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omeli [17]
4 years ago
10

In this diagram which equation could you prove to be true in order to conclude that the lines are parallel

Mathematics
2 answers:
Alinara [238K]4 years ago
7 0

Answer:

D)\frac{B}{A} =\frac{C}{D}

Step-by-step explanation:

Remember that parallel lines have the same slope, and the formula for slope is:

m=\frac{y2-y1}{x2-x1}

So now, if you insert the values of each line, you would get:

Point 2 line 1 (0,B)O

Point 1 line 1  (-A,0)

Insert into the formula:

m=\frac{y2-y1}{x2-x1}\\m=\frac{B-0}{0-(-A)}\\m=\frac{B}{A}

That is the slope of line 1.

Point 2 line 2 (D,0)

Point 1 line 2 (0,-C)

Insert into the formula:

m=\frac{y2-y1}{x2-x1}\\m=\frac{0-(-C)}{D-0}\\m=\frac{C}{D}

So the slope for the line number 2 is: m=\frac{C}{D}

If both slopes are equal we can prove that they are parallel, so the correct option would be D)\frac{B}{A} =\frac{C}{D}

KiRa [710]4 years ago
3 0
I would say that the correct answer is D. because y is always the numerator while x is the denominator for the equation y2 - y1/x2 - x1 which means if there is two y's or two x's on the same line you subtract the second one from the first one. If there is only one y and/or x and the other is 0 on the same line, it stays at y or x without subtracting y2 - y1 or x2 - x1. 

Since b is on the y coordinate and -a is on the x coordinate, you would make it b/a while -a is gonna be a positive since the lines are going up and to the right. Now, since c is the y coordinate and d is the x coordinate, make C the numerator and d the denominator since y is always the numerator and x is the denominator for these parallel line figures on the graph and the equation will be equaled to the fraction to the other fraction for parallel lines. 

So, your answer would be D. b/a = c/d 

Hope this helps, and is correct! 

<em>~ ShadowXReaper069</em>
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Solve system of equations 14x + y = -4 and y = 3x^2 - 11x - 4 
Keith_Richards [23]
Hello here is a solution : 

7 0
4 years ago
I need the solution
Nataly [62]

The only thing you can do here is to try to simplify, so I'll assume that's what you want to do.

\dfrac{b^2-a^2}{2a^2+ab-3b^2}

The numerator is a difference of squares, so it's easy to factorize:

b^2-a^2=(b-a)(b+a)

In the denominator, we factorize by grouping:

2a^2+ab-3b^2=2a^2-2ab+3ab-3b^2=2a(a-b)+3b(a-b)=(2a+3b)(a-b)

So we have

\dfrac{b^2-a^2}{2a^2+ab-3b^2}=\dfrac{(b-a)(b+a)}{(2a+3b)(a-b)}=-\dfrac{b+a}{2a+3b}

5 0
4 years ago
Determine the slope of each line, if any. Identify which of the lines is vertical or horizontal.
lys-0071 [83]

Answers:

  1. slope = 2
  2. slope = -3
  3. slope = 1
  4. slope = 1/4
  5. slope = 1/4

None of the five lines are horizontal or vertical

===============================================

Explanation:

Something like f(x) = 2x-5 is the same as y = 2x-5. This is because y = f(x).

Compare y = 2x-5 with y = mx+b. We see that m = 2 is the slope in this case.

The other four questions are handled the same way.

A horizontal line occurs when the slope is 0. Since none of the slopes are zero, none of the lines are horizontal.

A vertical line equation is of the form x = k, where k is any real number. The y variable is not involved. It is not possible to write a vertical line equation in the form y = mx+b because the slope m is undefined. So none of the given five equations are vertical lines.

Side note: something like y = x+6 is the same as y = 1x+6

4 0
3 years ago
Does the rule y=4 4^x represent a linear or an exponential function? Explain.
AlekseyPX

Answer:

Exponential

Step-by-step explanation:

This equation represents an exponential function because it will have exponential growth. The equation has a variable as an exponent which causes exponential growth. If it were linear there would be constant growth and a common difference.

6 0
3 years ago
Which operation results in an expression equivalent to ? 23y^4 - 6y^3 + 35y^2 - 20y
Tom [10]

Answer:

D. P + Q

Explanation:

Given:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Required:

Determine which operation will give an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

SOLUTION:

Perform each operation given to see which of them gives an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

Q - P:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Q - P = (5y² - 4y)(3y² + 7) - (8y⁴ + 6y³ + 8y)

(5y²(3y² + 7) -4y(3y² + 7)) - (8y⁴ + 6y³ + 8y)

(15y⁴ + 35y² - 12y³ - 28y) - (8y⁴ + 6y³ + 8y)

Open parentheses

15y⁴ + 35y² - 12y³ - 28y - 8y⁴ - 6y³ - 8y

Collect like terms

15y⁴ - 8y⁴ - 12y³ - 6y³ + 35y² - 28y - 8y

7y⁴ - 18y³ + 35y² - 36y

Therefore, P - Q does not give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

PQ:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

PQ = (8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y)

=(8y⁴(15y⁴ + 35y² - 12y³ - 28y) +6y³(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y) +8y(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y))

= 120y⁸ . . . . .

Note: You don't need to perform this operation further anymore. It would definitely not give us the equivalent expression we are looking for. The degree of the leading term (120y⁸) is way too greater than the degree of the leading term (23y⁴) of the equivalent expression we are looking for.

Thus, PQ cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P - Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P - Q = (8y⁴ + 6y³ + 8y) - (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y - 15y⁴ - 35y² + 12y³ + 28y

Collect like terms

8y⁴ - 15y⁴ + 6y³ + 12y³ - 35y² + 8y + 28y

-7y⁴ + 18y³ - 35y² + 36y

Therefore, P - Q cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P + Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P + Q = (8y⁴ + 6y³ + 8y) + (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y + 15y⁴ + 35y² - 12y³ - 28y

Collect like terms

8y⁴ + 15y⁴ + 6y³ - 12y³ + 35y² + 8y - 28y

23y³ - 6y³ + 35y² - 20y

Therefore, P + Q will result in an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

The answer is D.

7 0
3 years ago
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