What is the point of intersection when the system of equations below is graphed on the coordinate plane?
x-y=1
y-x=1
A) no intersection point
B)infinite intersection points
C)(1, 0)
D)(–1, 0)

The expression for the 1 less than the quotient of a number n and 6 is n/6 - 1.

It is defined as the combination of constants and variables with mathematical operators.

It is given that:

The word expression:

1 less than the quotient of a number n and 6:

Here n is the number:

A number is a mathematical entity that can be used to count, measure, or name things. For example, 1, 2, 56 etc. are the numbers.

The quotient of a number n and 6 = n/6

1 less than the quotient of a number n and 6:

n/6 - 1

If in the linear expression, one variable is present, then the expression is known as the linear expression in one variable.

Thus, the expression for the 1 less than the quotient of a number n and 6 is n/6 - 1.

Learn more about the expression here:

brainly.com/question/14083225

#SPJ2

Answer:

Last graph is the answer.

Step-by-step explanation:

The inequalities are y < [tex]\frac{2}{3}x[/tex] and y ≥ -x + 2

Now first inequality y < [tex]\frac{2}{3}x[/tex] will represent a dotted line passing through origin and a point (3, 2)

This inequality will be in the form a dotted line and shaded area will be below the line.

Second inequality is y ≥ -x + 2

Line representing the inequality will be solid in shape and shaded area will be above this line.

Common shaded area of both the inequalities matches with the last option.

Therefore, Last option (second row last figure) will be the answer.

Answer:

[tex]\frac{b^2}{2}[/tex]

Step-by-step explanation:

We are supposed to write an algebraic expression for given word phrase.

The square of b would be 'b' raised to second power: [tex]b^2[/tex].

Half the square of b would be [tex]b^2[/tex] divide by 2: [tex]\frac{b^2}{2}[/tex].

Therefore, our required expression would be [tex]\frac{b^2}{2}[/tex].

we know that

The distance 's formula between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

Step [tex]1[/tex]

Find the distance MN

[tex]M(-2,1)\\N(-1,4)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(4-1)^{2}+(-1+2)^{2}}[/tex]

[tex]d=\sqrt{(3)^{2}+(1)^{2}}[/tex]

[tex]dMN=\sqrt{10}\ units[/tex]

Step [tex]2[/tex]

Find the distance NL

[tex]N(-1,4)\\L(2,4)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(4-4)^{2}+(2+1)^{2}}[/tex]

[tex]d=\sqrt{(0)^{2}+(3)^{2}}[/tex]

[tex]dNL=3\ units[/tex]

Step [tex]3[/tex]

Find the distance LM

[tex]L(2,4)\\M(-2,1)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(1-4)^{2}+(-2-2)^{2}}[/tex]

[tex]d=\sqrt{(-3)^{2}+(-4)^{2}}[/tex]

[tex]dLM=5\ units[/tex]

Step [tex]4[/tex]

Find the perimeter of the triangle LMN

we know that

The perimeter of a triangle is equal to the sum of the three length sides

In this problem

[tex]Perimeter=MN+NL+LM[/tex]

substitute the values in the formula

[tex]Perimeter=(\sqrt{10}+3+5)\ units[/tex]

[tex]Perimeter=(8+\sqrt{10})\ units[/tex]

therefore

the answer is the option D

the perimeter of the triangle LMN is equal to [tex](8+\sqrt{10})\ units[/tex]

Answer:  The required measure of angle ABD is 148°.  

Step-by-step explanation:  Given that ray BD bisects angle CBE, ray BC is transparent to ray BA.

Also,

[tex]m\angle CBD=(3x+25)^\circ,~~m\angle DBE=(7x-19)^\circ,~~m\angle ABC=90^\circ.[/tex]

We are to find the measure of angle ABD.

Since ray BD bisects angle CBE, so we have

[tex]m\angle CBD=m\angle DBE\\\\\Rightarrow (3x+25)^\circ=(7x-19)^\circ\\\\\Rightarrow 3x+25=7x-19\\\\\Rightarrow 7x-3x=25+19\\\\\Rightarrow 4x=44\\\\\Rightarrow x=\dfrac{44}{4}\\\\\Rightarrow x=11.[/tex]

So, we get

[tex]m\angle CBD=(3\times11+25)^\circ=58^\circ.[/tex]

Therefore,

[tex]m\angle ABD=m\angle ABC+m\angle CBD=90^\circ+58^\circ=148^\circ.[/tex]

Thus, the required measure of angle ABD is 148°.  

Answer:

[tex]8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6[/tex]

Step-by-step explanation:

We want to simplify;

[tex](8r^6s^3-9r^5s^4+3r^4s^5)-(2r^4s^5-5r^3s^6-4r^5s^4)[/tex]

We expand the bracket to obtain;

[tex]8r^6s^3-9r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6+4r^5s^4[/tex]

We now group the like terms to obtain;

[tex]8r^6s^3-9r^5s^4+4r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6[/tex]

We now simplify to get;

[tex]8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6[/tex]

The correct answer is C

Answer:

c

Step-by-step explanation:

Answer: Hence, option 'A', 'B', 'D' , 'E' are correct.

Step-by-step explanation:

There are many different types of variations  as follows:

1) Direct variations in which the relationship between two different variables is direct.

2) Inverse variations in which the relationship between two different variables is opposite.

3) Joint variations in which the relationship between one variable varies directly with the product of two or more variables.

4) Combined variations in which one variable is directly proportion to second variable but inversely proportion with third variables.

Hence, option 'A', 'B', 'D' , 'E' are correct.

19.04 dollars.

The student is asking how to calculate the cost of driving 300 kilometers in a fuel-efficient car, with the cost given in dollars instead of euros. To calculate this, we need to perform several steps:

Step-by-step, this calculation is:

Therefore, it will cost approximately 19.04 dollars to drive 300 kilometers.

Proportional relationships are relationships with equal ratio. Mandy's conclusion is incorrect because the relationship is proportional, and it has a uniform ratio of 5.

To determine if the input and output are proportional, we simply divide the output (y) by the corresponding input (x).

i.e.

[tex]Ratio = \frac yx[/tex]

So, we have:

[tex]Ratio = \frac 51 = 5[/tex]

[tex]Ratio = \frac{10}{2} = 5[/tex]

[tex]Ratio = \frac{25}{5} = 5[/tex]

[tex]Ratio = \frac{50}{10} = 5[/tex]

For the four input and output data, the ratios are equal (i.e. 5).

This means that the relationship is proportional.

Hence, Mandy's conclusion is incorrect

Read more about proportional relationships at:

https://brainly.com/question/24312388

Mandy is correct about the proportional relationship as all the ratios of y to x in her table are equivalent to 5/1. However, there is an error with the last ratio listed as 10/50, which should be 50/10 to match the information in the table. After correcting this typo, it is clear that the relationship is proportional because all of the ratios simplify to the same value.

Mandy is indeed correct. For a set of ratios to represent a proportional relationship, they all must have the same value when simplified. From Mandy's table, the ratios formed from the inputs (x) and outputs (y) are 5/1, 10/2, 25/5, and 50/10. When simplified, these should give us a constant ratio if the relationship is proportional. Simplifying the ratios, we get:

Since all ratios simplify to the same number, 5/1, it indicates that the relationship between x and y is indeed proportional. However, the last ratio listed in the question, 10/50, is incorrect as it should be 50/10 to maintain consistency with the rest of the table. If we correct this ratio to 50/10, then all ratios confirm a proportional relationship.

To further validate this, a proportion can be created by setting two ratios equal to each other. For example, we can compare the first and the second ratio: 5/1 = 10/2. When you cross-multiply, the resulting equations, 5*2 = 10*1, are true, again confirming proportionality.

https://brainly.com/question/29765554

#SPJ3

area of a square = S^2

to find length of side take square toot of area

sqrt(59) = 7.68114

to nearest tenth = 7.7 inches

Triangle DPW was translated 1 unit to the left, then it was translated 5 units down and then it was rotated by 90 degrees counterclockwise.

Answer:

Option C - [tex]2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}[/tex]

Step-by-step explanation:

Given : Expression [tex]2\frac{1}{6}-5\frac{2}{3}[/tex]

To find : Perform the indicated operation in the expression ?

Solution :

Expression [tex]2\frac{1}{6}-5\frac{2}{3}[/tex]

Write mixed fraction into fraction,

[tex]2\frac{1}{6}-5\frac{2}{3}=\frac{13}{6}-\frac{17}{3}[/tex]

Taking Least common denominator,

[tex]2\frac{1}{6}-5\frac{2}{3}=\frac{13-34}{6}[/tex]

[tex]2\frac{1}{6}-5\frac{2}{3}=-\frac{21}{6}[/tex]

[tex]2\frac{1}{6}-5\frac{2}{3}=-\frac{7}{2}[/tex]

Write improper fraction into mixed fraction,

[tex]2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}[/tex]

Therefore, Option C is correct.