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IceJOKER [234]
3 years ago
14

Can a -150/10 be classified as both a rational number and a integer

Mathematics
1 answer:
irakobra [83]3 years ago
6 0
Yes! This is because -150/10 can be simplified to be -15, which is a rational number.

The word “rational” sounds like another math word you’ve heard of before. Do you know what it is?

Well, it’s “ratio”!! Ratios can be seen in the forms x:y and x/y.


ANY RATIONAL NUMBER HAS THE ABILITY TO BE WRITTEN AS A RATIO!! This will completely exclude numbers with super long decimal points (ex: 1.2345678809928374737272828...)

This number also meets the requirements of being an integer. An integer is any whole number (this excludes decimals and fractions)

I know it’s written as a fraction. However, the fraction could be simplified, making it -15, which means this is both a rational number and an integer!!
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Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t
Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1.f(x)=\frac{x^2+x-20}{x^2+4}

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0
3 years ago
3/7 - 10/7 Write fractions in simplest form
8_murik_8 [283]
<span>3/7 - 10/7
= - 7/7
= -1

answer
-1</span>
3 0
3 years ago
Big ideas math 15.3 question 14
KiRa [710]
Do you have a picture with the question?
6 0
3 years ago
Which fractions are equivalent to 1/8+5/8? Check all that apply.
dolphi86 [110]
The equivalent fractions are 3/4 & 12/16 hope this helps!!
7 0
3 years ago
Read 2 more answers
2. You are given the hyperbolic relation modeled by xy = -2. Do the following: a) Rewrite the relation such that the dependent v
N76 [4]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The given equation is :

  • xy =  - 2

1. The relationship such that dependent variable (y) is isolated is :

  • y =   \dfrac{ - 2}{x}

2. The table accompanying this equation :

  • x = -4 and y = 0.5

  • x = -1 and y = 2

  • x = - 0.5 and y = 4

  • x = 0 and y = undefined

  • x = 0.5 and y = -4

  • x = 1 and y = -2

  • x = 4 and y = -0.5

3. graph of the given equation is in attachment ~

4 0
3 years ago
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