Math questions pre algebra?

Question

My Math teacher gave us some math problems and told us to google it so see what other people came up with.... here goes....

Find the Domain
1. g(x)= 1/3x-3

2. f(x)= (the square root sign) -x-2

Solve...
1. x^3=x

2. x^2=-5x=24

Simplify
1. x^2+2x-3/x-5

2. x^3-27/x-3

3. (2x^-5y^-1)-3/(-5x^-3y^-4)-2

4. (x^-2y^3/-2x^-5y^7)^-3

Answer 1

1.

g(x) = 1 / 3x - 3

We can find the domain then

3x - 3 = 0

Add 3 to each side

3x - 3 + 3 = 3 + 0

3x = 3

Divide 3 to each side

3x / 3 = 3 / 3

x  = 1

There the domain is x = 1.

Comments

Domain of is all real numbers.

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Domain of is

Answer 2

 f(x) = the square root sign -x  - 2

f(x) =sqrt(-1(x + 2))

Substitute i^2 = -1 in the f(x)

f(x) = sqrt(i^2 ( x + 1)

f(x) = isqrt (x + 1)

We can find the Domain sqrt(x + 1) = 0

Take square each side

x + 1 = 0

Subtract from 1 to each side

x + 1 -  1 = 0 - 1

x + 0 = -1

x = -1

The domain is x = -1.

Comments

Domain of is

Answer 3

 Solve x³ = x

Subtract x to each side

x³ - x = 0 - x

x³ - x = 0

Take out common term x in x³ - x = 0

x(x² - 1) = 0

x(x² - 1²) = 0

Recall : a² - b² = (a - b)(a + b)

Substitute a = x  and b = 1 in the x² - 1²

x(x - 1)(x + 1) = 0

It have 3 roots

x = 0 or x - 1 = 0 or x + 1 = 0

x - 1 = 0

Add 1 to each side

x - 1 + 1 = 0 + 1

x - 0 = 1

x = 1

x + 1 = 0

Subtract 1 from each side

x + 1 - 1 = 0 - 1

x + 0 = -1

x = -1

The solution of the equation : -1 , 0 , 1.

 

 

Answer 4

 Solve x³ = x

Subtract x to each side

x³ - x = 0 - x

x³ - x = 0

Take out common term x in x³ - x = 0

x(x² - 1) = 0

x(x² - 1²) = 0

Recall : a² - b² = (a - b)(a + b)

Substitute a = x  and b = 1 in the x² - 1²

x(x - 1)(x + 1) = 0

It have 3 roots

x = 0 or x - 1 = 0 or x + 1 = 0

x - 1 = 0

Add 1 to each side

x - 1 + 1 = 0 + 1

x - 0 = 1

x = 1

x + 1 = 0

Subtract 1 from each side

x + 1 - 1 = 0 - 1

x + 0 = -1

x = -1

The solution of the equation : -1 , 0 , 1.

Answer 5

x² -5x = 24

Subtract 24 to each side

x² + 5x  -24= 24 - 24

x² + 8x - 3x  -24 = 0

Take out common term x  and -3 in the above equation

x(x + 8) -3(x +8) = 0

Take out common term x + 8 in the above equation

(x + 8)(x - 3) = 0

x + 8 = 0 or x - 3 = 0

If x + 8 = 0

Subtract 8 from each side

x + 8 - 8 = -8

x + 0 = -8

x = -8

If x - 3 = 0

Add 3 to each side

x -3 + 3 = 0 + 3

x - 0 = 3

x = 3.

The solution of the equation : -8 , 3.

Comments

Solutions of the equation x² -5x = 24 are x = -3 , 8.

Answer 6

Simplify 1. x^2 + 2x - 3 / x - 5 (x^2 + 3x - x - 3) / x - 5 Take out common term x and -1 in the above expression x(x + 3) -1(x + 3) / x - 5 Take out common term x + 3 x + 3[x - 1] / x - 5 (x + 3)(x - 1) / x - 5.

Answer 7

 Simplify

2.

x³  - 27 / x - 3

=x³ - 3³ / x - 3

Recall : (a³ - b³) = ( a - b)(a² + ab + b²)

Substitute x = a  and b = 3 in the above formula

(x - 3)(x² + 3x + 3²) / (x - 3)

= (x² + 3x + 9).

 

 

Answer 8

4.

(x^-2y³ / -2x^-5y⁷)^-3

Apply the power rule (a / b)^-3 = a^-3 / b^-3

(x^-2y³)^-3 / (-2x^-5y⁷)^-3

Apply the power rule (ab)^-3 = a^-3 b^-3

x⁶y^-9 / -2x¹⁵y^-21

Apply the power rule

y^-9+21 / -2x^15-6

y¹² / -2x⁹

-1/2(y¹² / x⁹)

-1/2(y¹² x^-9).

Answer 9

3.

(2x^-5y^-1) -3 / (-5x^-3y^-4) - 2

Apply the power rule

((2 / x⁵y) - 3) / (-5 / x³y⁴) - 2

(2 - 3x⁵y / x⁵y) / (-5 - 2x³y⁴ / x³y⁴)

Simplify

(x³y⁴ / x⁵y)(2 - 3x⁵y / -(5 + 2x³y⁴))

Apply the power rule

(y^4-1 / x^5-3)(2 - 3x⁵y / -(5 + 2x³y⁴))

-(y³ / x²)(2 - 3x⁵y / 5 + 2x³y⁴)

[-y³(2 - 3x⁵y) / x^2(5 + 2x³y⁴)] .

 

Answer 10

Siimplify

4) Given ( x ^-2 y ^ 3 / -2x  ^-5 y ^ 7 ) ^ -3

Apply the power rule (a / b) ^ -3 = a ^ -3 / b ^ -3

( x ^ -2 y ^ 3 ) ^ -3 / ( -2 x ^ -5 y ^ 7 ) ^ -3

Apply the power rule ( ab )^ -3 = a ^ -3 * b ^ -3

x ^ 6 y ^ -9 / ( ( -2 ) ^ -3 x ^ 15 y ^ -21 )

Apply the power rule

x ^ 6 y ^ -9 / ( -1/8) x ^ 15 y ^ -21

- 8 y ^ - 9 + 21 / x ^ 15 - 6

- 8 y ^ 12 / x ^ 9

- 8 y ^ 12 x ^ -9