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Contact [7]
3 years ago
11

How many solutions can be found for the equation −4x − 11 = 2(x − 3x) + 13?

Mathematics
2 answers:
Ratling [72]3 years ago
5 0

Answer:

The Equation is

- 4 x - 11 = 2 (x - 3 x) + 13

→ - 4 x - 11= 2 x - 2* 3 x + 13⇒⇒Using Distributive property of multiplication over subtraction which is , A * (B - C) = A*B - A*C

Bringing variables on one side and Constants on another side

→ - 4 x - 2 x + 6 x = 13 +11

→ - 6 x + 6 x = 24

0 =24

which is not possible.

Hence this equation has no solution.

If you look at the equation in the way

L HS= - 4 x - 11

A= - 4 x -11

R H S = 2 *(x - 3 x) + 13

B= 2 * (- 2 x) +13

 = - 4 x + 13

Both L HS and R H S, if looked individually represent lines in two dimensional plane, having equal slopes and different Y intercept.

So, the two lines are parallel.

Parallel lines do not intersect.

So, the equation has no solution.

Allushta [10]3 years ago
4 0
-4x - 11 = 2(x - 3x) + 13
-4x - 11 = 2x - 6x + 13
-4x - 11 = -4x + 13
-4x + 4x = 13 + 11
0 = 24...incorrect......no solutions for this problem
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If the diagonal of a square is 11.3 meters, approximately what is the perimeter of the square?
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Step-by-step explanation:

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A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

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