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marin [14]
2 years ago
11

Determine if the point is a solution to the system of linear equations

7B%20%20%5Clarge%20%5Cbegin%7Barray%7D%7B%7D%20%202x%20%2B%203y%20%3D%202%20%20%5C%5C%206y%20%3D%205%20-%204x%20%5Cend%7Barray%7D%7D%20" id="TexFormula1" title=" \huge \{{ \large \begin{array}{} 2x + 3y = 2 \\ 6y = 5 - 4x \end{array}} " alt=" \huge \{{ \large \begin{array}{} 2x + 3y = 2 \\ 6y = 5 - 4x \end{array}} " align="absmiddle" class="latex-formula"> ​
Mathematics
1 answer:
Kay [80]2 years ago
8 0

Answer:

The system has no solutions.

Step-by-step explanation:

If you stare at it for the bit you can tell the system is impossible. Let's see why:

First of all, rewrite the second equation by bringing all unknowns on the LHS:

4x+6y=5.

Then take twice the first equation, doubling every coefficient:

4x+6y=4 See how the LHS is the same? The system is asking you to find a quantity (4x+6y) which is at the same time equal to four AND five - emphasis on that "and" - which is obviously impossible, because it implies that 4=5. So the point - which you didn't provide - is not a solution. Nor is any other point.

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240.74

Step-by-step explanation:

y=±x*\frac{100+p}{100}

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28 in 21 in., 20 in.<br> Is this a right triangle?
ankoles [38]

Answer:

No, it is not a right triangle.

Step-by-step explanation:

The simplest way to determine is testing out the numbers with Pythagorian theorem.

If it complies with the theorem, it is a right triangle.

let's assume c = 28, b = 21, and a = 20

the longest side is the hypotenuse so side c (28 in) will be the hypotenuse.

According to the Pythagorian theorem, the square of the length of hypotenuse must equal to the sum of squares of other two sides.

check:

c^2 = 28^2 = 784

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because c^2 is not equal to a^2 + b^2, the triangle is not a right triangle.

7 0
2 years ago
3⁄2 c + 23 = 38<br><br> can you help me find out what c is pretty please
kolbaska11 [484]

Answer:

C=10

Step-by-step explanation:

To solve this problem, first, you have to isolate it on one side of the equation. Remember, isolate c on one side of the equation.

3/2c+23-23=38-23 (subtract 23 from both sides.)

38-23 (solve.)

38-23=15

3/2c=15

2*3/2c=15*2 (multiply 2 from both sides.)

15*2 (solve.)

15*2=30

3c=30

3c/3=30/3 (divide by 3 from both sides.)

30/3 (solve.)

30/3=10

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2 years ago
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