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Murrr4er [49]
3 years ago
13

I need help please!!

Mathematics
2 answers:
mina [271]3 years ago
5 0

Step-by-step explanation:

11

\frac{ - 1}{5}  \times  \frac{4}{7}

\:  \:  \:  \:  \:  \frac{ - 4}{35}

12

\frac{ - 1}{2}  \times  \frac{4}{5}

\frac{ - 2}{5}

13

\frac{ - 3}{2}  \times  \frac{7}{ - 10}

\frac{21}{20}

14

\frac{1}{2}  \times  \frac{7}{8}

\frac{7}{16}

15

\frac{ - 9}{5}  \times   \frac{1}{2}

\frac{ - 9}{10}

16

\frac{ - 32}{9}  \times  \frac{1}{3}

\frac{ - 32}{27}

17

\frac{ - 2}{1}   \div \frac{ - 19}{5}

\:  \:  \:  \:  \:  \: \frac{ - 2}{1}  \times  \frac{5}{ - 19}

\:  \:  \:  \:  \frac{ - 10}{19}

18

\frac{1}{9}  \div  \frac{ - 4}{3}

\frac{1}{9}  \times  \frac{3}{ - 4}

\frac{ - 1}{12}

19

\frac{13}{7}  \div  \frac{23}{4}

\frac{13}{7}  \times  \frac{4}{23}

\frac{52}{161}

20

\frac{ - 37}{10}  \div  \frac{9}{4}

\frac{ - 37}{10}  \times  \frac{4}{9}

\frac{ - 74}{45}

ella [17]3 years ago
5 0

Answer:

1l

Step-by-step explanation:

11.

-1/5÷7/4

Dividing a negative (-) and a positive (+) equals a negative (-)

- \frac{ 1}{5}  \div  \frac{7}{4}

To divide by fraction Multiply ✖ by the reciprocal of that fraction

-  \frac{1}{5}  \times  \frac{4}{7}

Multiply the fractions

-  \frac{4}{35}

Alternate form;

0.1142857

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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
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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

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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

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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.

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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

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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

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Step-by-step explanation:

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