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3 years ago

Write and graph an equation that is perpendicular to the equation x-3y=6 and through point (2,-3)

Mathematics

2 answers:

Korvikt [17]3 years ago

7 0

Answer:

y = mx + b is slope-intercept form of the linear equation.

y - y1 = m(x - x1) is point-slope form of the linear equation.

Ax + By = C (A ≥ 0) is standard form of the linear equation.

~~~~~~~~~~~

....

y – 5 = —— (x + 1)

1 1

y = —— x + —— + 5

3 3

1 16

(1) y = —— x + ——

3 3

Step-by-step explanation:

Akimi4 [234]3 years ago

6 0

Answer:

The answer to your question is y = -3x + 3

Step-by-step explanation:

See the graph below

Data

Perpendicular to x - 3y = 6

Point = (2, -3)

Process

1.- Find the slope of the line given

x - 3y = 6

-3y = -x + 6

y = -1/-3x + 6/-3

y = 1/3x - 2

slope = 1/3

2.- Find the slope of the new line

slope = -3

3.- Find the equation of the new line

y + 3 = -3(x - 2)

y + 3 = -3x + 6

y = -3x + 6 - 3

y = -3x + 3

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$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

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Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.