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2 years ago

A sequence is defined by the following: an = 4n – 1 What is a1?

Mathematics

2 answers:

Alexandra [31]2 years ago

6 0

an = 4n – 1

a1 means n=1

a1 = 4(1) -1

a1 = 4-1

a1 =3

Answer: 3

grigory [225]2 years ago

4 0

Answer:

Explain Your Answer:

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2 years ago

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

A bag contains 2 black balls, 4 yellow balls and 4 white balls. Event A is defined as drawing a white ball on the first draw and

arsen [322]

Answer: $P(B|A)=\dfrac{4}{9}$

Step-by-step explanation:

Given: A bag contains 2 black balls, 4 yellow balls and 4 white balls.

Event A is defined as drawing a white ball on the first draw.

Event B is defined as drawing a black ball on the second draw.

P(B|A) is expressed as the conditional probability of occurring B given that A.

i.e. it is the probability of happening B where A is already happended.

If a white ball is drawn at first, then the number of black ball(4) remains unchanged but the total number of balls (10) will become 9.

$Probability=\dfrac{Number\ of \ favorable \ outcomes}{Total\ outcomes}$

Then, $P(B|A)=\dfrac{4}{9}$

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3 years ago

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