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2 years ago

-3 2/7 as an improper fraction

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

5 0

Answer:

30/7

Step-by-step explanation:

There are

in 1

Therefore in 4 there are

Now

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2 years ago

Ned counts by fives . he starts at 80 . which number should he say next ?

hodyreva [135]

The next answer is 85.

Counting by five is the same as adding 5 to the number he starts with.

Therefore, 80 + 5 = 85

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2 years ago

For f(x) = 5x + 1

drek231 [11]

Answer:

A. $f(7) = 5(7) + 1\\\\f(7) = 36$

B. $f^{-1}(x) = \frac{x - 1}{5}$

C. $f^{-1}(7) =\frac{6}{5}$

D. $f(\frac{6}{5}) = 7$

Step-by-step explanation:

A. To solve the first part of the problem we must replace $x = 7$ in the function $f(x) = 5x + 1$

So:

$f(7) = 5(7) + 1\\\\f(7) = 36$

B. In part B we must find the inverse function of $f(x) = 5x + 1$

To find the inverse function do $y = f(x)$

$y = 5x +1$

Now clear the variable x.

$\frac{y - 1}{5} = x$

Replace x with y.

$y = \frac{x - 1}{5}$

Finally

$f^{-1}(x) = \frac{x - 1}{5}$

C. Now we take the inverse function found above and replace $x = 7$

$f^{-1}(7) = \frac{7 - 1}{5}\\\\f(7) = \frac{6}{5}$

D. Now we substitute $x = f(f^{-1}(7))$ in the original function.

$x = f( f^{-1}(7))\\\\f^{-1}(7) = \frac{6}{5}\\\\ x= f(\frac{6}{5} )\\\\f(\frac{6}{5}) = 5(\frac{6}{5}) + 1\\\\f(\frac{6}{5}) = 7$

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3 years ago

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Step-by-step explanation:

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3 years ago

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x plus two, over three, solve for f(g(x)) when x = 1

MaRussiya [10]

Easy
f(g(1))
evaluate g(1) then plug thatin for x in f(x)
g(1)=(x+2)/3
g(1)=(1+2)/3
g(1)=1

f(g(1))=
f(1)=(1)^2+3(1)+6
f(1)=1+3+6
f(1)=10
f(g(1))=10

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2 years ago

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