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3 years ago

Which one represents the equation log3 81 = 4 in exponential form?

Mathematics

2 answers:

MariettaO [177]3 years ago

6 0

Log(base 3)(81)=4
3^4=81

Tresset [83]3 years ago

5 0

Answer:

Exponential form is $log_{3}(3^{4})$ = 4.

Step-by-step explanation:

Given : $log_{3}(81)$ =4.

To find : Represent in exponential form .

Solution : We have given that $log_{3}(8)$ =4.

We can write as 81 = $3^{4}$ .

By the logarithm form $log_{a}(a^{n})$ =n.

Then ,

$log_{3}(3^{4})$ = 4

Therefore, Exponential form is $log_{3}(3^{4})$ = 4.

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Write the equation of a line perpendicular to y=3x+1and goes through the point ( 6,2) y=−13x+4 y=−13x−4 y=3x+4 y=3x−4

Korolek [52]

Answer:

$y=-\frac{1}{3}x+4$

Step-by-step explanation:

step 1

Find the slope of the perpendicular line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal

(the product of their slopes is equal to -1)

In this problem

we have

$y=3x+1$

The equation of the given line is $m=3$

the slope of the perpendicular line to the given line is

$m=-\frac{1}{3}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{3}$

$(x1,y1)=(6,2)$

substitute

$y-2=-\frac{1}{3}(x-6)$

Convert to slope intercept form

$y=mx+b$

Distribute right side

$y-2=-\frac{1}{3}x+2$

$y=-\frac{1}{3}x+2+2$

$y=-\frac{1}{3}x+4$

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3 years ago

What is the solution to the equation 9(x − 1) = 27? (1 point) x = 2.5 x = 0.5 x = −0.5 x = −3.5

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Step-by-step explanation:

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Solve -9(t - 2) = 4(1-15).

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Answer:

t = 74 / 9

Step-by-step explanation:

-9(t - 2) = 4(1 - 15)

-9t + 18 = 4 - 60

-9t + 18 = -56

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-9t = -74

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t = 74 / 9

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Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and

Olenka [21]

Answer:

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

$P(A \cap B) = P(A)*P(B)$

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)$

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

$P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)$

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}$

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

$P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}$

Probability that the series lasts exactly four games:

$p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})$

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

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3 years ago