Prove whether or not f(x)= 5x - 4 and g(x)= x+4/5 are inverses using composition of functions (PLEASE HELP)
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f(x)= 5x - 4 and g(x)= x+4/5 are inverses by composition of functions
To prove that two functions, f(x) = 5x - 4 and g(x) = (x + 4)/5, are inverses of each other, we need to show that their composition yields the identity function.
First, let's find the composition f(g(x)):
f(g(x)) = f((x + 4)/5)
= 5((x + 4)/5) - 4
= (x + 4) - 4
= x
Now, let's find the composition g(f(x)):
g(f(x)) = g(5x - 4)
= ((5x - 4) + 4)/5
= 5x/5
= x
Since both f(g(x)) and g(f(x)) simplify to x, we can conclude that f(x) = 5x - 4 and g(x) = (x + 4)/5 are indeed inverses of each other based on the composition of functions.
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Step-by-step explanation:
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Step-by-step explanation:
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Answers
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Step-by-step explanation:
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Answers
Answer:
136
Step-by-step explanation:
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a37= 460-9(36)
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[tex] \\ [/tex]
The first three terms of a sequence are 460, 451, 442[tex] \\ [/tex]
To Find :[tex] \\ [/tex]
37th term of sequence = ?[tex] \\ [/tex]
Solution :[tex] \\ [/tex]
Clearly these three terms are making an arithmetic sequence, with :
First term, a = 460 Common difference, d = 451 - 460 = - 9[tex] \\ [/tex]
Now, we have to find 37th term of sequence = [tex] \tt a_{37}[/tex]
[tex] \\ [/tex]
Now, we know that :
[tex] \large \underline{\boxed{\bf{a_{n} = a + (n-1)d}}}[/tex]
[tex] \\ [/tex]
[tex] \tt : \implies a_{37} = a + (37-1)d[/tex]
[tex] \tt : \implies a_{37} = a + 36d[/tex]
[tex] \\ [/tex]
Now, by substituting values of a and d :
[tex] \tt : \implies a_{37} = 460 + 36 \times (-9)[/tex]
[tex] \tt : \implies a_{37} = 460 - 324[/tex]
[tex] \tt : \implies a_{37} = 136[/tex]
[tex] \\ [/tex]
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Step-by-step explanation:
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Step-by-step explanation:
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The decimal number is 15.265.
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The decimals in the form of [tex]\dfrac{p}{q}[/tex] and p,q are co-primes.
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Have a great day/night
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Hey there!
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Lucas and Mitchell were asked to add the numbers 3.54 and 12.09.
Lucas stated that he completed the addition problem by using the following method:
First, I broke the numbers apart. I broke 3.54 into 3.00+0.50+0.04 and 12.09 into 12.00+0.09.
Then, I combined 0.04 and 0.09 to get 0.13.
Next, I combined 0.13 and 0.50 to get 0.63.
Then, I combined 3.00+12.00 to get 15.00.
Finally, I added the decimal to get 15.63.
Mitchell stated that he completed the addition problem by using the following method:
First, I added the whole numbers to get 3.00+12.00=15.00.
Next, I added the numbers in the tenths place, 0.50+0.00=0.50.
Then, I added the numbers in the hundredths place, 0.04+0.09=0.013.
I then added 0.50 and 0.013 to get 0.513.
Lastly, I added the decimal to the whole number to get 15.513.
Determine if Lucas and/or Mitchell had a flaw in their method. If they did, correct the flaw(s) so that both the methods used are correct. PLZ ANSWER THE QUESTION!
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