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2 years ago

(98+3)+2 simplified?

Mathematics

1 answer:

marin [14]2 years ago

7 0

Answer:

103

Step-by-step explanation:

98+3=102+2=103

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Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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2 years ago

Simplify the following expression:

Keith_Richards [23]

Answer:

Step-by-step explanation:

-2(2)-1+1(1)-1(-1)

Do multiplication

-4-1+1+1

Now add/subtract

-5+2

-3

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3 years ago

Work not needed answer fast

eduard

Answer:

PLSS YOUR IN SCHOOL RN

Step-by-step explanation:

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2 years ago

A man is now twice as old as his son. Fifteen years ago he was three times as old as his son was then. How old is the son now?

marysya [2.9K]

Answer:

20 is the answer to that question

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2 years ago

What is the solution to 3x + 30+ I= 10 + 2 + 5x + 2? O 6 018 9/2

noname [10]

Answer:

x=8.5

Step-by-step explanation:

3x+31=14+5x

17=2x

x=8.5

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3 years ago