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Sergeu [11.5K]
3 years ago
11

Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.

Mathematics
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

C

Step-by-step explanation:

a=12,b= 9, c=7

x=-9±√9²-4x12x7/2x12

=-9±√81-336/24

=-9±√-225/24

{x=9+í√225/24 or x=9-í√225/24}

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Two vertices of a rhombus are located at the points (0, 3) and (3, 0). Which of the following could be the other two vertices of
Nutka1998 [239]
Since a rhombus is symmetrical, if you have two points already you should be able to get the other two by mirroring.

In this case, take (0,3) and mirror it horizontally (not to be confused with vertically mirroring; there is a huge difference.) To horizontally mirror this point, fold your paper along the x-axis. You should get (0,-3) as your new point.

~The reason we mirrored (0,3) horizontally is because this point is on the positive y-axis, and we want to see where it'll be on the negative y-axis. Horizontally mirroring just helps us do that.

For your next point, take (3,0) and vertically mirror. This time, fold along the y-axis. Your new point should be (-3,0).

let me know if you have any questions:)
5 0
4 years ago
Given:f(x)=2x+5 and g(x)=
zhannawk [14.2K]

what is the value of g(x)

7 0
4 years ago
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natali 33 [55]

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7 0
3 years ago
5(x+4)+10-2x+3(2x+9)
qwelly [4]

Answer:

x = (-19)/3

Step-by-step explanation:

Solve for x:

10 - 2 x + 5 (x + 4) + 3 (2 x + 9) = 0

5 (x + 4) = 5 x + 20:

10 - 2 x + 5 x + 20 + 3 (2 x + 9) = 0

3 (2 x + 9) = 6 x + 27:

6 x + 27 + 5 x - 2 x + 10 + 20 = 0

Grouping like terms, 6 x + 5 x - 2 x + 10 + 20 + 27 = (-2 x + 5 x + 6 x) + (10 + 20 + 27):

(-2 x + 5 x + 6 x) + (10 + 20 + 27) = 0

-2 x + 5 x + 6 x = 9 x:

9 x + (10 + 20 + 27) = 0

10 + 20 + 27 = 57:

9 x + 57 = 0

Subtract 57 from both sides:

9 x + (57 - 57) = -57

57 - 57 = 0:

9 x = -57

Divide both sides of 9 x = -57 by 9:

(9 x)/9 = (-57)/9

9/9 = 1:

x = (-57)/9

The gcd of 57 and 9 is 3, so (-57)/9 = (-(3×19))/(3×3) = 3/3×(-19)/3 = (-19)/3:

Answer: x = (-19)/3

4 0
4 years ago
Jacob transformed quadrilateral FGHJ to F'G'H'J'.
Rudik [331]

Answer:

A. Reflection across the line x = 1

Step-by-step explanation:  

Please find the attachment.

We have been given that Jacob transformed quadrilateral FGHJ to F'G'H'J'.  We are asked to find which transformation Jacob used to reflect FGHJ to F'G'H'J'.

Since we know that while reflecting a figure, the line of reflection will lie between the original figure and reflected figure. Each point of the reflected figure will have the same distance from the line of reflection as the corresponding point of the original figure.              

Now let us see our given choices one by one.

A. Reflection across the line x = 1.

Upon looking at point G and G' we can see that both points are equidistant (2 units) from the line x=1. Other corresponding points of both quadrilaterals are also equidistant from line x=1, therefore, Jacob used the reflection across the line x=1 to transform quadrilateral FGHJ to F'G'H'J'.

B. Reflection across the line y = 1 .      

If Jacob had reflected quadrilateral across line y=1, the points of quadrilateral F'G'H'J' will lie in second and third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option B is not a correct choice.

C. Reflection across the line y-axis.

If Jacob had reflected quadrilateral across y-axis, the coordinates of points of quadrilateral F'G'H'J' will be G'(1,4), H'(1,2), J'(4,0) and F'(2,5). Therefore, option B is not a correct choice.

D.  Reflection across the x -axis.

If Jacob had reflected quadrilateral across x-axis, the points of quadrilateral F'G'H'J' will lie in third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option D is not a correct choice.

3 0
3 years ago
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