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Likurg_2 [28]
3 years ago
7

If a line perpendicular to 2y=x+5 that passes through (2,1) using the slope- intercept form

Mathematics
1 answer:
arlik [135]3 years ago
4 0
Slope of  2y = x + 5:-
This is y = 0.5 x + 2.5  in slope intercept form so its slope = 0.5

Slope of the line perpendicular to it = -1 / 0.5 = -2

it passes through (2 , 1) so we have

y- 1 = -2(x - 2)
y = -2x + 4 + 1

The answer is y = -2x + 5

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
juin [17]

Answer: (B) min = -14, max = -3

<u>Step-by-step explanation:</u>

The minimum is the y-value of the vertex of a positive (U-shaped) parabola. In the graph, the vertex is located at (4.5, -14). The y-value of the vertex is -14.

The maximum is the y-value of the vertex of a negative (∩-shaped) parabola. In the graph, the vertex is located at (1.5, -3). The y-value of the vertex is -3.

8 0
3 years ago
Geometry problem: Find the distance between the parallel lines y=-3/2x+1 and 2y+3x=10
s344n2d4d5 [400]

Answer:

d = 1.847

Step-by-step explanation:

To avoid any ambiguity, please enclose the slope -3/2 inside parentheses:

y = (-3/2)x + 1.  Next, solve the 2nd equation for y:  2y = - 3x + 10 => y = (-3/2)x + 5.  Notice how these two lines have the same slope (-3/2)?  Thus, the two given lines are parallel.

It's important to realize that the distance between these two lines is a line segment with perpendicular to both lines.  Thus, the segment representing this distance has the form y = (2/3)x + c.  Let's say that this segment passes through the y-intercept of y = (-3/2)x + 1; it thus passes through (0,1), and thus has the equation y = (2/3)x + 1.  

Find the point at which this y = (2/3)x + 1 intersects the line y = (3/2)x + 5.  Note that the xresultant point

Setting these two equations = to one another results in (2/3)x + 1 = (-3/2)x + 5.

Multiplying both sides by the LCD (6) will eliminate the fractions:

4x + 6 = -9x + 30.  Combining like terms:  13x = 24, and x = 24/13 = 1.846.

Substitute this x-value into   y = (2/3)x + 1 to find the y-coordinate of the point of intersection of y = (2/3)x + 1 and y = (-3/2)x + 5:

When x = 1.846, y = (2/3)(1.846) + 1 = 1.231.

Thus, the desired distance is that between (0,1) and (1.846, 1.231), and is:

d = √ [ (1.846 - 0)^2 + (1.231 - 1)^2 ] = √ [ (1.846)^2 + (0.053)^2 ]

This simplifies to                              d = √ [ 3.408 + 0.003 ] = √3.411, or

                                                          d = 1.847   (answer)

7 0
3 years ago
What are the coordinates of the vertices of the triangle under the translation (x, y) --&gt; (x + 3, y + 2)?
Afina-wow [57]
<span>The coordinates of the vertices of the triangle :
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</span>the translation (x, y) --> (x + 3, y + 2), the new coordinates of the vertices of the triangle will be:<span>
(2, 3), </span>(-1, 3), <span>(-1, 6)  

Answer:
</span><span>D. (2, 3), (-1, 3), (-1, 6)</span>
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Answer:

i think its the second and third

Step-by-step explanation:

im not completely sure though so please dont get mad if its wrong

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