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rodikova [14]
3 years ago
7

A system of equations has no solution. If the given equation (12x+7y=16) is one of the equations in the system, which of the equ

ations below could be the other?
Mathematics
1 answer:
lana [24]3 years ago
3 1

Answer:

<em>A.  12x + 7y = 20  and B.  12x + 7y = 25 </em>

Step-by-step explanation:

I think you question is lack of key information, allow me to add more to fullfil your question

<em>A system of equations has no solution. If the given equation (12x+7y=16) is one of the equations in the system, which of the equations below could be the other?</em>

<em>A. 12x + 7y = 20 </em>

<em>B. 12x + 7y = 25 </em>

<em>C. 12x + 9y = 20 </em>

<em>D. 12x + 9y = 25 </em>

My answer:

A. 12x + 7y = 20  and B.  12x + 7y = 25  

Because the two linear system are parallel to each other, they have the same slope: y = -7/12 without any intersection.

Hope it will find you well.

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Please help me thank youuu
FrozenT [24]

Answer:

  (c)  -3.8, 3

Step-by-step explanation:

The solutions to the equation f(x) = g(x) are the values of x where the function values are equal. These are the x-coordinates of the points where the graphs of y=f(x) and y=g(x) intersect each other.

<h3>Points of intersection</h3>

The points where the graphs cross can be estimated to be ...

  (-3.8, 6.5) and (3, -6)

The x-coordinates of these points are the solutions to the equation:

  x = -3.8, 3

8 0
1 year ago
14 1/2 - 2 1/3 what is the answer to this equation?
olchik [2.2K]

Answer:

12 1/6

It wants me to add more but there's nothing to add

7 0
1 year ago
Solve the inequality. -8c - 9 &lt; -17​
olya-2409 [2.1K]

Answer:

it would be c >1

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6 0
2 years ago
Read 2 more answers
Place these events on the probability
Aleksandr [31]

Answer:

  • P(tails)=1/2
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8 0
3 years ago
The figure shows △ABC . BD is the angle bisector of ∠ABC .
Misha Larkins [42]
From angle bisector theorem, we know that:

\dfrac{|AD|}{|DC|}=\dfrac{|BA|}{|BC|}\\\\\\\dfrac{|AD|}{|DC|}=\dfrac{8}{10}=\dfrac{4}{5}

Moreover:

|AD|+|DC|=6

so:

|DC|=6-|AD|

and

\dfrac{|AD|}{|DC|}=\dfrac{4}{5}\\\\\\\dfrac{|AD|}{6-|AD|}=\dfrac{4}{5}\qquad\qquad\text{cross multiplying}\\\\\\ 5\cdot|AD|=4\cdot\big(6-|AD|\big)\\\\5\cdot|AD|=24-4\cdot|AD|\\\\5\cdot|AD|+4\cdot|AD|=24\\\\9\cdot|AD|=24\qquad|:9\\\\\\|AD|=\dfrac{24}{9}=\boxed{\dfrac{8}{3}}

3 0
3 years ago
Read 2 more answers
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