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nikitadnepr [17]
3 years ago
10

Solve the following systems of linear equations using any algebraic method. If possible, check your solution.

Mathematics
1 answer:
julia-pushkina [17]3 years ago
5 0
-8x+18y=-12
24x-18y=18
16x=6
X=3/8
Y=3-12)/6
y=-3/2
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What relationship does the vertex of the parabola always have with the x-intercepts?
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<h3>The solutions to the equation are called the roots of the function</h3>

Step-by-step explanation:

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Which of the following are true statements.
Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

c^{2} =a^{2} +b^{2}----------(1)

Let c=41 and a = 7 and b=40

Put all the value in equation 1.

41^{2} =7^{2} +40^{2}

1681=49+1600

1681=1649

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let c=20 and a = 12 and b=16

20^{2} =12^{2} +16^{2}

400=144+256

400=400

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

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3 years ago
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