Which is equivalent to 3√8* ?

O x √8^3
O 8 ^3/x
O 8 ^x/3
O 8^3x

Answer:

(3(3) × 3(4) ) ÷ 3(2)

(9 × 12) ÷ 6

108 ÷ 6

18

each number under the leaves section would be a dog

answer: 11

Answer:

11

Step-by-step explanation:

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

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.

= Sum of sample mean absolute deviation / Number

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

= 2

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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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

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

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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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

Answer: two congruent segments.

Step-by-step explanation:

idlk im not smart

Answer:x>7

Step-by-step explanation:

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

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

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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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.

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

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

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

Let

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

Making n subject of the formula, we have

n = ㏒(D/L)/㏒2

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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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 !

Answer:

4/9, 49%, 4,9

Step-by-step explanation:

4/9=0,444444

49%=0,49

4,9=4,9

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]

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 expected enrolment in 7 years is  1845.

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

FV = P (1 + r) ^n

1500 x (1.03)^7 = 1845

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Answer:

-1

Step-by-step explanation:

3-1=2

2-1=1

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

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]

Answer:

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

Step-by-step explanation:

See attached image.

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.

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}

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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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-

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

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

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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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]

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

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

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

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