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Tanya [424]
2 years ago
6

Willing to give brainlyest if you can aswer these for me ASAP​

Mathematics
2 answers:
pav-90 [236]2 years ago
6 0

Answer:

x+5=12.          x = 7,

7x=21.            x = 3,

b-2/3=5/6.     b = 1.5,

3/4x=6x=8.    x = 1/3,

45=3(x+1).      x = 14,

22.5+1.5x.      x = 15,

20+(5•9).        $65

Step-by-step explanation:

 

vichka [17]2 years ago
5 0

Answer:

65

Step-by-step explanation:

its simple math

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Step-by-step explanation:

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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

4 0
3 years ago
Read 2 more answers
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USPshnik [31]
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