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2 years ago

Será a equação x² + 3x = x + 6 + x² do 2º grau?

Mathematics

1 answer:

goblinko [34]2 years ago

3 0

x² + 3x = x + 6 + x²

-x² -x²

3x = x + 6

-x -x

2x = 6

$\frac{2x}{2}$ = $\frac{6}{2}$

x = 3

Answer: x = 3

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2 years ago

Simplfiy the expression -5+i/2i

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2 years ago

I need help 6x-2y=10 x-2y=-5 solve by elimination

dem82 [27]

<h3>Explanation</h3>

Given the system of equations.

$\begin{cases} 6x - 2y = 10 \\ x - 2y = - 5 \end{cases}$

Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

$\begin{cases} 6x - 2y = 10 \\ - x + 2y = 5 \end{cases}$

There as we can get rid of the y-term by adding both equations.

$(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x = \frac{15}{5} \longrightarrow \frac{ \cancel{15}}{ \cancel{5}} = \frac{3}{1} \\ x = 3$

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

$x - 2y = - 5 \\ 3 - 2y = - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\ \frac{8}{2} = y \\ y = \frac{8}{2} \longrightarrow \frac{ \cancel{8}}{ \cancel{2}} = \frac{4}{1} \\ y = 4$

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

Answer Check by substituting both x and y values in both equations.

First Equation

$6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10 = 10 \longrightarrow \sf{true} \: \green{ \checkmark}$

Second Equation

$x - 2y = - 5 \\ 3 - 2(4) = - 5 \\ 3 - 8 = - 5 \\ - 5 = - 5 \longrightarrow \sf{true} \: \green{ \checkmark}$

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3>Answer</h3>

$\begin{cases} x = 3 \\ y = 4 \end{cases} \\ \sf \underline{Coordinate \: \: Form} \\ (3,4)$

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3 years ago

As his empire spread throughout the ancient world, Alexander founded roughly how many cities?

Studentka2010 [4]

About 70 Cities in Total.

For his more noteworthy grandness, Alexander established exactly 70 urban areas in the grounds he vanquished and requested them named after himself. Most celebrated, obviously, is Alexandria in Egypt. In India, when his dearest horse kicked the bucket, he requested a city to be constructed named Bucephala.

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Answer:

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Step-by-step explanation:

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