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3 years ago

-FAST PLEASE- How do I simplify expressions with 2 or more variables?

Mathematics

1 answer:

meriva3 years ago

3 0

Answer:

10+2x=4x

-2x -2x

10=2x

10/2 2/2

5=x

Step-by-step explanation:

So when you simplify an expression with two or more variables you always go for the lowest variable first. Then you can do your normal inverse steps.

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What is the equation of the line parallel to the line with the equation −x+3y=−3

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Answer:

y = 1/3x - 19/3

Step-by-step explanation:

3y = x -3

y = 1/3x -3

Slope = 1/3

Point= (1,-6)

y-intercept = -6 - (1/3)(1) = -6 - 1/3 = -19/3

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Edger Anderson earns $300 A week plus a 15 percent commission only on sales he makes after his first $1000 In sales.if Mr.Anders

Mama L [17]

525 is what i got as my answer good luck

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3 years ago

Read 2 more answers

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

3 years ago

Read 2 more answers

I need help for brainliest

devlian [24]

Answer:

Step-by-step explanation

the figure is composed of a square with an isosceles right triangle inside, we find the area of the square and we take away the area of the triangle.

Triangle area = lxl, right-angled triangle area side by side divided by two.

square area

8 * 8 = 64cm²

Triangle area

6 * 6 : 2 = 18 cm²

Area of the shaded region

64 - 18 = 46 cm²

7 0

2 years ago

The function f(x) = –x2 – 4x + 5 is shown on the graph. On a coordinate plane, a parabola opens down. It goes through (negative

sertanlavr [38]

The true statement about the function f(x) = -x² - 4x + 5 is that:

The range of the function is all real numbers less than or equal to 9.

<h3 /><h3>What is the domain and range for the function of y = f(x)?</h3>

The domain of a function is the set of given values of input for which the function is valid and true.

The range is the dependent variable of a given set of values for which the function is defined.

The domain of the function: f(x) = -x² - 4x + 5 are all real number from -∞ to +∞

For a parabola ax² + bx + c with the vertex $\mathbf{(x_v,y_v)}$

If a < 0, then the range is f(x) ≤ $\mathbf{y_v}$

If a > 0, then the range f(x) ≥ $\mathbf{y_v}$

Here; a = -1,

The vertex for an up-down facing parabola for a function y = ax² + bx + c is:

$\mathbf{x_v = -\dfrac{b}{2a}}$

Thus,

vertex $\mathbf{(x_v,y_v)}$ = (-2, 9)

Range: f(x) ≤ 9

Therefore, we can conclude that the range of the function is all real numbers less than or equal to 9.

Learn more about the domain and range of a function here:

brainly.com/question/26098895

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2 years ago