Which is equivalent to 3√8* ?

O x √8^3
O 8 ^3/x
O 8 ^x/3
O 8^3x

Answers

Answer 1

Answer:

O 8 ^x/3

Step-by-step explanation:

Remember that :

[tex]\sqrt[n]{x} = x^\frac{1}{n}[/tex]

………………………………

Then

[tex]\sqrt[n]{8^x} = (8^{x})^{\frac{1}{3}}[/tex]

      [tex]= 8^{x\times\frac{1}{3}[/tex]

      [tex]=8^{\frac{x}{3}[/tex]


Related Questions

How do you write an effective explanation to explain this problem?

Answers

The error happens in the last line.

The relevant log rule is ln(A) - ln(B) = ln(A/B)

The issue is that the 0 isn't ln(0), so that 0 cannot be thrown into the natural log like that.

What we can do is this set of steps

ln(3x) - 0

ln(3x) - ln(1)

ln(3x/1) ... applying the log rule mentioned

ln(3x)

But these steps are a needlessly overcomplicating things. We can simply go from ln(3x)-0 to ln(3x) in one step. Subtracting off zero doesn't change ln(3x) at all.

--------------------------

Let's continue the steps to solve for x.

ln(x^2) = ln(3x)

x^2 = 3x

x^2-3x = 0

x(x-3) = 0

x = 0 or x-3 = 0

x = 0 or x = 3

Those are the two possible solutions.

However, x = 0 isn't valid because it's not in the domain of ln(x^2). In other words, ln(0) isn't defined. The domain of y = ln(x) is x > 0

This means that x = 3 is the only solution.

On a coordinate plane, 2 parallelograms are shown. Parallelogram 1 has points (0, 2), (2, 6), (6, 4), and (4, 0). Parallelogram 2 has points (2, 0), (4, negative 6), (2, negative 8), and (0, negative 2). How do the areas of the parallelograms compare? The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2. The area of parallelogram 1 is 2 square units greater than the area of parallelogram 2. The area of parallelogram 1 is equal to the area of parallelogram 2. The area of parallelogram 1 is 2 square units less than the area of parallelogram 2

Answers

The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2

How to compare the areas?

The coordinates are given as:

Parallelogram 1: (0, 2), (2, 6), (6, 4), and (4, 0).

Parallelogram 2: (2, 0), (4, -6), (2, -8), and (0, -2)

Next, calculate the base and the height

For parallelogram 1, we have:

Base = (0, 2) and (2, 6)

Height = (6, 4), and (4, 0).

So, we have:

[tex]Base = \sqrt{(0 - 2)^2 + (2 -6)^2} = \sqrt{20}[/tex]

[tex]Height = \sqrt{(6 - 4)^2 + (4 -0)^2} = \sqrt{20}[/tex]

The area is:

Area = Base * Height

Area = √20 * √20

Area = 20

For parallelogram 2, we have:

Base = (0, -2) and (4, -6)

Height = (4, -6) and (2, -8)

So, we have:

[tex]Base = \sqrt{(0 -4)^2 + (-2 +6)^2} = \sqrt{32}[/tex]

[tex]Height = \sqrt{(4 - 2)^2 + (-6 +8)^2} = \sqrt{8}[/tex]

The area is:

Area = Base * Height

Area = √32 * √8

Area = 16

By comparing both areas, we have:

20 is greater than 16 by 4

This means that the area of parallelogram 1 is greater by 4 units

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The enrollment for a new program is 1,500 students. the program is expected to grow at a rate of 0.03. what is the expected enrollment in 7 years? round to the nearest whole number.

Answers

The expected enrolment in 7 years is  1845.

What is the expected enrolment?

The formula for calculating future value of the number of students is an exponential equation with this form:

FV = P (1 + r) ^n

FV = Future populationP = Present populationR = rate of growth of the populationN = number of years

1500 x (1.03)^7 = 1845

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ap:3, 2, 1....
what is its coomon difference ​

Answers

Answer:

-1

Step-by-step explanation:

3-1=2

2-1=1

A car travels ¹/4 mile in ³ minutes, what 4 is its speed in terms of miles per minutes? A car travels ¹ / 4 mile in ³ minutes , what 4 is its speed in terms of miles per minutes ?​

Answers

Answer:

1/12  mile per minute

Step-by-step explanation:

1/4 mile in 3 minutes

 then miles per minute   is   1/4 mile / 3 min = 1/12  mile/min

Simplify completely.
3 189x5y7
√/7x²y²
3

Answers

3xy multiplied by the cube root of y^2

Answer:

[tex]\large \boxed{\sf 3x\sqrt[3]{y^5}}[/tex]

Explanation:

[tex]\rightarrow \dfrac{\sqrt[3]{\sf 189x^5y^7}}{\sqrt[3]{\sf 7x^2y^2}}[/tex]

applying radical rule

[tex]\rightarrow \sf \sqrt[3]{\dfrac{\sf 189x^5y^7}{\sf 7x^2y^2}}[/tex]

simplify

[tex]\rightarrow \sqrt[3]{\sf 27x^3y^5}[/tex]

put cubic root

[tex]\rightarrow \sqrt[3]{\sf 27} \sqrt[3]{\sf {x^3} }\sqrt[3]{\sf y^5}}[/tex]

simplify

[tex]\rightarrow \sf 3x\sqrt[3]{y^5}[/tex]

This question was confusing 15 for whoever answers

Answers

Answer:

Refer to the attached image.  I'm not clear on one point.  Yes, it is confusing.

Step-by-step explanation:

See attached image.

2. analyze the data you gathered about the number of births in the united states in 2011.
a. what do you observe about michigan and florida? justify your conclusion by using the cartogram. (5points)
b. there were 4,008,000 births in the united states during 2011. what proportion of all national births are accounted for by michigan? what proportion of all national births are accounted for by florida? (5 points)
c. suppose a person born during 2011 in the united states is chosen randomly. is the person more likely to have been born in florida or michigan? explain. (5 points)
d. the population of the united states was 308,745,538 in 2011. there were 18,801,310 people in florida and 9,883,640 people in michigan. compute the probability that a randomly selected us resident in 2011 was living in florida. do the same for michigan. which is greater? (5 points)
e. the number of births in a region depends mostly on population size. there are also random fluctuations and regional differences. consider your answers about populations and birth rates. are your answers consistent with this fact? (5 points)

Answers

The figures show that the number of births for Florida and Michigan reduced

How to compute the probability?

From the data sourced, the number of births in Florida in 2010 was 214519 while in 2011, it was 213237. For Michigan, there was decrease from 114717 to 114159.

When a person born during 2011 in the United States is chosen randomly, the person is more likely to have been born in Florida.

The probability that a randomly selected us resident in 2011 was living in Florida will be:

= 18,801,310/308,745,538

= 0.061

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simplify the equation ( 3(3) × 3 (4) ) ÷ 3 (2)​

Answers

Answer:

(3(3) × 3(4) ) ÷ 3(2)

(9 × 12) ÷ 6

108 ÷ 6

18

Answer: 18

Step-by-Step Solution:

= (3(3) * 3(4)) / 3(2)
= (9 * 3(4)) / 3(2)
= (9 * 12) / 3(2)
= (108) / 3(2)
= 108/6
=> 18

which of the following is a Monomail

Answers

Reason:

sorry but here is no options so I am giving you a example so that you can understand by your own!

Answer;

5r+3

2x-6

X4+10

4y-1

Step by step answer;

So the method of identifying the Monomail is if you see a variable in a constant so it is a Monomail !

Mark as brainlest answer

We have two fractions, 1/6 3/8 ​ and we want to rewrite them so that they have a common denominator (and whole number numerators). What numbers could we use for the denominator? Choose 2 answers:

Answers

Answer:

See below

Step-by-step explanation:

The denominator has to be divisible by 6 and 8

  you could choose   24    48    72   96    etc

Evaluate the definite integral from pi/2 to put of cos theta/sqrt 1+ sin theta.​

Answers

Answer:

[tex]\textsf{B.}\quad -2(\sqrt{2}-1)[/tex]

Step-by-step explanation:

Given integral:

[tex]\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta[/tex]

Solve by using Integration by Substitution

Substitute u for one of the functions of [tex]\theta[/tex] to give a function that's easier to integrate.

[tex]\textsf{Let }u=1+\sin \theta[/tex]

Find the derivative of u and rewrite it so that [tex]d \theta[/tex] is on its own:

[tex]\implies \dfrac{du}{d \theta}=\cos \theta[/tex]

[tex]\implies d \theta=\dfrac{1}{\cos \theta}\:du[/tex]

Use the substitution to change the limits of the integral from [tex]\theta[/tex]-values to u-values:

[tex]\textsf{When }\theta=\pi \implies u=1[/tex]

[tex]\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2[/tex]

Substitute everything into the original integral and solve:

[tex]\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}[/tex]

How many dogs were in the sample?

Answers

each number under the leaves section would be a dog


answer: 11

Answer:

11

Step-by-step explanation:

3. Error Analysis A classmate began an indirect
proof as shown below. Explain and correct
your classmate's error.
Given: AABC
Prove: LA is obtuse.
Assume temporarily that LA is acute.

Answers

Given AABC i think

Help anyone? Simple problem ​

Answers

Answer:x>7

Step-by-step explanation:

Its saying a nuber greater than 7 and 3 plus 8 would be 11

If lies in the quadrant IV what can be the value of cos

Answers

When Ф lies in quadrant IV the value of cos Ф lies in [0, 1].

What is quadrant?

One fourth of the part of a circle is said to be the quadrant of the circle.

Given the angle Ф lies in quadrant IV.

As we know, cos Ф is positive only in first and fourth quadrant i.e when Ф lies in 0<Ф<90 and 270<Ф<360.

cosФ range from [-1,1]

So, cos Ф is positive when Ф lies in 1st and 4th quadrant.

&  cos Ф is negative when Ф lies in 2nd and 3rd quadrant.

If cos ∅ lies in the quadrant IV, anything between [0, 1] can be the value of cos ∅.

Hence, when Ф lies in quadrant IV the value of cos Ф lies in [0, 1].

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which of the following are NOT classified as a trapezoid? Select all that apply

a) rhombuses
b) parallelograms
c) rectangles
d) squares
e) quadrilaterals

Answers

Answer:

[tex]\boxed {a, b, c, d}[/tex]

Step-by-step explanation:

Rhombuses, parallelograms, rectangles, and squares cannot be classified as trapezoids as they have 2 pairs of parallel sides each. But quadrilaterals can be classified as trapezoids if they have a pair of parallel sides and a pair of non-parallel sides.

a) rhombuses

b) parallelograms

c) rectangles

d) squares

Match each box plot to the data set it represents.

Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
{20,18,8,18,10,8,4}
{12,14,16,20,4,8,10}
{12,16,4,20,8,6,8}

Answers

An expression can either be an equation or an inequality.

The correct classification of the terms are:

Expression; 4x+20

Inequality: 4x+20>40

Equation: 4x+20=40

We have given that,

{20,18,8,18,10,8,4}

{12,14,16,20,4,8,10}

{12,16,4,20,8,6,8}

How to determine the correct classification?

To determine the correct classification, we make use of the following highlights

When an expression has the =, then the expression is an equation.

When an expression has any of the following signs>, <, >=, <= or != , then the expression is an inequality.

When the expression is neither an equation nor an inequality, then it is just an expression.

Hence, the correct classifications are:

Expression; 4x+20

Inequality: 4x+20>40

Equation: 4x+20=40

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One lucky day, you meet a leprechaun who promises to give you fantastic wealth, but hands you only a penny before disappearing. You head home and place the
penny under your pillow. The next morning, to your surprise, you find two pennies under your pillow. The following morning, you find four pennies, and the morning after
that, eight pennies.
Suppose that you could keep making a single stack of the pennies. After how many days would the stack be long enough to reach a star that is about 3x10¹3 km
away? (Assume that 1 penny = 1.5 mm)

Answers

The number of  days would the stack be long enough to reach a star that is about 3 × 10¹³ km away is 64 days

How to find the number of days the penny would stack?

Since from the question, we see that the number of pennies double with each day. It forms a geoemetric progression with

first term a = L and common ratio , r = 2.

Since the number of pennies after n days equals N = 2ⁿ

Let

L = length of 1 penny = 1.5 mm = 1.5 × 10⁻³ m

So, after n days, the length of the stack of pennies is the geoemetric progression D = 2ⁿ × L

Number of days pennies would stack to reach star

Making n subject of the formula, we have

n = ㏒(D/L)/㏒2

Since D = the distance of the star = 3 × 10¹³ km = 3 × 10¹⁶ m, and L = length of 1 penny = 1.5 mm = 1.5 × 10⁻³ m

Substituting the values of the variables into the equation, we have

n = ㏒(D/L)/㏒2

n = ㏒(3 × 10¹⁶ m/1.5 × 10⁻³ m)/㏒2

n = ㏒(3/1.5 × 10¹⁹)/㏒2

n = ㏒(2 × 10¹⁹)/㏒2

n = ㏒2 + ㏒10¹⁹/㏒2

n = (19㏒10 + ㏒2)/㏒2

n = (19 + 0.3010)/0.3010

n = 19.3010/0.3010

n = 64.1

n ≅ 64 days

So, the number of  days would the stack be long enough to reach a star that is about 3 × 10¹³ km away is 64 days

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Without using mathematical tables or calculator, find, in surd form(radicals) the value of tan22.5​

Answers

Answer:

Step-by-step explanation:

tan 45=1

[tex]tan 2\alpha =\frac{2tan\alpha}{1-tan^2\alpha } \\put~\alpha =22.5\\2\alpha =2 \times 22.5=45\\tan~2\alpha =tan~45=1\\1=\frac{2tan \alpha }{1-tan^2\alpha} \\1-tan^2\alpha =2 tan\alpha \\ tan^2\alpha +2tan \alpha -1=0\\tan\alpha =\frac{-2 \pm\sqrt{4+4} }{2} \\=-1\pm\sqrt{2} \\\alpha =22.5\\so ~tan~\alpha ~is~positive.\\so ~tan~\alpha =\sqrt{2} -1[/tex]

The presence of a midpoint will result in what type of segments?

Answers

Answer: two congruent segments.

Step-by-step explanation:

idlk im not smart

The diagonals of a rhombus are: A. are the same length and intersect at different angles B. bisect each other at right angles C. bisect each other and intersect at different angles D. are the same length and intersect at a right angle

Answers

Answer:

Step-by-step explanation:

Comment

Either you know the answer to this or you don't. There is no point in eliminating the wrong answers until you get to the right one. It is better to get to the right one immediately.

A rhombus and its diagonals have many interesting properties.  This one is one of those properties. The diagonals are not the same length. The diagonals bisect each other at right angles.

Answer: B

Solve the equation for x by graphing.
-2x − 3 = 3x + 2

Answers

See attached photo. Have a nice day.

Let a population consist of the values ​cigarettes, ​cigarettes, and cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with​ replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a​ population?.

Answers

Based on the given mean absolute deviations, the sample mean absolute deviation will be 2 and the population mean absolute deviation will be 3.8.

What is the sample mean absolute deviation?

This can be found as:

= Sum of sample mean absolute deviation / Number

= (0 + 0.5 + 4.5 + 0.5 + 0 + 4 + 4.5 + 4 + 0) / 9

= 2

What is the population mean absolute deviation:

First find the population mean:

= (10 + 11 + 19) / 3

= 13.3

Population mean absolute deviation:
= 1/3 x ( (10 - 13.67) + (11 - 13.3) + (19 - 13.3))

= 3.8

We can therefore see that the sample mean deviations is a biased estimate of the population mean absolute deviation.

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Fraction
Puja Limbu did 8 out of 10 math problems and Raju lama did 11 out of 15 similar maths problems.Express the number of problems solved by each of them in fractions and identify who did better performance.

Answers

Using percentages, we have that Puja Limbu did 80% of the problems and Raju Lama did 73.3%, hence Puja Limbu had the better performance.

What is a percentage?

The percentage of an amount a over a total amount b is given by a multiplied by 100% and divided by b, that is:

[tex]P = \frac{a}{b} \times 100\%[/tex]

The percentage for Puja Limbu is:

[tex]P = \frac{8}{10} \times 100\% = 80\%[/tex]

The percentage for Raju Lama is:

[tex]P = \frac{11}{15} \times 100\% = 73.3\%[/tex]

80% > 73.3%, hence Puja Limbu had the better performance.

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Nadine had 20 minutes to do a three-problem quiz. She spent 10 7/10 minutes on question A and 6 2/5 minutes on question B. How much time did she have left for question C?

Answers

She has 2 minutes and 54 seconds left for question C.

20 minutes total

10 minutes and 42 seconds on question A
6 minutes and 24 seconds on question B
= 17 minutes and 6 seconds spent on both questions

20 minutes minus 17 minutes and 6 seconds = 2 minutes and 54 seconds

Smallest to greatest 4/9, 4.9, 49%

Answers

Answer:

4/9, 49%, 4,9

Step-by-step explanation:

4/9=0,444444

49%=0,49

4,9=4,9

What is the principal, if a person paid an amount of $432 after 1 year at 8% rate of interest per annum

Answers

Answer:

principal = $400

Step-by-step explanation:

amount(a)=principal(p) + interest(y).........(1)

Interest(y)=principal(p) x rate(r) x time(t)....(2)

In (1), p= a - y...........(3)

putting (3) in (2),

y = (a - y) x r% x t

y = (432 - y) x 8% x 1

y = 3456%-8y%

y + 0.08y = 3456%

1.08y = 34.56

y = 32

but a - y = p

432 - 32 = p

p = 400

principal = $400

If f(x) = 2x^2+3 and g(x)= 5x, evaluate the following: a. f(g(x)) b. g(g(3)) c. g(f(-2))

Answers

Answer:

A = 50x^2+3

B = 75

C = 55

Step-by-step explanation:

[tex]f(x)=2x^2+3\\g(x)=5x[/tex]

A. f(g(x))

[tex]2(5x)^2+3\\2(25x^2)+3\\=50x^2+3[/tex]

B. g(g(3))

[tex]5(5(3))\\5(15)\\=75[/tex]

C. g(f(-2))

[tex]g(2(-2)^2+3)\\g(8+3)\\g(11)\\5(11)\\=55[/tex]

1. At which point do Line CF and Line GF intersect? They intersect at point?

2. Look at Line AD and Like BE. Do these lines intersect?
(a) yes they will intersect at Point F?
(b) no they will never intersect?
(c) yes they will point at Point G
(d) yes they will intersect at Point F?

3. Look at Line BG and Line AC. Where do they intersect? They intersect at Point?

(Please hurry giving 50 points!) ​

Answers

Answer: i think no?

Step-by-step explanation:

AD and BE are both parallel lines (they are parallel to eachother), so they will never intersect


CF and GF intersect at point F (i think)


Bg and Ac intersect at point B (i think)


i dont want to give a definite answer in the event im wrong bc I just learned this like a few weeks ago-

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