3x+2y=2 in slope intercept form!!!

According to the following graph, the reflection point of (-3,4) across the line y = 2 is (-3,0).

The term reflection point in math is defined as

Here we need to find the reflection point of (-3, 4) across the line y = 2.

Here first we have t plot the point on the graph, and then we have to plot the line equation y = 2 on the graph,

Here in order to find the reflection of a point along the x-axis, then we have to keep the abscissa constant and see the reflection of ordinate along the x-axis.

Based on these rule, the resulting graph is obtained as follows.

Through the graph we have identified that the resulting reflection point is (-3,0).

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Answer:

52 bags and 2 extra

Step-by-step explanation:

Simply divide the no. of rolls by no. of rolls per bag

Answer:

12 bags

Step-by-step explanation:

158 rolls ÷ 13 rolls/bag = 12.15 bags

You will have 12 full bags plus about 2 extra rolls left over, which you can try to sell separately or as a "pack" of 2.  Or, you can sell one overstuffed bag with 2 extra rolls in it (for a total of 15 rolls) for a little extra on the price.

Answer:

7. .44

8. .009

Step-by-step explanation:

7. 44 ÷ 100 = .44

8. 9 ÷ 1000 = .009

27 2/10

   or

27 1/5

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

Therefore , the solution of the given problem of area comes out to be AB/DC =5 Imagine that your square is made out of more compact unit squares.

Area is the quantity that describes how much space .A surface or an object's shape occupy. Use a pencil to mark a square on the paper. Specifically, a 2-D figure. A shape's area on paper is referred to as its "Area" in this sentence. Imagine that your square is made out of more compact unit squares.

Since the heights of both trapezoids are equal, and the area of ABEF is twice the area of FECD, AB+ EF/2 = 2(DC+EF /2) AB + EF / 2= DC + EF,

AB + EF = 2DC + 2EF. EF is exactly halfway between AB and DC, so EF = AB + (AB+ DC)/2 = 2DC+ AB + DC, so 3/2 AB +  1/2DC = 3DC+AB, 1/2AB =5/2DC => AB/DC =5 Therefore , the solution of the given problem of area comes out to be AB/DC =5.

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By drawing a number line and then marking it as 0 and 1, and by dividing the distance in 3 equal parts we can graph 1/3 on the number line, where each part will be equal to the 1/3.

A number line is a diagram of a graded straight line used to represent real numbers in introductory mathematics. It is assumed that every point on a number line corresponds to a real number, and that every real number corresponds to a point.

A horizontal line with evenly spaced numerical increments is referred to as a number line. How the number on the line can be answered depends on the numbers present.

For drawing 1/3 on a number line, we will follow the following steps:

We will first draw a line and mark 0 and 1 on it.

In between 0 and 1, we will divide the total distance into 3 equal parts, where one part will represent (1/3)th portion.

Hence. by marking there we can locate 1/3 on the number line.

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The correct question may be like:

How do you plot 1/3 on a graph.

If the coach buys the packs of Gummi Bears at $0.60 and resales them at $1 , then the percent markup for the candy is (d) 67% .

The price for which the Coach buys Gummi Bears is = $0.60 ;

the price at which the Coach resales them is = $1 ;

So , the Markup is the difference in the price of purchase and price of resale of the candy ,

that means ; the [tex]Markup = \$1 - \$0.60[/tex] ;

= $0.4

So , the Percent Markup is = [tex]\frac{0.4}{0.6} \times 100[/tex]

= 66.66666...

≈ 67% .

Therefore , the percent Markup for the Candy is 67% .

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The only difference is a -2 in x in 1 transformation in 2 transformation. The function is shifted up by b units with f (x) + b.

Transformations can be divided into four categories: translation, reflection, rotation, and dilation. Rotate, reflect, or translate the geometric figures on a coordinate plane. The label given to a function, f, that maps to itself is the transformation, or f: X X. The pre-image X is transformed into the picture X after the transformation. It is possible to utilize any operation, or a combination of operations, in this transformation, including translation, rotation, reflection, and dilation.

given

J(2, 2), (3, 4), H(4,3), G(3,0) to J'(-1,2), I'(0, 4), H'(1, 3), G'(0, 0)

in 2 transformation as the only change is that -2 in x in 1 transformation

The function is shifted up by b units with f (x) + b.

The function is shifted downward by b units when f (x) b.

The function is moved left by b units when f (x + b) is used.

The function is moved right b units by the expression f (x b).

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y < 4/5x - 1/5 ,y > -4/5x + 1/5, y < 2x + 6, y > 2x + 6 are all in slope-intercept form

The slope intercept form in math is one of the forms used to calculate the equation of a straight line, given the slope of the line and intercept it forms with the y-axis. The slope intercept form is given as, y = mx + b, where 'm' is the slope of the straight line and 'b' is the y-intercept.

so the given equation are also in the form of slope intercept form

y < 4/5x - 1/5

y > -4/5x + 1/5

y < 2x + 6

y > 2x + 6

are all in slope-intercept form (y = mx + b) where m is the slope and b is the y-intercept.

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The equations which show direct proportion between a and b are:

a=2b and a=b/2.

Direct proportion or direct variation is the relation between two quantities where the ratio of the two is equal to a constant value. It is represented by the proportional symbol, ∝. In fact, the same symbol is used to represent inversely proportional, the matter of the fact that the other quantity is inverted here.

Example: x and y are two quantities or variables which are linked with each other directly, then we can say x ∝ y. When we remove the proportionality symbol, the ratio of x and y becomes equal to a constant, such as x/y = C, where C is a constant.

Given, two variables a and b

If two variables a and b are in direct proportion

then,

a/b = constant.

1. a = b²

a/b = b

which is not constant

⇒ a is not in direct proportion with b.

2. a = 2b

a/b = 2

which is constant

⇒a is in direct proportion with b.

3. a = b/2

a/b = 1/2

which is constant

⇒ a is in direct proportion with b.

4. a=b+2

a/b = 1+2/b

which is not constant

⇒ a is not in direct proportion with b.

5. a = 2/b

ab = 2

⇒a is inversely proportional to b.

Hence, In equation a=2b and a=b/2, a is directly proportional to b.

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The proportion and percentage is as follows;

[tex]$\begin{array}{cccc}\text { Score Interval } & f & \text { Proportion } & \text { Percentage } \\ 9-10 & 29 & 0.19 & 19 \% \\ 7-8 & 53 & 0.34 & 34 \% \\ 5-6 & 50 & 0.32 & 32 \% \\ 3-4 & 22 & 0.14 & 14 \% \\ 1-2 & 1 & 0.01 & 1 \%\end{array}$[/tex]

A percentage represents a ratio or fraction with a denominator that is always 100 as opposed to a proportion, which asserts the equivalent of two ratios or fractions.

Given

[tex]$\begin{array}{cccc}\text { Score Interval } & f & \text { Proportion } & \text { Percentage } \\ 9-10 & 29 & 0.19 & 19 \% \\ 7-8 & 53 & & \\ 5-6 & 50 & & \\ 3-4 & 22 & 0.14 & 14 \% \\ 1-2 & 1 & 0.01 & 1 \%\end{array}$[/tex]

To estimate the total frequency:

Total = sum f

Total = 29 + 53 + 50 + 22 + 1

Total = 155

The proportion (p) exists calculated by:

[tex]$p = \frac{f}{\text { Total }}[/tex]

The percentage (P) exists calculated by:

P = p × 100 %

For interval 7 - 8, we have:

p = 53 / 155 = 0.34

P = 0.34 × 100 % = 34 %

For interval 5 - 6, we have:

p = 50 / 155 = 0.32

P = 0.32 × 100 % = 32 %

So, the complete table is:

[tex]$\begin{array}{cccc}\text { Score Interval } & f & \text { Proportion } & \text { Percentage } \\ 9-10 & 29 & 0.19 & 19 \% \\ 7-8 & 53 & 0.34 & 34 \% \\ 5-6 & 50 & 0.32 & 32 \% \\ 3-4 & 22 & 0.14 & 14 \% \\ 1-2 & 1 & 0.01 & 1 \%\end{array}$[/tex]

The complete question is:

A statistics content developer at Aplia wanted to know whether study skills are related to memory quality. She invited student volunteers to perform an online memory task. The students saw a list of 60 words and were then asked to recognize a list of five brand new words that are related to words that were on the original list. Students were also asked to provide their GPAs.

Consider the following data set, which was collected from student volunteers in 2009. The table gives the frequency for the number of incorrectly identified related words. Use the dropdown menus to complete the table by filling in the missing values for the proportions and percentages.

Score Interval f Proportion Percentage

9â10 29 0.19 19%

7â8 53

5â6 50

3â4 22 0.14 14%

1â2 1 0.01 1%

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We would need to multiply the first equation by -4 to eliminate the y variable when solving the system of equations by elimination.

The first equation is: 3x + 4y = 5

To eliminate the y variable when solving the system of equations by elimination, we need to multiply the first equation by -4. This is because when two equations are multiplied by the same number, any terms that have the same variable will be eliminated when the equations are added together.

Formula:

3x + 4y = 5

-4(3x + 4y = 5)

3x + 4y = 5

-12x - 16y = -20

Thus, we would need to multiply the first equation by -4 to eliminate the y variable when solving the system of equations by elimination.

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The pharmacist utilized a 10% percentage mixture, according to the solution, in a quantity of 2 ounces.

To calculate the percentage, we must first determine the value's ratio to the sum of its parts. We must then multiply the result by 100.

Comparing two quantities while rebasing the second quantity to 100 is the most fundamental use of percentages. If the proportion of employed women who are women is of relevance to us, let's say.

We are aware that m = CV, where

V = volume and c = concentration

Let

M is equal to 6.5%, m' is equal to 7.9%, and m = 3.7%.

It is known that m + m' = M.

where cv + c'v' = CV

Since cv + c'v' = CV, the formula is:

c = 3.7% concentration = 0.037;

v = volume of 3.7% = 1 ounce;

c = 7.9% concentration = 0.079;

v' = volume of 7.9%;

C = 6.5% concentration = 0.065;

and V = volume of 6.5% = v + v'

We obtain by altering the values of the variables in the equation

3.7% × 1/3.7% + 7.9 %v' = 6.5%(v + v')

3.7% + 7.9 %v' = 6.5%(1 + v')

3.7% + 7.9 %v' = 6.5% + 6.5%v'

7.9%v' - 6.5%v' = 6.5% - 3.7%

1.4%v' = 2.8%

v' = 2.8%/1.4%

2 ounces are v'.

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Answer: 1/2

Step-by-step explanation:

The question is asking you to find how many centimeters in the scale drawing equal 1 foot on the actual building. The answer will be 1/2 because 1 centimeter is equal to 2 feet in reality. But since we want to know the answer of how many centimeters is in 1 foot, we will divide that in half, to get an answer of 1/2 centimeter, or A.

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Answer:

n = - 1 , n = 3

Step-by-step explanation:

n² - 2n - 3 = 0 ( add 3 to both sides )

n² - 2n = 3

to complete the square

add ( half the coefficient of the x- term)² to bpth sides

n² + 2(- 1)n + 1 = 3 + 1

(n - 1)² = 4 ( take square root of both sides )

n - 1 = ± [tex]\sqrt{4}[/tex] = ± 2 ( add 1 to both sides )

n = 1 ± 2

Then

n = 1 - 2 = - 1

n = 1 + 2 = 3

Answer: The Gradient of line A and B are

2 and -1

Step-by-step explanation:

For the given two points A(x1, y1) and B(x2, y2)

The gradient of the line AB is y2-y1/x2-x1

So the gradient of line A is

5-1/2-0

2

the gradient of line B is

5-0/0-5

-1

An inequality to find how many days he will have to go kayaking. is; d ≥ 4

We are told that Jay wants to go paddle boarding at least 8 hours each week. This means greater than or equal to 8. That is the minimum number of hours each week is 8 hours.

Now, we are told that he averages 2 hours per day. If the number of days is given by d, then we have the inequality as;

2d ≥ 8

Divide both sides by 2 to get;

d ≥ 4

That is the domain of the number of days required to go kayaking

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Answer:

2

Step-by-step explanation:

0.2(x+50)-6=0.4(3x+20) multiply both sides by 10 to clear the decimals

2(x + 50) = 4(3x + 20) distribute across the terms in the parentheses

2x + 100 = 12x + 80 subtract 80, 2x from both sides

20 = 10x divide both sides by 10

x=2

The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]

[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]

Eigenspace corresponding to lambda=5,1,4

The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]

for lambda=5

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]

Reducing the matrix A-5I by elementary row operations

[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]

[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]

[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]

The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix

[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis

[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]

we reduce the matrix A-I by elementary row operations as follows.

[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]

[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]

Thus, the solutions x of (A-I) X=0 satisfy

[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]

[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]

The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]

case:3

lambda=4

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]

we reduce the matrix A-4I by elementary row operations as follows.

[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]

[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]

Thus, the solutions x of (A-4IX)=0 satisfy

[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]

[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]

The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]

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The solution indicates 40 positive three-digit integers of this kind are available.

The numbers used for counting as well as organizing are known as natural numbers in mathematics. Cardinal numbers and ordinal numbers are two different types of numbers that are used for counting and ordering, respectively.

Left of zero on the number line, a negative number is an integer. It is not zero, though.

Any solution x will mod 23 will also have x+23n as a solution, for some integer n.

Since;

900/23 = 39 3/23,

we know:

There are 39 or 40 three-digit integers of this form.

As it happens,

100 is the smallest 3-digit solution. So, there are 40 three-digit numbers that are of the form 100 +23n,

hence 40 solutions to the equation.

The equation reduces, mod 23, to

10x = 11

Its solutions are x = 23n +8.

This shows There are 40 three-digit positive integers of this form.

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On solving the provided question, we can say that the required probability is = 13/204.

Probability theory, a subfield of mathematics, gauges the likelihood of an occurrence or a claim being true. An event's probability is a number between 0 and 1, where approximately 0 indicates how unlikely the event is to occur and 1 indicates certainty. A probability is a numerical representation of the likelihood or likelihood that a particular event will occur. Alternative ways to express probabilities are as percentages from 0% to 100% or from 0 to 1. the percentage of occurrences in a complete set of equally likely possibilities that result in a certain occurrence compared to the total number of outcomes.

probability of 1st card =  1/4

since the card is not replaced

total number remaining cards = 51

second card 13/51

the required probability is = 13/204

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Answer:

[tex]\sqrt[3]{x^{9} x}[/tex]

Step-by-step explanation:

When you are multiplying powers that have the same bases (in this case x) you add the exponents. If you do not see an exponent, the exponent is 1.

9 + 1 = 10

When the treasurer is solving a problem using the expression shown. -23.50 ÷ 5, the problem that the treasurer is lost likely solving is D. A total of -$23.50 in dues is owed to the club. There are 5 club members who each owe the same amount in dues. What is the balance,in dollars, of each members account?

From the information, the club treasurer is solving a problem using the expression shown. -23.50 ÷ 5. An expression show the relationship between the variables.

In this case, a total of $23.50 in dues is owed to the club. There are 5 club members who each owe the same amount in dues.

The correct option is D.

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we can find out the polynomial by the degree of the Polynomial.

Polynomial

A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using mathematical operations such as addition, subtraction, multiplication and division

Degree of polynomial

The degree of a polynomial is defined as the highest exponent of a monomial within a polynomial. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial.

Types of polynomial

Depending upon the number of terms, polynomials are divided into the following categories:

1-Monomial

2-Binomial

3-Trinomial

4-Polynomial containing 4 terms (Quadronomial)

5-Polynomial containing 5 terms (pentanomial ) and so on.

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The factor of expression 3x² 12xy is 3x². To find the factor, we need to divide 12xy by 3x². We start by dividing the coefficients, 12 divided by 3 is 4. Then we divide the x terms, x divided by x is 1. Finally, we divide the y terms, y divided by y is 1. Therefore, the factor of 3x² 12xy is 3x².

The factor of  expression 3x² 12xy is 3x². To find the factor, we need to divide 12xy by 3x². We start by dividing the coefficients, 12 divided by 3 is 4. Then we divide the x terms, x divided by x is 1. This means that the x part of the factor is 3x. Next, we divide the y terms, y divided by y is 1. This means that the y part of the factor is y. When we combine the two parts, 3x and y, we get the factor of 3x². Therefore, the factor of 3x² 12xy is 3x². This means that 3x² is a factor of 12xy, which can be seen by multiplying 3x² by 4y, which results in 12xy. This shows that 3x² is a factor of 12xy.

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Step 3 is Define the problem Goals.

there are 8 steps, which are as follows,

1- Define the Problem

2- Clarification of the Problem

3- Define the problem Goals

4- Identify main Cause of the Problem

5- Develop a Action Plan

6- Execute that  Action Plan

7- Analyze the Results

8- Continuous Improvements

Problem solving

Problem solving is the method of determining the problem and then clarify the doubts and prepare a action plan and then execute the action plan and find out the desired results and if the results are not appropriate then improve the action plan and your way of doing until you get the desired results.

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a. The ratio of rows of corn to beans is 13 : 12 or 13/12 or 13 to 12.

b. The ratio of rows of lettuce to the total number of rows in the garden is 5 : 52 or 5 to 52 or 5/52.

a. The ratio using the word 'to' is corn to beans.

In its most basic form, a ratio is a comparison between two comparable quantities.

There are two types of proportions One is the direct proportion, whereby increasing one number by a constant k also increases the other quantity by the same constant k, and vice versa.

If one quantity is increased by a constant k, the other will decrease by the same constant k in the case of inverse proportion, and vice versa.

We know a ratio between a and b can be written as a : b, a/b or a to b.

From the given information the ratio of rows of corn to beans is,

corn/beans = 13/12 Or 13 : 12 Or 13 to 12.

The ratio of rows of lettuce to the total number of rows in the garden is,

lettuce/total number of rows in the garden = 5/52 Or 5 : 52 Or 5 to 52.

The ratio using the word 'to' is corn to beans.

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The answer of the equation "2x^2 -  4x - 6 =  0" is -1 and 3.

The equation is solved using factorization:

2x^2 -  4x - 6 =  0        , given equation

2x^2 -  6x + 2x - 6 =  0     , factorizing

2x(x - 3) + 2 ( x - 3 ) = 0    , taking 2x common from the first two terms and 2 common from the last two terms

(2x + 2) (x - 3) = 0

now solving further

( 2x + 2 ) = 0   &   ( x - 3 ) =0

2x = (0 - 2)    ,          x  = (0 +  3)

2x = (-2)        ,         x = 3

x = (-2/2)

x = -1

Thus, the equation gives the solution as x = -1 and x = 3.

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The corresponding equation of the proportional relationship will be;

⇒ y = - 1/4x

Any relationship that is always in the same ratio and quantity which vary directly with each other is called the proportional.

Given that;

The graph of a proportional relationship contains the point (-12, 3).

Now,

Let the equation of the proportion is,

⇒ y = kx

Where, 'k' is constant of proportion.

Since, The graph of a proportional relationship contains the point (-12, 3).

Hence, Substitute x = 0 12 and y = 3 in above equation we get;

⇒ 3 = k × - 12

⇒ 3 / (-12) = k

⇒ k = - 1/4

Thus, The equation of the proportion is,

⇒ y = - 1/4x

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Answer:

The total cost of the science class trip to the zoo is $1200

Step-by-step explanation:

The cost of the science class trip to the zoo is $1200, and the cost of Mr. Sexton's class trip to the state park is $1000.