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Which expression is equivalent to (6^−2 • 3^5)^−2?

three raised to the third power divided by six raised to the fourth power
negative three raised to the third power divided by six raised to the fourth power
six raised to the fourth power divided by three raised to the tenth power
negative six raised to the fourth power divided by three raised to the tenth power

Correlation is a bi-variate analysis that measures the strength of association between two variables and the direction of the relationship.

What is bi-variate analysis?

Bivariate analysis is one in every of the only kinds of quantitative (statistical) analysis. It involves the analysis of 2 variables (often denoted as X, Y), for the aim of deciding the empirical relationship between them.

Main Body:

The correlation coefficient's value ranges from +1 to -1. the entire degree of correlation between the 2 variables is indicated by a worth of one. The association between the 2 variables are weaker because the parametric statistic price approaches zero.

Hence correlation investigate linear association between two variables.

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The average speed of the train is 18m/sec and the length of the train is 72 m.

let  x represent the length of the train.

and 'S' be the speed of the train.

Given, that the train passes through 360 m long tunnel  completely in 24 sec.

⇒ 360+x/24 = s

⇒ 24s = 360+x ...eq(1)

And also given that it passes through another tunnel 216 m long completely in 16 sec.

⇒ 216+x/24 = s

⇒ 16s = 216 + x .... eq(2)

⇒ subtract eq(1) from eq(2)

⇒ 24s = 360 + x

   16s = 216 + x

8s = 144

s = 18 m/sec

length of the train: 24s = 360+x

24(18) = 360+x

432 = 360+x

x = 432-360

x = 72m

Hence the length of the train is 72 m.

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One-half of the number is 18.

Given that

One-third of a number is 12.

Let "x" be the number

Given that,

One-third of a number is 12

Which means,

[tex]\frac{1}{3} *x = 12[/tex]

x = 3*12

x = 36

Thus the number is 36

Work out one-half of the number

one half of the number = [tex]\frac{x}{2}[/tex]

one half of the number = [tex]\frac{36}{2}[/tex]

one half of the number = 18

Hence the answer is, one-half of the number is 18.

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Answer:

129 degrees.

Step-by-step explanation:

The Sum of all the interior angles of an a-sided polygon is always (n−2) × 180 degrees.

As this is a seven-sided polygon, the sum of its interior angles is (7−2) × 180 degrees = 5 × 180 degrees = 900 degrees.

Step 1: Find the sum of six of the angles is 130 degrees, 118 degrees, 135 degrees, 128 degrees, 126 degrees, and 134 = 771 degrees

Hence the seventh angle is x = 900 degrees − 771 degrees= 129 degrees.

The answer of the question is A. Cluster Analysis

Data reduction is a stage of qualitative data analysis techniques. Data reduction is simplification, classification, and unnecessary hoarding of data in such a way that the data can produce meaningful information and facilitate drawing conclusions.

Data reduction is done to provide more specific representation and make it easier for researchers to collect data then look for additional data if needed.

Various dimension reduction methods are used to reduce the complexity of data before further processing. One of data reduction techniques is Cluster Analysis. Cluster analysis is used to group or classify data with various types, for example in types according to the concept, and certain categories according to the research being conducted.

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Kirk will drive 15 kilometer east  to get back to his home

The problem will be solved using basic addition operator and subtraction

Let the given information be represented as follows

he drives 15 kilometers west = 15 W

45 kilometers east = 45 E

75 kilometers west = 75 W

Kirk will drive East to get home = ?

since the distances are all in the horizontal direction we add up the west directions and subtract the east direction

15 W + 75 W = 45 E + ?

90 W = 45 E + ?

? = 90 - 45

? = 45 E

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Answer:

Step-by-step explanation:

3y+6-4x=0

3y-4x=0-6

3y-4x=-6

y-4x=6/3

y-4x=2

y-x=2/4

y-x=0.5

8 AM to 3 PM is 7 hours = 420 mins

Find how many rotations:
420 / 36 = 11.66666 rotations

Using ratio

1 rotation : 360 degrees

11.66666 : x degrees

x = 11.66666 / 1 * 360 = 4200 degrees

The function f(x) = -[tex]5x^5 -5760-2880x+520x^2+260x^3-10x^4[/tex] end behavior is As, x → infinity , f(x) → - infinity ; as x → - infinity , f(x) → infinity .

Given,

The function is given as:

f(x) = -[tex]5x^5 -5760-2880x+520x^2+260x^3-10x^4[/tex]

To find the bend behavior of f(x)

Now, According to the question:

Based on the given condition:

Analyze the end behavior of

f(x) = -[tex]5x^5 -5760-2880x+520x^2+260x^3-10x^4[/tex]

As, x → infinity , f(x) → - infinity ;

as x → - infinity , f(x) → infinity .

Hence, The function f(x) = -[tex]5x^5 -5760-2880x+520x^2+260x^3-10x^4[/tex] end behavior is As, x → infinity , f(x) → - infinity ; as x → - infinity , f(x) → infinity .

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Answer:

(-5,-4)

as 90 degree rotation counter clockwise would land in the third quaderant where both x and y are negative

The equation of the parabola, in vertex form, that represents the ball's distance, y, from the hole, x seconds after the ball is hit is

y = (-2m/s^2)t^2 + (-2m/s)t + 6m

ax² + bx + c = 0 is a quadratic equation, which is a second-order polynomial equation in a single variable. a 0. It has at least one solution because it is a second-order polynomial equation, which is guaranteed by the algebraic fundamental theorem. The answer could be simple or complicated.

Given,

Assume for the moment that the hole is located at y = 0m, where x represents the time variable.

We are aware of:

The vertices are (-1s, 8m).

x in seconds and y in meters, I presume

The ball enters the hole at time x = 1s, hence we also have the following point:

(1s, 0m).

The vertex of the quadratic equation y = ax2 + bx + c is located at:

x = -b/2a,

Next, we have:

-1 = -b/2a.

Next, we have equations for trees:

8m = a(-1s)2, b(-1s), and c

0m = (a(1s)2) + (b(1s)2) + (c)

-1s = -b/2a.

The third equation should have one variable isolated first, and then it should be substituted in one of the other two:

1s2a = b.

Therefore, we can change b to bi 1s2a in the first two equations.

8m = 1s×2a×1s - a×1s2a1s + c

0m = 1s×2a×1s + a×1s×2a + c

We can combine the two equations to get:

8m is equal to a(1s2 - 2s2) + c, or a×1s2 - c.

0m is equal to a(1s2 + 2s2) + c, or a×3s2 + c.

We may quickly remove c from the second equation and then substitute it in the first equation:

c = -a3s^2

The initial equation is now:

8m is equal to -a×1s2 - a×3s2 - a×4s2.

a = 8m/-4s^2 = -2m/s^2.

We can now determine the values of c and b using a.

c = -(-(-2m/s2)*3s2)*3s2 = 6m.

B is equal to 1s*2a = 1s*(-2m/s2) = -2m/s.

The equation is then:

y = (-2m/s^2)t^2 + (-2m/s)t + 6m

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Answer:

x y

-2 -9

0 1

2 -9

4 -19

explanation:

you simply replace the x with the figure given in the x table or columns or rows

Answer:

so -2024y0000 division is s

The equation of the required perpendicular line is y = (-5/4)x - 30.

We are given an equation. The equation is linear in nature. The equation represents a straight line. The equation of the given straight line is y = (4/5)x + 23. We need to find the equation of the straight line that is perpendicular to the given line and passes through the point that has coordinates (-40, 20).

The slope of the given line is 45. Let the slope of the required perpendicular line be represented by the variable "m". The slope of the perpendicular line must be the negative reciprocal of the slope of the given line.

m = -1/(4/5)

m = -5/4

The equation of the required line is of the form y = mx + c.

y = (-5/4)x + c

We know that this line passes the point (-40, 20).

20 = (-5/4)(-40) + c

20 = 50 + c

c = -30

Hence, the equation of the required line is y = (-5/4)x - 30.

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Hey there! I explained and wrote everything on paper since it's hard to type so you can view the pic uploaded! Hope it helps, if you have any doubts, please feel free to ask :)

With deceleration of 6 ft/s² and initial speed of 66 ft/s, the airplane will stop after travelling 363 ft.

Use the equation of motion:

v² = u² + 2 · a · s

Where:

v = final velocity

a = acceleration

u = initial velocity

s = distance

Deceleration is a negative acceleration.

Parameters given from the problem:

u = 66 ft/s

a = - 6 ft/s²

v = 0  since the airplane stops.

Plug these parameters into the formula:

v² = u² + 2 · a · s

0 = 66² + 2 · (-6)· s

s = 66² / 12

s = 363 ft.

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Answer:

D

Step-by-step explanation:

The slope if before the x so in this case it's -3.

The y-intercept is at the end so it's -2.

Answer:

D. slope = -3 and y-intercept = -2

Step-by-step explanation:

Slope Intercept Form:

The variables:

y = -3x -2

The equation of line passing through the point (-3, -1) and is perpendicular to the line y = -x/2 - 5 is y = 2x + 5

perpendicular lines are defined as two lines that meet or intersect each other at right angles (90°).

In such lines the product of their gradients is -1.

if y = -x/2 - 5

then comparing the equation with y = mx + c

it means slope(m) = -1/2

so the slope(p) of the second line perpendicular to the above line will be

p x m = -1

p = -1/m

p = -1 ÷ -1/2

p = 2

so if x = -3, y = -1 and substituting these values into the equation of line , y = mx + c becomes

-1 = 2 x (-3) + c

-1 = -6 + c

collect like terms

c= 6 - 1

c = 5

Equation of the second line becomes,

y = 2x + 5

In conclusion, the equation of the lines passing through point(-3, -1) is y = 2x + 5

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Answer and Explanation:

m∠1 = 43° because opposite angles are congruent

m∠2 = 137° because ∠2 and 43° are supplementary, so m∠2 + 43° = 180°

m∠3 = 137° because opposite angles are congruent, so m∠3 = m∠2

m∠5 = 43° because alternate angles are congruent

m∠6 = 137° because ∠5 and ∠6 are supplementary, so 43° + m∠6 = 180°

m∠7 = 137° because alternate exterior angles are congruent, so m∠7 = m∠2

m∠8 = 43° because alternate exterior angles are congruent, so m∠8 = m∠1

Answer:

angle 1: 43

angle 2: 137

Step-by-step explanation:

Lets find angle 1 first :)

Since oposite diagonal angles are the same, this will make angle 1 43 degrees :)

Now lets find angle 2!

Since 1 was 43 degrees, and a flat line is 180 degrees we must subtract!

Lets do that now!

180 - 43 = 137

Angle 2 will be 137 degrees!

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x=14/3 and y=-28 when  x = -6. and  y = 20 for equation y = 6x +8

x=30 and y=-4 for when  x = -6. and  y = 20 for equation  y = 2/3x

x=-15 and y=11,  when  x = -6. and  y = 20 for equation   y = -x + 5

x=2 and y=-2 ,  when  x = -6. and  y = 20 for equation  y = 6.5x + 7

Two or more expressions with an Equal sign is called as Equation.

For equation y = 6x +8

When x=-6

y=6(-6)+8=-36+8=-28

when y =20

20=6x+8

28=6x

x=28/6=14/3

x=14/3 and y=-28 when  x = -6. and  y = 20 for equation y = 6x +8

For equation y = 2/3x

When x=-6, y=2/3(-6)=-4

y=20, 20=2/3x

60=2x

x=30

x=30 and y=-4 for when  x = -6. and  y = 20 for equation  y = 2/3x

For equation y = -x + 5

When x=-6, y=6+5=11

When y=20,

20=-x+5

15=-x

x=-15

x=-15 and y=11,  when  x = -6. and  y = 20 for equation   y = -x + 5

For equation y = 6.5x + 7

When x=-6

y=6.5(-6)+7

y=-39+7=-2

When y=20

20=6.5x+7

13=6.5x

2=x

x=2 and y=-2 ,  when  x = -6. and  y = 20 for equation  y = 6.5x + 7

Hence,

x=14/3 and y=-28 when  x = -6. and  y = 20 for equation y = 6x +8

x=30 and y=-4 for when  x = -6. and  y = 20 for equation  y = 2/3x

x=-15 and y=11,  when  x = -6. and  y = 20 for equation   y = -x + 5

x=2 and y=-2 ,  when  x = -6. and  y = 20 for equation  y = 6.5x + 7

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Answer: 6x(-x+3)

Step-by-step explanation:

You can factor out the 6 and x. -6x^2 divided by 6x = -x, and 18x divided by 6x = 3. So the answer is 1. 6x(-x+3)

Answer:

-6x^2+18x

-6x^2-18x

6x^2+18x

-6x^2+108x

Step-by-step explanation:

1.

6x(-x+3)

6x(-x)+6x(3)

-6x^2+18x

2.

-6x(X+3)

-6x(X)+-6x(3)

-6x^2-18x

3.

x(6x+18)

x(6x)+x(18)

6x^2+18x

4.

6(-x^2+18x)

6(-x^2)+6(18x)

-6x^2+108x

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The numbers are inserted into the tree and shown in the below diagrams.

Given that,

Put these numbers into a 2-3-4 tree: 10, 15, 20, 40, 60, 80, 90, 100, 30, 70, and 50.

Binary Tree Data Structure: What is it?

A Binary Tree is a Tree data structure with two children or less. We typically refer to them as the left and right child since each element in a binary tree can only have two children.

Representation of a Binary Tree

A pointer to the topmost node of a binary tree serves as its representation. The root's value is NULL if the tree is empty.

The following components are found in a binary tree node:

Data

Pointer to left child in data

Pointer to the right-hand child

Thereforer, The numbers are inserted into the tree

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The question was incomplete. Check below the complete question.

insert 10, 15, 20, 40, 60, 80, 90, 100, 30, 70, and 50 into a 2-3-4 tree and select the correct answer

The graph of the system of inequalities can be seen in the image at the end.

Here we have the following system of inequalities:

y < 2x - 5

y > -3x + 1

So first we need to graph the two lines:

y = 2x - 5

y = -3x + 1

With dashed lines (because the symbols < and > are used).

And then we shade all the region below the line:

y = 2x - 5

And we shade the region above the line:

y = -3x + 1

So we will get the graph below:

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Answer:

=====================

Slope-intercept form is:

Given line in standard form:

Convert this to slope-intercept form:

Answer:

[tex]\boxed{y=\dfrac{3}{4}x-\dfrac{3}{4}}[/tex]

[tex]\implies \textsf{Slope}=\dfrac{3}{4}[/tex]

[tex]\implies \textsf{$y$-intercept}=-\dfrac{3}{4}[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{6.3 cm}\underline{Slope-intercept form of a linear equation}\\\\$y=mx+b$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $b$ is the $y$-intercept.\\\end{minipage}}[/tex]

Given equation:

[tex]-3x+4y+3=0[/tex]

To write the given equation in slope-intercept form, rearrange the equation to isolate y.

Add 3x to both sides of the equation:

[tex]\implies -3x+4y+3+3x=0+3x[/tex]

[tex]\implies 4y+3=3x[/tex]

Subtract 3 from both sides of the equation:

[tex]\implies 4y+3-3=3x-3[/tex]

[tex]\implies 4y=3x-3[/tex]

Divide both sides of the equation by 4:

[tex]\implies \dfrac{4y}{4}=\dfrac{3x-3}{4}[/tex]

[tex]\implies y=\dfrac{3}{4}x-\dfrac{3}{4}[/tex]

Compare the rearranged equation with the slope-intercept formula:

[tex]\implies \textsf{Slope}=\dfrac{3}{4}[/tex]

[tex]\implies \textsf{$y$-intercept}=-\dfrac{3}{4}[/tex]

Answer:

No real zeros.

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Discriminant of the Quadratic Formula}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\When $b^2-4ac > 0 \implies$ two real zeros.\\When $b^2-4ac=0 \implies$ one real zero.\\When $b^2-4ac < 0 \implies$ no real zeros.\\\end{minipage}}[/tex]

The value of the discriminant shows how many zeros the function has.

Given quadratic function:

[tex]f(x)=5x^2+5x+21[/tex]

Therefore:

Substitute the values into the discriminant and solve:

[tex]\begin{aligned}\implies b^2-4ac&=(5)^2-4(5)(21)\\& = 25 - 20(21)\\&=25-420\\&=-395\end{aligned}[/tex]

Therefore, as -395 < 0, there are no real zeros.

You invest $892 at the end of each year in an investment with annual rate of return 6%. Then  the investment worth after 15 year will be 14182.8 dollar

we know that

SI = prt

where

SI= simple interest

A= amount

P= principal

r= rate

t= number of year

according to the question

SI= 892* 0.06*1

 = 53.52

 A =  892+53.52

    = 945.52

after 15 years amount will be

A= 945.52*15

  = 14182.8

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The simplified expression to represent the number of members this year is 1.15m

An expression simply refers to the mathematical statements which have at least two terms which are related by an operator and contain either numbers, variables, or both.

From the information, the membership in the volunteers Club increased by 15% from last year. If the number of members last year was m.

The number of members will be:

= Old members+ Increase in members

= m + (15% × m)

= m + 0.15m

= 1.15m

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Answer:it depends on how many sides their are

Step-by-step explanation:

just factual

Answer:

below

Step-by-step explanation:

Rate =  4/5 mi / 2/5 hr =  2 mi /hr

3/4 hr for lunch

20 miles / 2 miles/hr   + 3/4 hr =  10 3/4 hr to hike 20 miles with a lunch