x-4y=-2
Determine if the given ordered pair, (2, 1), satisfies the given equation.
O YES
O NO

Answers

Answer 1

Answer:

Yes.

Step-by-step explanation:

Replace x with 2 and y with 1 and see if the equation is true

x - 4y = -2

2 - 4(1) = -2

2 - 4 = -2

-2 = -2

The statement is true so the point is on the line.


Related Questions

Dilate this figure by a scale factor of 1.5. Is the image a stretch or compression?

Answers

Answer:

compression

Step-by-step explanation:

If b>1 then the graph will compress

Select whether the pair of lines is parallel, perpendicular, or neither.
y = 5, y = −3

Answers

The given pair of lines are parallel lines.

What are parallel lines?Parallel lines in geometry are coplanar, straight lines that don't cross at any point. In the same three-dimensional space, parallel planes are any planes that never cross. Curves with a fixed minimum distance between them and no contact or intersection are said to be parallel. A line with the same slope is said to be parallel. To put it another way, these lines won't ever cross. At the interception, perpendicular lines always intersect and form right angles.

So, graph the lines on the given equation:

y = 5y = -3

(Refer to the graph attached below)

We can easily look at the graph and tell that the given lines of the equations are parallel to each other.

Therefore, the given pair of lines are parallel lines.

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Use inverse trigonometric functions to solve the following equations. If there is more than one solution, enter all solutions as a comma-separated list (like "1, 3"). If an equation has no solutions, enter "DNE".Solve tan(θ)=1 for θ (where 0≤θ<2π).θ=Solve 7tan(θ)=−15 for θ (where 0≤θ<2π).θ=

Answers

Starting with the equation:

[tex]\tan (\theta)=1[/tex]

take the inverse tangent function to both sides of the equation:

[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]

Yet another value can be found for this equation to be true since the period of the tangent function is π:

[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]

Starting with the equation:

[tex]7\tan (\theta)=-15[/tex]

Divide both sides by 7:

[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]

Take the inverse tangent to both sides of the equation:

[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]

The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:

[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]

We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:

[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]

Therefore, for each equation we know that:

[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex]

Starting with the equation:

[tex]\tan (\theta)=1[/tex]

take the inverse tangent function to both sides of the equation:

[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]

Yet another value can be found for this equation to be true since the period of the tangent function is π:

[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]

Starting with the equation:

[tex]7\tan (\theta)=-15[/tex]

Divide both sides by 7:

[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]

Take the inverse tangent to both sides of the equation:

[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]

The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:

[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]

We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:

[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]

Therefore, for each equation we know that:

[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex]

write the equation of a line through ( -3, 5), parallel to y = -x.

y= ___ x + ___

Answers

Answer:

We took it at grade 7 oh god

1 (Express 360° into centesimal system.)​

Answers

Answer:  400 gradians

Reason:

100 gradians = 90 degrees

4*100 gradians = 4*90 degrees

400 gradians = 360 degrees

In a​ state's lottery, you can bet ​$3 by selecting three​ digits, each between 0 and 9 inclusive. If the same three numbers are drawn in the same​ order, you win and collect ​$500. Complete parts​ (a) through​ (e).

Answers

Using the Fundamental Counting Theorem, it is found that:

a) 1000 combinations are possible.

b) The probability of winning is of 0.001.

c) If you win, the net profit is of $497.

d) The expected value of a $3 bet is of -$2.5.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n trials, each with [tex]n_1, n_2, \cdots, n_n[/tex] possible results, each thing independent of the other, the number of results is given as follows:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

In the context of this problem, three digits that can be repeated are chosen, hence the parameters are:

[tex]n_1 = n_2 = n_3 = 10[/tex]

Hence the number of combinations is:

N = 10³ = 10 x 10 x 10 = 1000.

The order also has to be correct, hence the there is only one winning outcome and the probability is:

p = 1/1000 = 0.001.

You bet $3, and if you win you collect $500, hence the net profit is of:

500 - 3 = $497.

Then the distribution of earnings are as follows:

P(X = -3) = 0.999 -> losing.P(X = 497) = 0.001. -> winning.

Hence the expected value is:

E(X) = -3 x 0.999 + 497 x 0.001 = -$2.5.

Missing information

The problem is given by the image at the end of the answer.

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Use the following function rule to find g(r + 2). Simplify your answer.
g(k)= k-4
g(r + 2) =

Answers

Using the function rule, the value of g(r +2)  is r-2

How can we evaluate and simply g(r + 2) using the function rule?

Given the function rule g(k) = k-4.

Then g(r + 2)  can be found by replacing k with r+2 in the function g(k):

g(r+2) = r + 2 - 4

           = r - 2

Therefore, g(r+2) gives r - 2 when evaluated using the function rule.

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Two numbers have a sum of 35 and a product of 250. what are the numbers

Answers

Answer:

25 and 10

Step-by-step explanation:

because 25+10=35 and 25*10=250

What is 541,000 rounded to the nearest 10,000

Answers

Answer:

540,000

Step-by-step explanation:

Since you are rounding to the 10,000s place you must look at the 1,000s place to decide what to round to. Since in the 1,000s place is a 1, you will round the number down (1 < 5).

robert john and paul start to run from the same point in the same direction around a circular track.
if they take 126 seconds 154 seconds and 198 seconds respectively to complete one round along the track, when will they next meet again at the starting point?

Answers

They will next meet again after 1386 seconds or 23.1 minutes from the starting point.

Given that Robert, John, and Paul begin running in the same direction from the same starting point around a circular track If they take 126 seconds, 154 seconds, and 198 seconds respectively to complete one round of the track.

What is HCF?

The highest common factor is defined as the set of numbers with the highest common multiple. The highest positive integer with more than one factor in the set is HCF.

To determine whether they will next meet again at the starting point.

We have to calculate the LCM of the numbers 126, 154, and 198

⇒ Prime factors of 126 = 2 × 2 × 3 ×7

⇒ Prime factors of 154 = 2 × 11 ×7

⇒ Prime factors of 198 = 2 × 3 × 3 ×11

So the LCM of the numbers 126, 154, and 198 will be as:

⇒ 2 × 3 × 3 × 7 × 11 = 1386

Therefore, they will meet after 1386 seconds or 23.1 minutes.

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State if the three side lengths form an acute obtuse or a right triangle

Answers

Given three side lengths form an acute obtuse or a right triangle 17, 21, & 28

Check Right triangle

[tex]\begin{gathered} Hyp^2=opp^2+adj^2 \\ 28^2=17^2+21^2 \\ 28^2\text{ = 289 +441} \\ 784\text{ }\ne\text{ 730} \end{gathered}[/tex]

Not a Right triangle

An obtuse triangle is a triangle with one obtuse angle and two acute angles. Since a triangle's angles must sum to 180° in Euclidean geometry.

Not an Obtuse triangle

[tex]\begin{gathered} \sin \text{ A= }\frac{opp}{hyp} \\ \sin \text{ A = }\frac{17}{28} \\ A=sin^{-1}\text{ }\frac{17}{28} \\ A=37.4^0\text{ (less than 90)} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ B = }\frac{adj}{hyp} \\ \cos \text{ B = }\frac{21}{28} \\ B=cos^{-1}\frac{21}{28} \\ B=41.4^0\text{ (less than 90)} \end{gathered}[/tex]

[tex]\begin{gathered} \tan \text{ C = }\frac{opp}{adj} \\ \tan \text{ C = }\frac{17}{21} \\ C=tan^{-1\text{ }}\frac{17}{21}\text{ } \\ C=38.9^0\text{ (less than 90) } \end{gathered}[/tex]

Hence it is acute angle because all angles are less than 90°

solve the equation, and enter the solutions from least to greatest. If there is only one solution, enter “n.a” for the second solution. (picture of equation listed below)

Answers

Answer

x = 1 or x = n.a.

Step-by-step explanation

[tex]\frac{1}{x}+\frac{1}{x-10}=\frac{x-9}{x-10}[/tex]

Multiplying by (x - 10) at both sides of the equation:

[tex]\begin{gathered} (x-10)(\frac{1}{x}+\frac{1}{x-10})=\frac{x-9}{x-10}(x-10) \\ \text{ Distributing and simplifying:} \\ \frac{x-10}{x}+\frac{x-10}{x-10}=x-9 \\ \frac{x-10}{x}+1=x-9 \end{gathered}[/tex]

Multiplying by x at both sides of the equation:

[tex]\begin{gathered} x(\frac{x-10}{x}+1)=x(x-9) \\ \text{ Distributing and simplifying:} \\ \frac{x(x-10)}{x}+x=x^2-9x \\ x-10+x=x^2-9x \\ 2x-10=x^2-9x \end{gathered}[/tex]

Subtracting 2x and adding 10 at both sides of the equation:

[tex]\begin{gathered} 2x-10-2x+10=x^2-9x-2x+10 \\ 0=x^2-11x+10 \end{gathered}[/tex]

We can solve this equation with the help of the quadratic formula with the coefficients a = 1, b = -11, and c = 10, as follows:

[tex]\begin{gathered} x_{1,2}=\frac{-b\pm{}\sqrt{b^2-4ac}}{2a} \\ x_{1,2}=\frac{11\pm\sqrt{(-11)^2-4\cdot1\operatorname{\cdot}10}}{2\operatorname{\cdot}1} \\ x_{1,2}=\frac{11\pm\sqrt{81}}{2} \\ x_1=\frac{11+9}{2}=10 \\ x_2=\frac{11-9}{2}=1 \end{gathered}[/tex]

The solution x = 10 is not possible because it makes zero the denominator in 2 of the rational expressions of the original equation. In consequence, it must be discarded.

Find the difference between the product of 2.5 and 7.5and the sum of 2.7 and 9.55

Answers

Difference = subtraction
Sum = addition
Product = addition

2.5 x 7.5 = 18.75
2.7 + 9.55 = 12.25

12.25 - 18.75 = 6.5

Your answer is 6.5

Apologies if anything is incorrect.

H=4(x+3y)+2 make x the subject

Answers

When the formula is changed to x, the equation is x = 1/4(H - 2) - 3y

How to change the subject of formula to x?

From the question, the equation is given as

H=4(x+3y)+2

Subtract 2 from both sides of the equation

So, we have the following equation

H - 2 =4(x+3y)+2 -2

Evaluate

So, we have the following equation

H - 2 =4(x+3y)

Divide by 4

So, we have the following equation

1/4(H - 2) = x +3y

Subtract 3y from both sides

x = 1/4(H - 2) - 3y

Hence, the solution for x is x = 1/4(H - 2) - 3y

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The simplified solution for the x is given as x = [H - 2]/4 - 3y.

As mentioned in the question, for an equation H=4(x+3y)+2, we have to transform the equation in the subject of x.

What is simplification?

The operational processes in mathematics for the functions to make the function or expression.

Here,
Simplification of the given function in a manner to the transposition of the variables from left to right,
H=4(x+3y)+2
H - 2 = 4(x+3y)
{H - 2}/4 = (x + 3y)
x = [H - 2]/4 - 3y

Thus, the simplified solution for the x is given as x = [H - 2]/4 - 3y.

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In a class of 25 students, 15 of them have a cat, 16 of them have a dog and 3 of them have neither. Find the probabillity that a student chosen at random has both

Answers

9/25

Step-by-step explanation:

students who have either cat or dog =25-3=22

students who have both= 15+16-22= 9

probabillity that a student chosen at random has both= 9/25

The equation y =1/2x represents a proportional relationship. What is the constant of proportionality? A. 1/2B.XC. 0D. 2

Answers

You have the following equation:

y = 1/2 x

In order to determine what is the constant of proportionality, consider the general fom of proportional relatioship, given by:

y = kx

by comparing the previous equation with y=1/2x, you can notice that k=1/2. Then, the constant of proportionality is k = 1/2

A. 1/2

Please help soon thanks

Answers

Answer:

x = -7

Step-by-step explanation:

Hi!

Ok we know a few things here :

x + 37 + x + 67 + 90 = 180

2x + 104 = 90

2x = -14

x = -7

Hope this helps!

Have a great day! :)

you are given the set of letters {a, b, c, d, e}. what is the probability that in a random five-letter string (in which each letter appears exactly once, and with all such strings equally likely) the letters a and b are next to each other?

Answers

The probability in a random 5-letter string that the letters A and B will be next to each other is 0.4

For A and B to be next to each other in the string of letters, consider the AB combination as a single letter. Then there are 4 letters AB, C, D, E.

There are 4! number of ways to arrange these 4 letters.

4!= 24 ways

We know AB can be arranged in 2 ways ( AB or BA)

Thus there are 2*4! = 2 * 24 = 48 ways of arranging the letters such that AB are next to each other in the random Five-letter string

The total number of letters in the set is 5 and hence the total number of ways the letters can be arranged without A and B necessarily being next to each other  is 5!= 120

The probability that in the random five-letter string( in which each letter appears only once with all such strings equally likely) the letters A and B are next to each other= 48/120 = 0.4

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Number 6. For questions 5-7, (a) use synthetic division to show that x is a zero.(b) find the remaining factors of f(x).(c) use your results to find the complete factorization of f(x).(d) list all zeros of f(x).(e) graph the function.

Answers

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given polynomials

[tex]f(x)=x^3+6x^2-15x-100[/tex]

One of the zeroes is:

[tex]\begin{gathered} x=-5 \\ \text{this implies that:} \\ (x+5)=0 \end{gathered}[/tex]

STEP 2: Use synthetic division to divide the polynomials

[tex]\frac{x^3+6x^2-15x-100}{x+5}[/tex]

Write the coefficients of the numerator

[tex]1\:\:6\:\:-15\:\:-100[/tex][tex]\begin{gathered} \mathrm{Write\:the\:problem\:in\:synthetic\:division\:format} \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ Carry\:down\:the\:leading\:coefficient,\:unchanged,\:to\:below\:the\:division\: \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ 6-5=1 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -15-5=-20 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ \left(-20\right)\left(-5\right)=100 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -100+100=0 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:0}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{The\:last\:carry-down\:value\:is\:the\:remainder} \\ 0 \end{gathered}[/tex]

The last carry-down value is the remainder and it is 0 (zero)

Since the remainder is a zero, hence, x=-5 is a zero

Step 3: Answer question b

To get the factors, the remainder of the division in step 2 is given as:

The remaining factors of f(x) is:

[tex]x^2+x-20[/tex]

STEP 4: Answer Question c

[tex]\begin{gathered} roots=(x+5)(x^2+x-20) \\ Factorize\text{ the other root to have:} \\ Using\text{ factorization methods:} \\ (x^2+x-20)=(x^2+5x-4x-20) \\ x(x+5)-4(x+5)=0 \\ (x-4)(x+5)=0 \end{gathered}[/tex]

The complete factorization will give:

[tex](x+5)(x-4)(x+5)[/tex]

STEP 5: Answer question d

The zeroes of f(x) will be:

[tex]\begin{gathered} zeroes\text{ of f\lparen x\rparen=?, we equate the roots to 0} \\ zeroes\Rightarrow x=-5,4,-5 \end{gathered}[/tex]

zeroes are: -5,4,-5

STEP 6: Plot the graph

HELP ME PLSS OMLLL IMA CRY

Answers

Answer:

1. 15 pieces of candy

2. 48 cows

3. 81 ham subs

4. 24 boys

Step-by-step explanation:

Assuming you just want answers, but lmk

a candy manufacturer produces bags of jelly beans. the weight of a bag of jelly beans is normally distributed with a mean of 12 ounces and a standard deviation of 0.4 ounces. if a random sample of 4 bags of jelly beans is selected what is the probability that the sample average will be greater than 11.6?

Answers

Answer:

2%

Step-by-step explanation:

IMAGE ATTACHED,please help me with this one please.​

Answers

Answer:

Step-by-step explanation:

$17.64 for 147 text messages

Therefore 1 text message will cost $17.64 ÷ 147 Text messages = $0.12 per text message

I'm not good with graphs so I really need help solving and understanding this

Answers

4x + 3y = -24 (option D)

Explanation:

To determine the orrect option, we would find the equation of the line.

Equation of line: y = mx + c

m = slope, c = y-intercept

Using the slope formula:

[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]

We would use any two points on the graph.

Using points: (-6, 0) and (0, -8)

[tex]\begin{gathered} x_1=-6,y_1=0,x_2=0,y_2\text{ = -}8 \\ m\text{ = }\frac{-8-0}{0-(-6)} \\ m\text{ =}\frac{-8}{0+6}=\frac{-8}{6}=-\frac{4}{3} \end{gathered}[/tex]

The y intercept is the point where the line crosses the y axis and the value of x is zero. It crosses the line at y = -8

The equation of line becomes:

y = -4/3 x + (-8)

y = -4/3x - 8

Multiply both sides by 3:

3(y) = 3(-4/3 x) - 3(8)

3y = -4x - 24

3y + 4x = -24

4x + 3y = -24 (option D)

Analyze each situation. Identify a reasonable

domain and range. Is the domain continuous of discrete? A car has a 15-gallon gas tank and gets a

maximum average mileage of 24 miles per gallon

Answers

The reasonable domain and range are 0 ≤ x ≤ 15 and 0 ≤ y ≤ 360 and the domain is continuous

How to determine the domain?

From the question, we have

Gallon gas tank = 15 gallons

This means that the domain is from 0 to 15

This can be represented as 0 ≤ x ≤ 15

For the range, we have

0 * 24 ≤ y ≤ 15 * 24

Evaluate

0 ≤ y ≤ 360

This means that

Range: 0 ≤ y ≤ 360

Lastly, the domain of the situation is continuous

This is so because the gas in the tank can take decimal values

Take for instance, the gas tank can be 13.79 gallons full

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Kadeesha invested
$
900
$900 in an account that pays 1.5% interest compounded annually. Assuming no deposits or withdrawals are made, find how much money Kadeesha would have in the account 11 years after her initial investment. Round to the nearest tenth (if necessary).

Answers

The amount of money Kadeesha have in her account after 11 years is $1059.3.

What is the compound interest?

Compound interest is the interest on savings calculated on both the initial principal and the accumulated interest from previous periods.

The formula used to find the compound interest = [tex]A=P(1+\frac{r}{100})^{nt}[/tex]

Given that, principal=$900, rate of interest=1.5% and time period =11 years.

Now, amount =900(1+1.5/100)¹¹

= 900(1+0.015)¹¹

= 900(1.015)¹¹

= 900×1.177

= $1059.3

Therefore, the amount of money Kadeesha have in her account after 11 years is $1059.3.

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4x5y² how would I solve this

Answers

multiply the numbers

Answer is 20y2

discuss the advantages and disadvantages of using sampling to reduce the number of data objects. would simple random sampling (without replacement) be a good approach to sampling? why or why not? what kind of sampling method that you would like to use? g

Answers

The advantage and disadvantages are sampling.

We have to find the advantages and disadvantages of sampling.

The advantage is:

It provides an opportunity to do data analysis with a lower likelihood of carrying an error.

Researchers can analyze the data that is collected with a smaller margin of error thanks to random sampling. This is permitted since the sampling procedure is governed by predetermined boundaries. The fact that the entire procedure is random ensures that the random sample accurately represents the complete population, which enables the data to offer precise insights into particular topic matters.

The disadvantage is:

No extra information is taken into account.

Although unconscious bias is eliminated by random sampling, deliberate bias remains in the process. Researchers can select areas for random sampling in which they think particular outcomes can be attained to confirm their own bias. The random sampling does not take into account any other information, however sometimes the data collector's supplementary information is retained.

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Find the distance between the points (-16, 20) and (-16, -14).2007(-16, 20)161284-20-16-12-8-4048121620-4-812(-16, -14)-16-20units

Answers

Point A

(-16, 20)

Point B

(-16, -14)

The Distance Formula itself is actually derived from the Pythagorean Theorem which is a

[tex]\begin{gathered} d=\sqrt[]{(y2-y1)^2+(x2-x1)^2} \\ d=\sqrt[]{(-14-20)^2+(-16-(-16}))^2 \\ d=\sqrt[]{(-34)^2} \\ d=34 \end{gathered}[/tex]

The distance would be 34 units

write equations for the horizontal and vertical lines passing through the point (-7,7)​

Answers

The vertical line's equation is x = -7, and the horizontal line's equation           Is y = 7.

Where the given coordinate is (--7, 7).

What do the vertical and horizontal lines represent?A vertical line is a straight, up and down line in a coordinate plane that is parallel to the y-axis. The horizontal line, on the other hand, is parallel to the x-axis and goes straight, left, and right.

Here,

The value of x, is -7, will never change for a vertical line. This holds true for any y value. As a result, x = -7 is the equation for a vertical line passing through this point.

Similarly, the value of y, is 7, will never change for a horizontal line. This holds true for any x value. As a result, the equation for a horizontal line passing through this point is as follows: y =7

As a result, the vertical line's equation is x = -7 and the horizontal line's equation is y = 7.

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You are designing a poster with an area of 625 cm2 to contain a printing area in the middle and have the margins of 4cm at the top and bottom and 7cm on each side. Find the largest possible printing area. Round your answer to the nearest four decimal places.

Answers

Explanation

Step 1

diagram

so

Step 2

let

a)total area= side1*side 2

[tex]side1*side2=625[/tex]

and

[tex]printing\text{ area=\lparen side1-14\rparen\lparen side2-8\rparen}[/tex]

Step 3

solve

let

[tex]\begin{gathered} side1\text{ = x} \\ side\text{ 2 =}y \\ xy=625 \\ y=\frac{625}{x}\Rightarrow equation(1) \end{gathered}[/tex]

and in teh second equation we have

[tex]\begin{gathered} pr\imaginaryI nt\imaginaryI ng\text{area=}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times1-14}\operatorname{\rparen}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times2-8}\operatorname{\rparen} \\ A=(x-14)(y-8) \\ replace\text{ the y value} \\ A=(x-14)(\frac{625}{x}-8) \\ A=\frac{625x}{x}-\frac{8750}{x}-8x+112) \\ A=-\frac{8750}{x}+625-8x+112 \\ dA=\frac{8750}{x^2}-8 \\ \frac{8750}{x^2}=8 \\ x^2=\frac{8750}{8} \\ x=\sqrt[\placeholder{⬚}]{\frac{8750}{8}} \\ x=33 \end{gathered}[/tex]

replace to find y

[tex]\begin{gathered} y=\frac{625}{x}\operatorname{\Rightarrow}equat\imaginaryI on(1) \\ y=\frac{625}{33} \\ y=18.9 \end{gathered}[/tex]

so

the maximum area is

[tex]\begin{gathered} x-14=33-14=19 \\ y-8=18.9=10.9 \end{gathered}[/tex]

so, the greates area is

[tex]18.9*19=359.1[/tex]

therefore, the answer is

359.1 square cm

I hope this helps you

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