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VARVARA [1.3K]
3 years ago
12

Find the midpoint of the segment with the following endpoints. (7,-6) and (4, -10)​

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

(5.5,-8)

Step-by-step explanation:

(7 + 4)/2 = 5.5

(-6 + -10)/2 = -8

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Consider the equation 2y – 4x = 12. Which equation, when graphed with the given equation, will form a system with one solution?
lbvjy [14]

Answer:

  –y – 2x = 6 . . . choice 1

Step-by-step explanation:

Consider the system obtained using the other answer choices:

  -y +2x = 12 . . . choice 2

The result of multiplying the given equation by -1/2 is ...

  -y +2x = -6 . . . . a line parallel to that of answer choice 2. The system would have no solution

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  y = 2x +6 . . . choice 3

Solve the given equation for y by adding 4x and dividing by 2. That will give ...

  y = 2x +6 . . . . a line identical to that of answer choice 3. The system would have an infinite number of solutions.

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  y = 2x +12 . . . choice 4

As we found in the previous discussion, the given line can be written as ...

  y = 2x + 6 . . . . a line parallel to that of answer choice 4. The system would have no solution.

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The attached graph shows the given line and that of answer choice 1.

5 0
3 years ago
Read 2 more answers
What are the like terms in the expression 6x + 9 – 4x – y - 8
KiRa [710]
The like terms is 2x-1-y
5 0
3 years ago
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Find the slope of this line
FromTheMoon [43]

Answer:

slope is 1

Step-by-step explanation:

0--2=2

1--1=2

2/2-1

6 0
3 years ago
HELP PLS I WILL MARK BRAINLIEST ANSWER ASAP HELP PLEASE!!!
hodyreva [135]

Answer:

hope this helps you

4 0
3 years ago
Please find the answer , it is a very importatnt question
Anna [14]

Step-by-step explanation:

The given expression is :

\dfrac{2-\sqrt5}{2+3\sqrt5}=\sqrt5 x+y ....(1)

Taking LHS of the above expression

\dfrac{2-\sqrt5}{2+3\sqrt5}

Rationalizing both denominator and numerator.

\dfrac{2-\sqrt5}{2+3\sqrt5}\times \dfrac{2-3\sqrt5}{2-3\sqrt5}=\dfrac{4-6\sqrt5-2\sqrt5+15}{2^2-(3\sqrt5)^2}\\\\=\dfrac{19-8\sqrt5}{4-45}\\\\=\dfrac{19-8\sqrt5}{-41}\\\\=\dfrac{\sqrt5 \times 8}{41}+(\dfrac{-19}{41})

Equation (1) becomes,

\dfrac{\sqrt5 \times 8}{41}+(\dfrac{-19}{41})=\sqrt 5\ x+y

Equating both sides,

x=\dfrac{8}{41} and y=\dfrac{-19}{41}

Hence, this is the required solution.

7 0
3 years ago
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