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3 years ago

How are greater than and greater than or equal to similar? How are they different

1 answer:

5 0

Greater than means it can only be greater than the value give for example x is greater than 10, x can only be a number greater than 10. Greater than or equal to is similar because the answer can be greater than the value given or equal to it. For example x is greater than 10, x could be equal to 10 or any value greater than it.

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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

3х2 = 147<br> Solve by undoing

Dmitry_Shevchenko [17]

Answer: +/- 7

Step-by-step explanation:

3x² = 147

To solve for x divide through by three first.

3x² = 147

x² = 49, now we take the square root of both side by trying to apply laws of indices.

√x² = √49

The square root will neutralize the effect of the square because

√a = a¹/² so (x²)¹/², and (x²)¹/² =

x²×¹/² = x, therefore the solution is

x = +/- 7.

3 0

3 years ago

Can someone help me with math? I DO NEED help with questions 13, 11, 9, 8, 7, 5, 3. PLEASE HELP ME!!!!! I need these answered by

jek_recluse [69]

Absolute value is just how many integers rhat number is away from 0. i say this because it’s not too hard of a thing to learn and it’s very important in the higher grades.

3 is 2 6/7
5 is a greater than symbol >
7 is 1/3 and negative 1/3 so -1/3
8 is 14 and -14
9 is 21 and -21
11 is | -75 |
i’m not sure about 13 but i believe it could be | -8 |
i hope this helped

8 0

2 years ago

Hey can you please help? I am struggling a lot don’t know how to solve this

Ket [755]

Answer:

64.8

Step-by-step explanation:

Pythagoreon Theroem

a^2 + b^2 = c^2 in this situation it would actually be the opposite, seeing where the "x" is (on a leg) c^2 - b^2 = a^2. First take the square of 67 and 17

67 squared is 4489 and 17 squared is 289 then subtract both results

4489-289 = 4200 take the square root of this number

64.8074 (Round to the tenths 64.8)

8 0

2 years ago

A recycling bin is in the shape of a right rectangular prism. The bin is 12 meters long, 512 meters wide, and 612 meters tall. W

Colt1911 [192]

The volume of a rectangular prism can be found using the formula V = WHL.

W is the width, H is the height, and L is the length.

V = 512×612×12
V = 3,760,128

The volume of the bin is 3,760,128

3 0

3 years ago

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