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olga_2 [115]
2 years ago
11

Please helppppppppppp​

Mathematics
2 answers:
Eduardwww [97]2 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Substitute the given values of x into f : x → \frac{3}{7} x + 4

x = - 14 → (\frac{3}{7} × - 14) + 4 = (3 × - 2) + 4 = - 6 + 4 = - 2

x = 28 → ( \frac{3}{7} × 28) + 4 = (3 × 4) + 4 = 12 + 4 = 16

x = \frac{7}{8} → (\frac{3}{7} × \frac{7}{8} ) + 4 = \frac{3}{8} + 4 = \frac{3}{8} + \frac{32}{8} = \frac{35}{8}

x = - \frac{2}{9} → (\frac{3}{7} × - \frac{2}{9} ) + 4 = (\frac{1}{7} × - \frac{2}{3}) + 4 = - \frac{2}{21} + \frac{84}{21} = \frac{82}{21}

Hatshy [7]2 years ago
6 0

Answer:

See below in bold.

Step-by-step explanation:

1. 3/7 * -14 + 4

   = -6 + 4

    = -2.

2. 3/7 * 28 + 4

   = 12 + 4

    = 16.

3 3/7 * 7/8 + 4

   =  3/8 + 4

    =   35/8 or 4 3/8.

4. 3/7 * -2/9 + 4

   = -6/63 + 4

   = -2/21 + 4

    = -82/21 or -3 19/21.

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Dan has the same number of nickels and dimes in his piggy bank. If the total amount of those coins is $0.90, how many of each co
Karolina [17]

Hello There!

<u><em>n - d = 0</em></u>

<u><em>5n+10d = 90</em></u>

<u><em>----------------------</em></u>

<u><em>n-d = 0</em></u>

<u><em>n+2d = 18</em></u>

<u><em>-------------------</em></u>

<u><em>Subtract and solve for "d":</em></u>

<u><em>3d = 18</em></u>

<u><em>d = 6 (# of dimes)</em></u>

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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

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x^2+x+29=0

It cannot be factorize .

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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

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