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enot [183]
3 years ago
7

Hola, me podrían ayudar a simplificar está ecuación? 5x^2-7x(3x-8)+3x^2-8x+3

Mathematics
1 answer:
KengaRu [80]3 years ago
8 0

Answer:

-13x^2+48x+3

Step-by-step explanation:

=8x^2-21x^2+56x-8x+3

=  -13x^2+48x+3

(Recuerde hacer preguntas en inglés la próxima ve)

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-((-7) +(4)(-12)+(-3)(11)+(-2))
lakkis [162]

Answer:

<em>90</em>

Step-by-step explanation:

-((-7) + 4 × (-12) + (-3) × 11 + (-2))

-(-7 + 4 × (-12) + (-3) × 11 + (-2))

-(-7 - 48 - 33 - 2)

-(-90) → 90

Final Answer - 90

Hope this helped! :)

3 0
3 years ago
Read 2 more answers
Solve for c. a(b − c)=d
Lilit [14]
A(b - c) = d
ab - ac = d
-ac = -ab + d
ac = ab - d
ac/a = (ab - d)/a
c = (ab - d)/a

Your answer will be A. c = (ab - d)/a. Hope this helps!
6 0
3 years ago
Read 2 more answers
What is 9/16 x 2/3 (show ur work) thanks
Ilya [14]

Answer:

\displaystyle{ \frac{9}{16} \times  \frac{2}{3}  }

\displaystyle{ \frac{9 \times 2}{16 \times 3} }

\displaystyle{ \frac{18}{48} }

\displaystyle{ \frac{ \cancel2 \times  \cancel3 \times 3}{2 \times  \cancel2 \times 2 \times 2 \times  \cancel3} }

\displaystyle{ \frac{3}{2 \times 2 \times 2} }

\displaystyle{ \frac{3}{8} }

6 0
2 years ago
Which of the following is a solution of x2 − 8x = –27?
Sergeu [11.5K]
Its 4 plus i square root of eleven 
4 0
3 years ago
Read 2 more answers
Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr
Bad White [126]

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

x = \frac{-b +- R }{2*a}

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

x = \frac{-b +- C*i }{2*a}

We have two complex solutions.

If D = 0

√0 = 0

then:

x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}

We have only one real solution (or two equal solutions, depending on how you see it)

3 0
3 years ago
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