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adoni [48]
3 years ago
6

Evaluate f(x) = sqrt(5x + 4) for x = 12

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer is 8

Work Shown:

f(x) = sqrt(5*x + 4)

f(12) = sqrt(5*12 + 4) ... replace every x with 12

f(12) = sqrt(60 + 4)

f(12) = sqrt( 64 )

f(12) = 8

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Morgarella [4.7K]

Answer:

  A)  The functions are not inverses of each other.

Step-by-step explanation:

f(g(x))=\sqrt{(x^2+3)-3}=\sqrt{x^2}=|x|\ne x

The result of f(g(x)) is not always x, so the functions are not inverses of each other.

In general, a quadratic (or any even-degree polynomial) such as g(x) cannot have an inverse function because it does not pass the horizontal line test.

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3 years ago
5/9 × 8/11 what is the answer
nikitadnepr [17]

Answer:

\frac{5}{9}  \times  \frac{8}{11}  =  \frac{5 \times 8}{9 \times 11}  =  \frac{40}{99}  = 0.4040

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Hannah loved mini-Snickers bars. She enjoyed making treats with them. She used 1/2 of her supply to make a blizzard. She froze 1
tensa zangetsu [6.8K]
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3 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)

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lora16 [44]

Answer

~~~~~~~~~~~~

slope: 1

y intercept: 4

equation: y=x+4

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