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3 years ago

Can someone tell me whats 7/3 x 2/7?

Mathematics

2 answers:

Trava [24]3 years ago

6 0

Answer:

2/3

Step-by-step explanation:

You can cross simplify so you can cancel out the 7s because they are opposite of one another.

Romashka-Z-Leto [24]3 years ago

4 0

Answer: 8 1/6

Hope this helps!

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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

Is the given number a solution of an equation or inequality 3x-4=10;5

kotykmax [81]

The given number is not a solution of the equation i.e. the number is an inequality

<h3>How to determine the solution?</h3>

The equation is given as:

3x- 4 = 10

The number is given as:

This means that

x = 5

Substitute x = 5 in 3x- 4 = 10

3 * 5- 4 = 10

Evaluate the product

15- 4 = 10

Evaluate the difference

11 = 10

The above equation is not true i.e. the equation is actually an inequality

Hence, the given number is not a solution of the equation i.e. the number is an inequality