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son4ous [18]
3 years ago
9

While on summer vacation, Mr. Duda traveled about 230 miles to a city near Dallas in 3 hours. How many miles

Mathematics
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

76 2/3 (Rounded = 77)

Step-by-step explanation:

Step 1: Divide 230 by 3.

Step 2: Get your answer.

Answer - 76.66666666...

Hope this helps!

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Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]
Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

  • f(x) = \frac{3x-2}{x+1}
  • g(x)=x+5
  • f[ -4 ] = ?
  • g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

g(\frac{3x-2}{x+1}) =  (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) =  \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) =  13

Therefore, the value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

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2 years ago
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Natali5045456 [20]

Answer:

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Step-by-step explanation:

Answer is C

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For which value of a does Limit of g(x) as x approaches alpha not exist?​
Hunter-Best [27]

The value of a where the Limit of g(x) as x approaches alpha not exist are -1 and 1

<h3>Limit of a function</h3>

The limit of a function is the limit of a function as x tends to a value.

From the given graph, you can see that the function g(x) goes large at the point where the arrows orange and purple point down from the x-coordinates -1 and 1.

Hence the value of a where the Limit of g(x) as x approaches alpha not exist are -1 and 1

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f(h(x))= 2x -21

Step-by-step explanation:

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

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