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The height of the triangle =4sqrt(2)

Area =[4sqrt(2)]^2 / 2 =16 sq. units - the area of triangle

Just like in the previous problem you posted, factor out the GCF of each term in the parentheses.

In this case, we are utilizing the rule of a sum of cubes. The rule is the following:

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

To remember the signs, you can remember this acronym. "soap." The acronym dictates the signage sequentially. 

Same sign

Opposite sign

Always

Positive

\(a=x\\ b=\sqrt[3]{9769}\)

9769 is not a perfect cube, so this is the best we can do.

\((x+\sqrt[3]{9769})(x^2-x\sqrt[3]{9769}+\sqrt[3]{9769}^2)\)

4/[4+5] x 90 = 40 degrees

40 x 1.10 =44 degrees

90 - 44 = 46 degrees - the other angle.

1 - 46/50 =8% - decrease in the 2nd angle.

Let  x  be the side length of the triangle. Here's a quick drawing:

From the Pythagorean theorem...

(x/2)² + (√6)²  =  x²

x²/4 + 6  =  x²                Multiply through by  4 .

x² + 24  =  4x²               Subtract  x²  from both sides.

24  =  3x²                       Divide both sides by  3  then take the positive square root of both sides.

x  =  √8

And...

area of triangle, in square units  =  (1/2)(x)(√6)   =   (1/2)(√8)(√6)   =   (1/2)(4√3)   =   2√3

Here you go:

\(\hspace{7mm}23\frac{5}{9}=23.\overline{5}\\ 9\hspace{1mm}\overline{)212}\\ \hspace{-5mm}-18\\ \hspace{6mm}\overline{\hspace{4mm}}\\ \quad\hspace{3mm}32\\ \hspace{2mm}-27\\ \hspace{8mm}\overline{\hspace{4mm}}\\ \quad\quad\hspace{1mm}5\)

ok thanks CPhill 

∛ 5   is irrational  .....we can only approximate its value

what's the cube root of 5 though

Let the sides of the isosceles triangle =1, then:

Hypotenuse=sqrt(2)

Area of triangle = [1 x 1] / 2 =1/2

Area of the square =sqrt(2)^2 =2

Area of pentagon =2 + 1/2 =2 1/2 sq. units.

{1/2} / {2 1/2} =1/5 - ratio of area of triangle to the pentagon.

Using

a^3  +  b^3   =   (a  + b) (a^2  - ab + b^2)   ...   we have...

x^3   +   5    =

( x  +  ∛ 5  )   (  x^2    -  x ∛ 5   +   [∛ 5 ]^2  )

Let's call the length of each leg of the right triangle  " t " .

And let's call the side length of the square  " s " .

The triangle is an isosceles right triangle, so

t² + t²  =  s²

2t²  =  s²

And...

area of triangle  =  (1/2)(t * t)   =  t²/2

area of pentagon  =  area of square + area of triangle

area of pentagon  =          s²             +         t²/2                Substitute  2t²  in for  s² .

area of pentagon  =          2t²            +         t²/2

So...

\(\frac{\text{area of triangle}}{\text{area of pentagon}}=\frac{\frac{t^2}{2}}{2t^2+\frac{t^2}{2}}=\frac{\frac12}{2+\frac12}=\frac12\cdot\frac25=\frac15=\frac{20}{100}\)

The area of the triangle is  20%  of the area of the pentagon.  

Great job, hectictar !. Thank you.

     \(\frac{1}{1+\frac{1}{\sqrt2}}\)

                     Rewrite the  1  in the denominator as  \(\frac{\sqrt2}{\sqrt2}\) .

=  \(\frac{1}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt2}}\)

                     Add the fractions in the denominator together.

=  \(\frac{1}{\frac{\sqrt2+1}{\sqrt2}}\)

                                That is the same as...

=  \(1\div\frac{\sqrt2+1}{\sqrt2}\)

                                Invert the second fraction and multiply.

=  \(1\,\cdot\,\frac{\sqrt2}{\sqrt2+1}\)

=  \(\frac{\sqrt2}{\sqrt2+1}\)

                                Multiply the numerator and denominator by  \(\sqrt2-1\) .

=  \(\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}\)

=  \(\frac{2-\sqrt2}{2-1}\)

=  \(\frac{2-\sqrt2}{1}\)

=  \(2-\sqrt2\)

I honestly do not know the limitations of the profile picture, but for reference my picture is a square at 128x128 pixels.

 3x^2-29x+56 by x-7.

                3x     -     8

x -  7   [    3x^2     -   29x      +       56  ] 

                3x^2     -   21x

              __________________________

                               -8x         +       56

                               -8x         +       56

                             _________________

                                                       0

Yes it does but you need to findthe value of x first.

I do not think the asker understood this.

She seemed to think they could be found independently of one another. 

okay...I will do my best!

You did not completely distribute the 3x to every term in your divisor, which is a common error that can result from a lack of concentration or attentiveness. You seem to understand the process (based on your above work), so continue practicing until you master it. For long division, I would recommend checking your math frequently during long division to ensure no errors are made. Once one error is made, it flows through until the end. Also, it may be difficult to spot an error after finishing because of sensory overload. 

So how did I get 7x instead of 21x?

\(3x-8\) looks good to me!

-8x/x=-8, -8*(x-7)=-8x+56, -8x-(-8x)=0, 56-56=0, 3x-8 R=0?

Graphing is confusing especially if you are new at it. 

y=x

Now graph those points on a number plane.    Conman13 has already done it for you. :)

y=-x     (The negative sign has to go in front of the x value to get the y yalue.

Now graph those points on a number plane.

The only time that y=x=-x is when x=0  and y=0 so that i the only point of intersection for these 2 graphs.

I immediately see an issue here!

\(\hspace{26mm}3x\\ x-7\overline{)3x^2-29x+56}\\ \hspace{12mm}3x^2-\textcolor{red}{21x}\\ \hspace{10mm}\overline{\quad\quad\quad\quad\quad}\\ \hspace{15mm}0\hspace{1mm}-\hspace{3mm}8x\)

See if you can finish it from here.