“Use the properties to rewrite this expression with the fewest terms possible:3+7(x - y) + 2x - 5y”

Answers

Answer 1
[tex]-5y+2x+7(x-y)+3[/tex]

Expanding 7(x - y) in the above expression gives

[tex]-5y^{}+2x+7x-7y+3[/tex]

adding the like terms (2x+ 7x) and (-5y-7y) gives

[tex](-5y-7y)+(2x+7x)+3[/tex][tex]\rightarrow\textcolor{#FF7968}{-12y+8x+3.}[/tex]

The last expression is the simplest form we can convert our expression into.


Related Questions

Analyze the equations in the graphs to find the slope of each equation the y-intercept of each equation in the solution for the system of equations equation 1: y = 50x + 122

Answers

Given:

[tex]y=50x+122\ldots\text{ (1)}[/tex][tex]y=1540-82x\ldots\text{ (2)}[/tex]

The general equation is

[tex]y=mx+c[/tex]

m is a slope and c is the y-intercept.

From equation (1),

[tex]\text{Slope = 50 and y intercept is 122}[/tex]

From equation (2)

[tex]\text{Slope = -82 and yintercept is }1540[/tex]

From equation (1) and (2)

Substitute equation (2) in (1)

[tex]1540-82x=50x+122[/tex][tex]50x+82x=1540-122[/tex][tex]132x=1418[/tex][tex]x=\frac{1418}{132}[/tex][tex]x=44[/tex]

Substitute in (2)

[tex]undefined[/tex]

Fifteen strips, 11/4" wide, are to be ripped from a sheet of plywood. If 1/8" is lost with each cut, how much of the plywood sheet is used in making the 15 strips? (Assume 15 cuts are necessary.)

Answers

The size of the plywood sheet used is;

[tex]\frac{37}{8}^{\doubleprime}[/tex]

Here, we want to get the size of the part of the plywood sheet lost

From the question, we are told that 1/8 inches is lost

The size lost would be;

[tex]\frac{1}{8}\times\text{ 15 = }\frac{15}{8}[/tex]

This is the size that was lost

To get the total part of the plywood used, we simply add the width of all the strips to the amount of the plywood lost

We have this as;

[tex]\frac{11}{4}\text{ + }\frac{15}{8}\text{ = }\frac{22+15}{8}\text{ = }\frac{37}{8}[/tex]

The position of an open-water swimmer is shown in the graph. The shortest route to the shoreline is one that is perpendicular to the sh Ay 10 00 6 water 4 shore |(2, 1) swimmer 19 -2 2 1 3 4 5X N -2 An equation that represents the shortest path is y=

Answers

Answer:

Explanation:

From the graph, we ca

Find the prime factorization of the following number write any repeated factors using exponents

Answers

Notice that 100=10*10, and 10=2*5. 2 and 5 are prime numbers; therefore,

[tex]\begin{gathered} 100=10\cdot10=(2\cdot5)(2\cdot5)=2\cdot2\cdot5\cdot5=2^2\cdot5^2 \\ \Rightarrow100=2^2\cdot5^2 \end{gathered}[/tex]

The answer is 100=2^2*5^2

Please find the square root. Round your answer to the nearest tenth. [tex] \sqrt{58 } = [/tex]

Answers

Determine the square root of 58.

[tex]\begin{gathered} \sqrt[]{58}=7.615 \\ \approx7.6 \end{gathered}[/tex]

So answer is 7.6.

4/7 X 1/2 = in fraction

Answers

Consider the given expression,

[tex]P=\frac{4}{7}\times\frac{1}{2}[/tex]

The product of fractions is obtained in the form of a fraction whose numberator is the product of numerators of fractions, and the denominator of the product is the product of denominators of the given fractions,

[tex]\begin{gathered} P=\frac{4\times1}{7\times2} \\ P=\frac{4}{14} \end{gathered}[/tex]

Thus, the product of the given fractions is 4/14 .

Write an addition equation and a subtraction equation
to represent the problem using? for the unknown.
Then solve.
There are 30 actors in a school play. There are
10 actors from second grade. The rest are from third
grade. How many actors are from third grade?
a. Equations:
b. Solve

Answers

The Equation is 10 + x= 30 and 20 actors are from third grade.

What is Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side. It demonstrates the equality of the relationship between the expressions printed on the left and right sides. LHS = RHS is a common mathematical formula.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given:

There are 30 actors in a school play.

There are 10 actors from second grade.

The rest are from third grade.

let the actors in third grade is x.

Equation is:

Actors from second grade + Actors from third grade = Total actors

10 + x= 30

Now, solving

Subtract 10 from both side

10 +x - 10 = 30 - 10

x = 20

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Zachary is designing a new board game, and is trying to figure out allthe possible outcomes. How many different possible outcomes arethere if he spins a spinner with three equal-sized sections labeledWalk, Run, Stop, spins a spinner with four equal-sized sections labeledRed, Green, Blue, Orange, and spins a spinner with 5 equal-sizedsections labeled Monday, Tuesday, Wednesday, Thursday, Friday?

Answers

ANSWER

60 possible outcomes

EXPLANATION

If he spins the 3-section spinner, there are 3 possible outcomes: Walk, Run, Stop.

If he spins the 4-section spinner, there are 4 possible outcomes: red, green, blue, orange.

If he spins the 5-section spinner, there are 5 possible outcomes: Monday, Tuesday, Wednesday, Thursday, Friday.

If he has to spin the three spinners, the total possible outcomes is the product of the possible outcomes of each spinner: 3x4x5 = 60.

Determine the value of each limit for the function below.f(x)=x/(x-2)^2(a) lim f(x). (b) lim f(x)x---2^-. x---2^+

Answers

We will have the following:

a)

[tex]\lim _{x\rightarrow2^-}\frac{x}{(x-2)^2}=\infty[/tex]

b)

[tex]\lim _{x\rightarrow2^+}\frac{x}{(x-2)^2}=\infty[/tex]

Han and clan are stuffing enveloppes Han can stuff 20 envelopes in one minute and Clare can stuff 10 envelopes in one minute. They start working together on a pile of 1000 envelopes. How long does it take them to finish the pile.

Answers

uff = Given

Han can stuff 20 envelopes in one minute

Clare can stuff 10 envelopes in one minute

Together they start working on a pile off 1000 envelope.

Find

How long does it take them to finish the pile.

Explanation

as we have given

in one minute , Han can stuff = 20 envelope

in one minute , Clare can stuff = 10 envelope

together in one minute , they can stuff =

[tex]\begin{gathered} 20+10=30 \\ \\ \end{gathered}[/tex]

we know that the number of time it will take to finish stuffing would be number of envelope / joint rate = 1000/30

so , time taken to finish the pile =

[tex]\begin{gathered} \frac{1000}{30} \\ \\ \frac{100}{3} \\ \\ 33\frac{1}{3} \\ or \\ 33min20sec \end{gathered}[/tex]

Final Answer

Hence , the time taken by them to finish the pile is 33 minutes 20 seconds

Find the value of each variable.All answers must be in simplest radical form

Answers

Radical

x = √10 • tan 45° = √10• 1 = √10

then

x= √10

y= √x^2 + 10 = √ 10 +10 = √20

Then answer is

x=√10

y= √20

Given that the two triangles are similar find the unknowns length of the side labeled in

Answers

Answer:

The unknown length of the side labeled n is 10.5 units

Explanation:

Given:

Two similar triangles with one unknown

To find:

the unknown length of the side labelled n

For two triangles to be similar, the ratio of their corresponding sides will equal

[tex]\begin{gathered} side\text{ with 36 corresponds to side with 27} \\ side\text{ with 14 corresponds to side with n} \\ The\text{ ratio:} \\ \frac{14}{n}\text{ = }\frac{36}{27} \end{gathered}[/tex]

[tex]\begin{gathered} crossmultiply: \\ 14(27)\text{ = 36\lparen n\rparen} \\ 36n\text{ = 378} \\ \\ divide\text{ both sides by n:} \\ \frac{36n}{36}\text{ = }\frac{378}{36} \\ n\text{ = 10.5} \end{gathered}[/tex]

The unknown length of the side labeled n is 10.5 units

if the area of polygon A is 72 and Q is a scaled copy and the area of Q is 5 what scale factor got 72 to 5

Answers

A area= 72

Q area =5

So, if we multiply the A area by the square of the scale factor ( since they are areas) we obtain area Q:

72 x^2 = 5

Solving for x:

x^2 = 5/72

x = √(5/72)

x= 0.26

Which of the following are solutions to the following solutions to the following solutions?

Answers

We have to find the solutions to the equation:

[tex]|x+4|=8[/tex]

The absolute value function is in fact a piecewise function, so it may have two solutions.

We consider for the first solution that the argument inside the absolute function is positive, that is x + 4 > 0. Then, we will have:

[tex]\begin{gathered} x+4=8 \\ x=8-4 \\ x=4 \end{gathered}[/tex]

Now, we consider that the the argument is negative and is made positive by the absolute value function (it will shift the sign, which can be represented by a multiplication by -1). This means that x + 4 < 0, and the solution will be:

[tex]\begin{gathered} -(x+4)=8 \\ -x-4=8 \\ -x=8+4 \\ -x=12 \\ x=-12 \end{gathered}[/tex]

We can see it in a graph as:

Answer: the solutions are x = 4 and x = -12.

The first 19 terms of the sequence 9, 2, -5, -12,... find the sum of the arithmetic sequence

Answers

To find the sum of the ari

A machine worked for 4hours and used 6kilowatts of electricity.What is the rate ofenergy consumed inkilowatts per hour?*Enter your answer as a decimal

Answers

4 hours ---> 6 kilowatts

1 hour -----> x kilowatts

[tex]\begin{gathered} 4\times x=1\times6 \\ 4x=6 \\ \frac{4x}{4}=\frac{6}{4} \\ x=\frac{3}{2}=1.5 \end{gathered}[/tex]

answer:

1.5 kilowatts per hour

how do you find the domain in a range of number 2?

Answers

The domain is all the x values included in the function, while the range are all the y values included in the function.

Based on the graph:

Answer:

• Domain:

[tex](-\infty,\text{ }\infty)[/tex]

• Range:

[tex](0,\infty)[/tex]

Find the complement requested angle of 10% A/ 350B/20C/170D/80

Answers

The complementary angles are angles in which the sum of them is equal to 90º

So: 90º-10º=80º

So, the complementary angle is 80º

An object moves in simple harmonic motion with period 6 seconds and amplitude 4cm. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a negative direction. Give the equation modeling the displacement d as a function of time t.

Answers

The general function for describing the displacement from the mean position in harmonic motion is:

[tex]d(t)=A\cdot\sin (\frac{2\pi}{T}\cdot t+\phi)\text{.}[/tex]

Where:

• A is the amplitude,

,

• T is the period,

,

• φ is initial phase displacement.

From the statement, we know that:

• the amplitude is 4 cm,

,

• at time t = 0 its displacement d from the rest is 0 → d(t = 0) = 0,

,

• initially, it moves in a negative direction.

s

x+y+z=12x+4y+2z = -6-x+9y-3z=-49 Can someone please help me solve this system of equation?

Answers

Let's begin by listing out the information given to us:

[tex]\begin{gathered} x+y+z=1 \\ 2x+4y+2z=-6 \\ -x+9y-3z=-49 \end{gathered}[/tex]

To solve this 3 variable equation, let's eliminate one of the variables

add equation 1 & 3, we have:

[tex]\begin{gathered} x-x+y+9y+z-3z=1-49 \\ 10y-2z=-48 \\ Make\text{ z the }subject,we\text{ have:} \\ -2z=-10y-48 \\ divide\text{ through by -2} \\ z=5y+24 \end{gathered}[/tex]

Substitute z into equation 1, 2 & 3

[tex]\begin{gathered} x+y+5y+24=1 \\ x+6y=1-24 \\ x+6y=-23 \end{gathered}[/tex]

[tex]\begin{gathered} 2x+4y+2\left(5y+24\right)=-6 \\ 2x+4y+10y+48=-6 \\ 2x+14y=-6-48 \\ 2x+14y=-54 \end{gathered}[/tex]

[tex]\begin{gathered} -x+9y-3\left(5y+24\right)=-49 \\ -x+9y-15y-72=-49 \\ -x-6y=-49+72 \\ -x-6y=23 \end{gathered}[/tex]

Solve as a simultaneous equation, we have:

[tex]\begin{gathered} x+6y=-23 \\ 2x+14y=-54 \\ \text{Multiply the top equation by 2 \& subtract it from the bottom equation} \\ 2\cdot(x+6y=-23)\Rightarrow2x+12y=-46 \\ 2x+14y=-54-(2x+12y=-46) \\ 2x-2x+14y-12y=-54-(-46) \\ 2y=-8 \\ y=-4 \end{gathered}[/tex]

Substitute y = -4 into x + 6y = -23, we have:

[tex]\begin{gathered} x+6\left(-4\right)=-23 \\ x-24=-23 \\ x=-23+24 \\ x=1 \end{gathered}[/tex]

Substitute y = -4 into z = 5y + 24, we have:

[tex]\begin{gathered} z=5\left(-4\right)+24 \\ z=-20+24 \\ z=4 \end{gathered}[/tex]

what's the answer for proportions 7/9=b/b-10

Answers

Answer:

-35

Step-by-step explanation:

[tex]\frac{7}{9}[/tex] = [tex]\frac{b}{b - 10}[/tex]  multiply both sides by 9(b -10)

[tex]\frac{9(b - 10)}{1}[/tex]  [tex](\frac{7}{9})[/tex] = [tex]\frac{9(b -10)}{1}[/tex] [tex](\frac{b}{b-10})[/tex]  On the right side of the equation, the 9's cancel out and on the right side of the equation the (b -10) cancels out to leave

7(b -10) = 9b  Distribute the 7

7b - 70 = 9b  Subtract 7b from both sides

-70 = 2b  Divide both sides by 2

-35 = b

Number 14. Directions in pic. And also when you graph do the main function in red and the inverse in blue

Answers

Question 14.

Given the function:

[tex]f(x)=-\frac{2}{3}x-4[/tex]

Let's find the inverse of the function.

To find the inverse, take the following steps.

Step 1.

Rewrite f(x) for y

[tex]y=-\frac{2}{3}x-4[/tex]

Step 2.

Interchange the variables:

[tex]x=-\frac{2}{3}y-4[/tex]

Step 3.

Solve for y

Add 4 to both sides:

[tex]\begin{gathered} x+4=-\frac{2}{3}y-4+4 \\ \\ x+4=-\frac{2}{3}y \end{gathered}[/tex]

Multply all terms by 3:

[tex]\begin{gathered} 3x+3(4)=-\frac{2}{3}y\ast3 \\ \\ 3x+12=-2y \end{gathered}[/tex]

Divide all terms by -2:

[tex]\begin{gathered} -\frac{3}{2}x+\frac{12}{-2}=\frac{-2y}{-2} \\ \\ -\frac{3}{2}x-6=y \\ \\ y=-\frac{3}{2}x-6 \end{gathered}[/tex]

Therefore, the inverse of the function is:

[tex]f^{-1}(x)=-\frac{3}{2}x-6[/tex]

Let's graph both functions.

To graph each function let's use two points for each.

• Main function:

Find two point usnig the function.

When x = 3:

[tex]\begin{gathered} f(3)=-\frac{2}{3}\ast3-4 \\ \\ f(3)=-2-4 \\ \\ f(3)=-6 \end{gathered}[/tex]

When x = 0:

[tex]\begin{gathered} f(0)=-\frac{2}{3}\ast(0)-4 \\ \\ f(-3)=-4 \end{gathered}[/tex]

For the main function, we have the points:

(3, -6) and (0, -4)

Inverse function:

When x = 2:

[tex]\begin{gathered} f^{-1}(2)=-\frac{3}{2}\ast(2)-6 \\ \\ f^{-1}(2)=-3-6 \\ \\ f^1(2)=-9 \end{gathered}[/tex]

When x = -2:

[tex]\begin{gathered} f^{-1}(-2)=-\frac{3}{2}\ast(-2)-6 \\ \\ f^1(-2)=3-6 \\ \\ f^{-1}(2)=-3 \end{gathered}[/tex]

For the inverse function, we have the points:

(2, -9) and (-2, -3)

To graph both functions, we have:

ANSWER:

[tex]\begin{gathered} \text{ Inverse function:} \\ f^{-1}(x)=-\frac{3}{2}x-6 \end{gathered}[/tex]

Yoko plans to watch 2 movies each month. Write an equation to represent the total number of movies n that she will watch in m months.

Answers

Answer:

2m because 2 times the months will tell us how many she has watched for example in 2 months she will watch 4 because 2*2 is 4

Find the tangent of each angle that is not the right angle. Drag and drop the numbers into the boxes to show the tangent of each angle. B 76 tan ZA tan ZB 2.45 0.38 0.93

Answers

From the trignometric ratio of right angle triangle :

The ratio for the tangent of any angle of right angle triangle is the ratio of the side Opposite to that angle to the adjacent side of that angle :

[tex]\tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}}[/tex]

In the given triangle :The side opposite to the angle A is BC and the adjacent side AC

So,

[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan A=\frac{BC}{AC} \end{gathered}[/tex]

In the figure : we have AC = 76, BC = 31 and AB = 82.1

Substitute the value and simplify :

[tex]\begin{gathered} \tan A=\frac{BC}{AC} \\ \tan A=\frac{31}{76} \\ \tan A=0.407 \\ \tan A=0.41 \end{gathered}[/tex]

Thus, tan A = 0.41

Now, the side opposite to the angle B is AC and the adjacent side is BC

thus :

[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan B=\frac{AC}{BC} \end{gathered}[/tex]

In the figure : we have AC = 76, BC = 31 and AB = 82.1

Substitute the value and simplify :

[tex]\begin{gathered} \tan B=\frac{AC}{BC} \\ \tan B=\frac{76}{31} \\ \tan B=2.451 \end{gathered}[/tex]

tan B = 2.451

Answer :

tanA = 0.41

tanB = 2.45

The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 170 cm2 , what is the length of the diagonal?The length of the diagonal is cm.Give your answer to 2 decimal places.Submit QuestionQuestion 25

Answers

The formula to find the area of a rectangle is:

[tex]\begin{gathered} A=l\cdot w \\ \text{ Where} \\ \text{ A is the area} \\ l\text{ is the length} \\ w\text{ is the width} \end{gathered}[/tex]

Since the rectangle area is 170cm², we can write the following equation.

[tex]170=l\cdot w\Rightarrow\text{ Equation 1}[/tex]

On the other hand, we know that the width of the rectangle is 6 less than twice its length. Then, we can write another equation.

[tex]\begin{gathered} w=2l-6\Rightarrow\text{ Equation 2} \\ \text{ Because} \\ 2l\Rightarrow\text{ Twice length} \\ 2l-6\Rightarrow\text{ 6 less than twice length} \end{gathered}[/tex]

Now, we solve the found system of equations.

[tex]\begin{cases}170=l\cdot w\Rightarrow\text{ Equation 1} \\ w=2l-6\Rightarrow\text{ Equation 2}\end{cases}[/tex]

For this, we can use the substitution method.

Step 1: we replace the value of w from Equation 2 into Equation 1. Then, we solve for l.

[tex]\begin{gathered} 170=l(2l-6) \\ \text{Apply the distributive property} \\ 170=l\cdot2l-l\cdot6 \\ 170=2l^2-6l \\ \text{ Subtract 170 from both sides} \\ 0=2l^2-6l-170 \end{gathered}[/tex]

We can use the quadratic formula to solve the above equation.

[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \\ \text{ For }ax^2+bx+c=0 \end{gathered}[/tex]

Then, we have:

[tex]\begin{gathered} a=2 \\ b=-6 \\ c=-170 \\ l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ l=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(2)(-170)}}{2(2)} \\ l=\frac{6\pm\sqrt[]{1396}}{4} \\ \end{gathered}[/tex]

There are two solutions for l.

[tex]\begin{gathered} l_1=\frac{6+\sqrt[]{1396}}{4}\approx10.84 \\ l_2=\frac{6-\sqrt[]{1396}}{4}\approx-7.84 \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}[/tex]

Since the value of l can not be negative, the value of l is 10.84.

Step 2: We replace the value of l into any of the equations of the system to find the value of w. For example, in Equation 1.

[tex]\begin{gathered} 170=l\cdot w\Rightarrow\text{ Equation 1} \\ 170=10.84\cdot w \\ \text{ Divide by 10.84 from both sides} \\ \frac{170}{10.84}=\frac{10.84\cdot w}{10.84} \\ 15.68\approx w \end{gathered}[/tex]

Now, the long side, the wide side and the diagonal of the rectangle form a right triangle.

Then, we can use the Pythagorean theorem formula to find the length of the diagonal.

[tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Where} \\ a\text{ and }b\text{ are the legs} \\ c\text{ is the hypotenuse} \end{gathered}[/tex]

In this case, we have:

[tex]\begin{gathered} a=10.84 \\ b=15.68 \\ a^2+b^2=c^2 \\ (10.84)^2+(15.68)^2=c^2 \\ 117.51+245.86=c^2 \\ 363.37=c^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{363.37}=\sqrt[]{c^2} \\ 19.06=c \end{gathered}[/tex]

Therefore, the length of the diagonal of the given rectangle is 19.06 cm rounded to 2 decimal places.

The value of an IBM share one day was $ 74.50 more than the value of an AT&T share.

Answers

An algebraic expression we can use to compare the price of IBM shares as being $74.50 more than AT&T shares is x + 74.50, where x is the value of AT&T shares.

What is an algebraic expression?

An algebraic expression consists of variables, terms, constants, and mathematical operations, including addition, subtraction, multiplication, division, and others.

The five algebraic expressions include monomial, polynomial, binomial, trinomial, multinomial.

We can also describe algebraic expressions as falling under the following categories:

Elementary algebraAdvanced algebraAbstract algebraLinear algebraCommutative algebra.

An example of an algebraic expression is 2x + 3y.

Let the value of AT&T share = x

Let the value of IBM share = x + 74.50

Thus, we can, algebraically, conclude that AT&T's share price is x while the price of IBM's share is x + 74.50 on that particular day.

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Find the length to the nearest whole number of the diagonal (hypotenuse) of a square with 30 cm on a side. Round answers to the nearest tenth if necessary. Your answer

Answers

Notice that we can draw a triangle in the square , and that the length of the square's diagonal is the same as the length of the triangle's hypotenuse. The triangle is a right triangle therefore it satisfies the Phytagorean Theorem. To calculate for it's hypotenuse , we will use:

[tex]c^2=a^2+b^2[/tex]

where c is the hypotenuse, and a, b are the other legs of the triangle.

[tex]\begin{gathered} c^2=30^2+30^2 \\ c^2=1800 \\ c=\sqrt[]{1800} \\ c=42.43 \end{gathered}[/tex]

Since the hypotenuse of the triangle is 42.43 cm. Therefore, the square's diagonal is also 42.43 cm

Answer:

The square's diagonal is 42.43 cm

cell phone company A charges a fee of $50 per month plus an additional $0.10 for every minute talked. cell phone company B computes its monthly charge by using the equation y=$0.05 + $75 where y is the total cost and X is the number of minutes talked.

Answers

We will first write A equation

Let x be the number of minutes

y = 0.10x + 50

Comparing the above with y=mx + b where m is the rate of change

m = 0.10

Company B

y = 0.05x + 75

comparing with y =mx + b

rate of change (m) = 0.05

Hence, company A has a higher rate of change at $0.10

5. What is the correlation coefficient for the given data?

Answers

Let us plot the data on the graph to obtain its correlation coefficient.

5) 'r' is known to be the symbol for the correlation coefficient.

Hence, the correlation coefficient from the graph is

[tex]r=0.9741[/tex]

6) There is a strong correlation for the data set because the r-value is larger than 0.7 which is close to +1.

How do I get my answer?

Answers

Answer:

[tex] \frac{2}{9 {d}^{14} } [/tex]

Step-by-step explanation:

[tex] \frac{ {4d}^{ - 5} }{18 {d}^{9} } = \frac{4}{18} \times \frac{ {d}^{ - 5} }{ {d}^{9} } = \frac{2}{9} {d}^{ - 14} = \frac{2}{ {9d}^{14} } [/tex]

Other Questions
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