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3 years ago

I need help....!!!!!

Mathematics

1 answer:

Vikki [24]3 years ago

8 0

C. Because the full length is 44. Now the equation is 6x-5=44. Add 5 to both sides. 6x=49. 49/6= 8.16666666666 so you round and get 8.17

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Can u help me I have 2 minutes to do this

Triss [41]

Answer:

0.2, 0.6,0.25,0.75,0.82

Step-by-step explanation:

so you u have to do it put least to greatest

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3 years ago

Read 2 more answers

Marie made the box plot below to display a set of data. Which set of data could the box plot represent?

Assoli18 [71]

Answer:

see below

Step-by-step explanation:

First of all, you want to find the data set the matches the extreme values of 5 and 35. That eliminates the 2nd and 4th choices.

Then you want to find the data set that has a median of 15. The first data set has a middle value (median) of 20, so that choice is eliminated.

The data set of the 3rd choice matches the box plot extremes, median, and quartile values.

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2 years ago

An animal park has lions,tiger an zebras.2/5 of the animals are lions and 3/10 of the animals are zebras. what percentage are ti

maksim [4K]

Answer:

3/10

Step-by-step explanation:

You have to make them the same fraction.

2/5 = 4/10

3/10 =3/10

3/10 + 4/10 = 7/10

10/10 - 10/10 =3/10

3/10 is the amount of tigers.

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1 year ago

What is the slope indicated in the table below? Х 10 20 40 Y 20 17 14 11 30

Leya [2.2K]

Answer:

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Step-by-step explanation:

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2 years ago

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