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sveta [45]
2 years ago
13

5C%3A%20answer" id="TexFormula1" title="3(m + 10) - 2(m - n) \: what \: is \: the \: answer" alt="3(m + 10) - 2(m - n) \: what \: is \: the \: answer" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Umnica [9.8K]2 years ago
8 0

Answer:

5m+30+2n

Step-by-step explanation:

3m+30 and - 2m+2n

5m+30+2n

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –
Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

(x+3)^2+(x+3)-2=0

Let

u=(x+3)

Rewrite the equation

(u)^2+(u)-2=0

Complete  the square

u^2+u=2

u^2+u+1/4=2+1/4

u^2+u+1/4=9/4

rewrite as perfect squares

(u+1/2)^2=9/4

square root both sides

(u+1/2)=\pm\frac{3}{2}

u=(-1/2)\pm\frac{3}{2}

u=(-1/2)+\frac{3}{2}=1

u=(-1/2)-\frac{3}{2}=-2

the solutions are

u=-2,u=1

<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

(u)^2+(u)-2=0

so

a=1\\b=1\\c=-2

substitute in the formula

u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}

u=\frac{-1\pm\sqrt{9}} {2}

u=\frac{-1\pm3} {2}

u=\frac{-1+3} {2}=1

u=\frac{-1-3} {2}=-2

the solutions are

u=-2,u=1

<em>Find the solutions of  the original equation</em>

For u=-2

-2=(x+3) ----> x=-2-3=-5

For u=1

1=(x+3) ----> x=1-3=-2

therefore

The solutions of the original equation are

x=-5 and x=-2

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Answer:

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Step-by-step explanation:

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I think it's B.............
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