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saw5 [17]
2 years ago
13

Which equation represents a line that is parallel to the line y = - 4x +5?

Mathematics
2 answers:
Vikentia [17]2 years ago
6 0
Y= -4x+3

slope is still the same so it would be parallel to that line. Only the y-intercept is different so it would just pass through a different point
alekssr [168]2 years ago
4 0

option 1 is the answer.

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Is this equation an infinite solution, no solution, or one solution?? 7(2+5x) = 3x + 14
shusha [124]

Answer:

One Solution

Step-by-step explanation:

Once you solve the equation, you get 0 which satisfies for X, therefore it has a solution.

7 0
3 years ago
Which does NOT represent an obtuse angle
Darya [45]
Hey there!

An obtuse angle is defined as an angle that is wider than 90° but less than 180°. For example, a 68° would not be considered obtuse, but a 107° would. As long as the degree is between 90 and 180, the angle is obtuse. Any other angle (either between 0 and 90 OR 180 and 360) is not obtuse. 

You can use this information to answer the question based on the angles you have been given. 

Hope this helped you out! :-)
5 0
3 years ago
The midpoint of AB is M(1, -4). If the coordinates of A are (-3,-6), what are the coordinates of B?
kramer

Answer:

The coordinates of B is (5,-2)

Step-by-step explanation:

In order to find the coordinates of B, you have to use midpoint formula. Then substitute the following values into the formula :

m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )

Let (x1,y1) be A (-3,-6),

Let (x2,y2) be B,

Let midpoint be M (1,-4),

(1, - 4) =  (\frac{ - 3 + x2}{2} , \frac{ - 6 + y2}{2} )

by \: comparison, \:

\frac{ - 3 + x2}{2}  = 1

- 3 + x2 = 2

x2 = 5

\frac{ - 6 + y2}{2}  =  - 4

- 6 + y2 =  - 8

y2 =  - 2

6 0
3 years ago
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tangare [24]
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3 0
3 years ago
Is this an example of a horizontal asymptote?
8_murik_8 [283]

Answer:

Step-by-step explanation:

Hi there,

The graph indicated is showing a horizontal asymptote. In fact, it is showing both a horizontal and a <em>vertical </em>asymptote.

To tell which type it is, notice where the graph "shoots off" and almost forms an imaginary straight line in one direction. Using this logic, the horizontal asymptote will be exactly horizontal, parallel to x-axis, and vertical asymptote will be exactly vertical, parallel to y-axis.

With this graph, we notice the horizontal asymptote is at y=0, where the x-axis is. The vertical asymptote is bit more difficult to determine graphically, but can definitely say it is past x=-10. We could determine it if we had the function, but that is not necessary for this question.

Study well, and persevere. If you liked this solution, leave a Thanks or give a rating!

thanks,

6 0
3 years ago
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