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Darina [25.2K]
3 years ago
10

(5a^2+6a+2)-(7a^2-7a-5)

Mathematics
1 answer:
krok68 [10]3 years ago
8 0

For this case we must simplify the following expression:(5a^ 2 + 6a + 2) - (7a^ 2-7a-5) =

We eliminate the parentheses taking into account that:

- * + = -\\- * - = +

So:

5a ^ 2 + 6a + 2-7a ^ 2 + 7a + 5 =

We add similar terms taking into account that:

Equal signs are added and the same sign is placed.

Different signs are subtracted and the major sign is placed.

5a ^ 2-7a^2 + 6a + 7a +2 +5 =\\-2a ^ 2 + 13a + 7

Answer:

-2a ^ 2 + 13a + 7

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What is [(7x^3)(y^2)]^2/7 in radical form?
GuDViN [60]
Remember
x^ \frac{m}{n} = \sqrt[n]{x^m}
and
(zy)^m=(z^m)(y^m)
and
(x^m)^n=x^(mn)

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8 0
3 years ago
Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
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3 years ago
Write an equation of a line in point-slope form that has a slope of -2 and passes through (5, -1).
svetlana [45]
I hope this helps you

7 0
3 years ago
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