All categories

3 years ago

Please write the answer

Mathematics

1 answer:

Travka [436]3 years ago

6 0

$\sf \dfrac{3^x - 5 \times 3^{(x-2)}}{3^{(x-3)}} \\ \\ \longrightarrow \sf \dfrac{ {3}^{x} }{ {3}^{(x - 3)}} - \frac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{[x - (x - 3)]} - \dfrac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ {3}^{(x - 3)} } \times \dfrac{ {3}^{x}}{9} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ \bigg(\dfrac{ {3}^{x} }{ {3}^{3} }\bigg) } \times \dfrac{ {3}^{x}}{9}\\ \\ \longrightarrow \sf {3}^{3} - \dfrac{ ({3}^{3})( 5)}{ {3}^{x} } \times \dfrac{ {3}^{x}}{9}\\ \\ \sf \longrightarrow {3}^{3} - (5 \times 3) \\ \\ \longrightarrow \sf \: 27 - 15 \\ \\ \longrightarrow \leadsto{\underline{\boxed{\sf{ \pink{ 12}}}}}$

You might be interested in

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

8 0

3 years ago

Select the correct answer from each drop-down menu.The center of the circle is at O. Segment AB is tangent to circle O at point

IRISSAK [1]

Based on the tangent theorem, m∡ADO = 90°.

Based on the angles of intersecting chords theorem, m∡CGF = 1/2(mCF + mDE).

What is the Angles of Intersecting Chords Theorem?

The angles of intersecting chords theorem states that when two chords in a circle intersect, the angle formed inside the circle at the point of intersection has a measure that is half of the sum of the measures of the intercepted arcs that are formed by the angle and its vertical angle.

<h3>What is the Tangent Theorem?</h3>

According to the tangent theorem, if a line is tangent to a circle, the segment forms a right angle at the point of tangency with the radius of the circle.

Since line AB is tangent to circle O at point D, therefore, based on the tangent theorem:

The measures of angle ADO = 90°

Chord CE and DF intersect in the circle, therefore, based on the angles of intersecting chords theorem:

The measure of CGF = 1/2(mCF + mDE).

Learn more about the tangent theorem on:

brainly.com/question/9892082

#SPJ1

4 0

1 year ago

Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union

quester [9]

Answer:

Correct option is d

Step-by-step explanation:

The objective is to estimate the mean number of hours per month a union member is absent from work and from the question the sample size is large so in order to be accurate in this calculation we require A large sample confidence interval for μ.

6 0

4 years ago

Why is a linear function not be adequate for describing the supply and demand fuctions?

taurus [48]

Supply and demand functios often describe a curve rather than a straight line.

7 0

4 years ago

Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobi

quester [9]

Answer:

450 km

Step-by-step explanation:

Equations

We can define 3 variables: a, b, d. Let "a" and "b" represent the speeds of the cars leaving cities A and B, respectively. Let "d" represent the distance between the two cities. We can write three equations in these three variables:

1. The relation between "a" and "b":

a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B

2. The relation between speed and distance when the cars leave at the same time:

d = (a +b)·5 . . . . . . distance = speed × time

3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is ...

distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance

4.5a + (150/b)(a +b) = d

Solution

Substituting for d, we have ...

4.5a + 150/b(a +b) = 5(a +b)

4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses

5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side

Now, we can substitute for "a" and solve for b.

5b^2 + 0.5b(b-10) -150(b -10 +b) = 0

5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses

11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2

(11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor

The solutions to this equation are ...

b = 60/11 = 5 5/11 . . . and . . . b = 50

Since b must be greater than 10, the first solution is extraneous, and the values of the variables are ...

b = 50
a = b-10 = 40
d = 5(a+b) = 5(90) = 450

The distance between A and B is 450 km.

_____

Check

When the cars leave at the same time, their speed of closure is the sum of their speeds. They will cover 450 km in ...

(450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h

When car A leaves 4 1/2 hours early, it covers a distance of ...

(4.5 h)(40 km/h) = 180 km

before car B leaves. The distance remaining to be covered is ...

450 km - 180 km = 270 km

When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in ...

(270 km)/(90 km/h) = 3 h

In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.

5 0

4 years ago

Read 2 more answers

Please write the answer​

Please write the answer