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Rudik [331]
3 years ago
14

HELP HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
Anton [14]3 years ago
4 0

The answer is C. A fraction is basically another way of writing division, and  six divided by three is two.

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Hey, please help me thanks
rjkz [21]

Answer:

240

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Mike received a 10% raise each month for 3
s2008m [1.1K]

Answer:

his salary after three months would be 1300

Step-by-step explanation:

10 percent of 1000 is a hundred and multiply that by 3, it would equal 300 and add that to the 1000.

7 0
3 years ago
Divide the polynomials:<br> x^2-3x+9/x-2
Kruka [31]

The value of dividing x^2-3x+9 by x-2 is x + 3

<h3>Ways of dividing polynomials</h3>

There are several ways to divide polynomial functions; some of these ways include

  • By factorization
  • By long division
  • By synthetic division
  • By using technology such as graph

<h3>How to divide the polynomials?</h3>

The expression for the polynomial division is given as:

x^2-3x+9/x-2

To divide polynomial functions, we make use of the division by factorization method

Start by expanding the numerator of the polynomial division

x^2-3x+9/x-2 = x^2 + 3x - 6x + 9/x-2

Factorize the equation

x^2-3x+9/x-2 = x(x + 3) - 2(x + 3)/x - 2

Factor out x + 3 from the numerator

x^2-3x+9/x-2 = (x- 2)(x + 3)/x - 2

Cancel out the common factors

x^2-3x+9/x-2 = x + 3

Hence, the value of dividing x^2-3x+9 by x-2 is x + 3

Read more about polynomial division at:

brainly.com/question/25289437

#SPJ1

5 0
2 years ago
Consider the system of quadratic equations \begin{align*} y &amp;=3x^2 - 5x, \\ y &amp;= 2x^2 - x - c, \end{align*}where $c$ is
shtirl [24]

Hello, we need to solve this system, c being a real number.

\begin{cases}y &= 3x^2-5x\\y &= 2x^2-x-c\end{cases}

y=y, right? So, it comes.

3x^2-5x=2x^2-x-c\\\\3x^2-2x^2-5x+x+c=0\\\\\boxed{x^2-4x+c=0}

We can compute the discriminant.

\Delta=b^2-4ac=4^2-4c=4(4-c)

If the discriminant is 0, there is 1 solution.

It means for 4(4-c)=0  4-c=0  \boxed{c=4}

And the solution is

x_2=x_1=\dfrac{4}{2}=2

If the discriminant is > 0, there are 2 real solutions.

It means 4(4-c) > 0 <=> 4-c > 0 <=> \boxed{c

And the solution are

x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{4-c}}{2}=2-\sqrt{4-c}\\\\x_2=2+\sqrt{4-c}

If the discriminant is < 0, there are no real solutions.

It means 4(4-c) < 0 <=> 4-c < 0 <=> \boxed{c>4}

There are no real solutions and the complex solutions are

x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{i^2(c-4)}}{2}=2-\sqrt{c-4}\cdot i\\\\x_2=2+\sqrt{c-4}\cdot i

Thank you.

8 0
4 years ago
Please help me I begging youI just need to do this then I’m done I really want to go to sleep
PilotLPTM [1.2K]
Slope: 3, y-intercept: 4, equation: y = 3x + 4
6 0
3 years ago
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