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2 years ago

List the potential solutions to 2 ln x = 4 ln 2 from least to greatest.

Mathematics

2 answers:

Contact [7]2 years ago

6 0

X = -4 and X = 4
then the answer to the solution is 4

svp [43]2 years ago

3 0

We have to find the potential solutions to $2 ln x = 4 ln 2$ from least to greatest.

Using the properties of ln function.

$a \times ln b = ln b^{a}$

Therefore, we get

$ln x^{2} = ln 2^{4}$

$ln x^{2} = ln 16$

taking antilog on both the sides, we get

$x^{2} = 16$

So, $x = \pm 4$

Therefore, the potential solutions to 2 ln x = 4 ln 2 from least to greatest is -4 and 4.

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Jennifer has a square bedroom. The area of her room is 144 ft2. What is the length of one side of Jennifer’s room?

Jobisdone [24]

Answer:

Option B. $12\ ft$

Step-by-step explanation:

we know that

The area of a square is equal to

$A=x^{2}$

where

x is the length side of the square

In this problem we have

$A=144\ ft^{2}$

substitute and solve for x

$144=x^{2}$

suqre root both sides

$x=12\ ft$

7 0

2 years ago

The volleyball team at West View High School is comparing T-shirt companies where they can purchase their practice shirts. The t

mina [271]

10.5x = 7.5x + 30
10.5x - 7.5x = 30
3x = 30
x = 30/3
x = 10

10.5(10) = 105
7.5(10) + 30 = 75 + 30 = 105

they need to purchase 10 t-shirts from each company for a total of $105 for each company

5 0

2 years ago

Read 2 more answers

A pencil has density of .875 g/ml and a mass of 3.5 grams what is the volume?

SSSSS [86.1K]

Answer:

4 ml

Step-by-step explanation:

m = D/ v where m is mass and D is density and V is volume

.875 = 3.5 / V

Multiply each side by V

.875 V = 3.5

Divide each side by .875

V = 3.5/.875

V =4

8 0

2 years ago

Read 2 more answers

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

2 years ago

Which of the following lists of three numbers could form the side lengths of a triangle? <img src="https://tex.z-dn.net/?f=A.%20

yan [13]

Answer:

B. 122, 257, 137

Step-by-step explanation:

The triangle inequality rule states that the sum of any two sides of a triangle must be greater than the third side. If A, B and C are sides of a triangle then A + B > C, A + C > B, B + C > A

Testing for the options:

1) 10, 20, 30

10 + 20 (= 30) is not greater than 30. It cannot form a triangle

2) 122, 257, 137

122 + 137 (259) is greater than 257. It can form a triangle

3) 8.6, 12.2, 2.7

8.6 + 2.7 (11.3) is not greater than 12.2. It cannot form a triangle

4) 1/2, 1/5, 1/6

1/5 + 1/6 (11//30) is not greater than 1/2. It cannot form a triangle

8 0

2 years ago