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kherson [118]
2 years ago
8

Solve the system of equations -3x-5y=-16 and 4x+6y=16

Mathematics
1 answer:
Ann [662]2 years ago
7 0

Answer:

x=-56 y=80

Step-by-step explanation:

solved with delta way

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What is the point of intersection when the system of equations below is graphed on the coordinate plane?
lora16 [44]

Answer:

<h2>not exist</h2>

Step-by-step explanation:

The coordinates of the intersection of the line are the solution of the system of equations.

\underline{+\left\{\begin{array}{ccc}x-y=1\\y-x=1\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad0=2\qquad\bold{FALSE}

The system of equations has no solution. Therefore, the lines are parallel (the intersection does not exist).

5 0
3 years ago
What expression is equivalent
Citrus2011 [14]

Answer:

-4x^2, -3x, -5

Step-by-step explanation:

-4x^2 + 2x - 5 (1 + x)

-4x^2 + 2x - 5 + 5x

-4x^2 - 3x - 5

3 0
3 years ago
Ariel collected information about mockingbirds for a science project.
ipn [44]

Answer:

420/30=14 beats per second, meanin 1680 beats is your answer.

Step-by-step explanation:

8 0
3 years ago
Is 0.416 terminating or repeating
ollegr [7]

terminating................. because the number does not repeat

4 0
3 years ago
If it takes four men three hours to dig two holes, how long does it take one man to dig three holes?
damaskus [11]

Answer:

it will take one man 18 hours to dig the three holes

Step-by-step explanation:

To solve this problem, we can set up a simple relation as follows:

4 men dig two holes in 3hours

4 men will dig 1 hole in x hours.

2 holes => 3 hours

1 hole => x hours

x = (1 X 3)/2 = 1.5 hours

Therefore, 4 men will dig 1 hole in 1.5 hours.

From this, we can find how long it will take 1 man to dig 1 hole

if 4men dig a hole in 1.5 hours

1 man will dig 1 hole in x hours

4 X 1.5 = 6 hours

1 man will dig 1 hole in 6 hours

finally, if 1 man digs a hole in 6 hours, it will take the man 6 hours X 3 to dig 3 holes = 18hours

it will take one man 18 hours to dig the three holes

8 0
2 years ago
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