, we can apply logarithm with base 10 (log) in both sides in order to isolate X. Remember that log10=1.

$[tex]y=10^{(x+\frac{2}{3}) }\\\\log y=log 10^{(x+\frac{2}{3}) }\\log y = (x+\frac{2}{3}) . log10\\\frac{log Y}{log10} = (x+\frac{2}{3})\\\frac{log Y}{1} = (x+\frac{2}{3})\\log Y-\frac{2}{3}=x$ [/tex]