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3 years ago

Does anyone know how to do #11 ? I need your help again :(

Mathematics

1 answer:

Klio2033 [76]3 years ago

6 0

Answer:

Step-by-step explanation:

Four consecutive counting numbers would look like this

x + 1

x + 2

x + 3

Do you see that these numbers are 1 larger than the one before it. So if x is 20 the numbers would be 20 21 22 23

Now take the 4 general numbers and add them

x + x + 1 + x+2 + x + 3

4x + 6

Two will go into 4x to give 4x/2 = 2x

Two will go into 6 to give 6/2 = 3

Always happens. There are no exceptions. Try x = 20 again

20 + 21 + 22 + 23 = 86

Can this number be divided by 2? Yes it can 86/2 = 43

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$\hbox{Domain:}\\
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x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
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x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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3 years ago

I need this done please

erastova [34]

Answer:

thanks thanks thanks thanks thanks for your points

8 0

2 years ago

What is the perimeter of a rectangle with sides 2/3x+10 and 1/3x+5

gogolik [260]

Answer:

The perimeter of the rectangle is 2x+ 30

Step-by-step explanation:

Sides of the rectangle are 2/3x+10 and 1/3x+5

Now, Perimeter of the Rectangle = 2(Length + Breadth)

Here, sum of the sides = $(\frac{2}{3}x + 10 ) + (\frac{1}{3}x + 5 ) = (\frac{2}{3}x +\frac{1}{3}x ) + ( 10+ 5 )$

adding like terms with like terms, we get

( $\frac{2x + x}{3}) + 15 = \frac{3x}{3} + 15$

Hence, Sum = (x + 15)

Now, 2(Length + Breadth) = 2(x+15) = 2x + 30

hence, the perimeter of the rectangle = 2x+ 30

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3 years ago