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Written clearly in the answer.

Guest said, see if f(x) = x works.

It means, see if f(x) = x is a valid solution.

You could just superimpose them to see it is not true:

Thanks Melody! I think I have the answer now.

****  deleted ****    Misread the question....Sorry !     ..

Thank you so much Alan!!!!!!!!!!!!!!!

As follows:

Given that all the circles have equal radius of 10 cm, what is the area of the shaded blue part?

S=(10+10)*(10+10)

H(t)  =  -16·t² + 162

Find the value of t that makes H(t) = 0 ...

-16·t² + 162  =  0

          -16·t²  =  -162

                t²  =  -162/ -16

                t²  =  162/ 16

                 t  =  sqrt(162/16)     (ignore the negative value)

Let  a  be the length across the top of the red rectangle.

Let  b  be the length across the top of the blue rectangle.

Let  c  be the height across the left side of the red rectangle.

Let  d  be the height across the left side of the yellow rectangle.

Therefore, the area of the

   red rectangle is:      a·c  =  12

   blue rectangle is:     b·c  =  15

   yellow rectangle is:  a·d  =  8

   green rectangle is:  b·d  =  ??

a·c  =  12   --->   a  =  12/c

a·d  =  8     --->   a  =  8/d

Combining these two equations:  12/c  =  8/d   --->   12d  =  8c   --->   3d  =  2c   --->   c  =  3d/2

  b·c  =  15   --->   c  =  15/b

Combining  c  =  15/b  with  c  =  3d/2   --->   15/b  =  3d/2   --->   3bd  =  30   --->   b·d  =  10

In a math contest, every correct answer is worth 3 points and every wrong answer is worth -2 points. If Ralph got 30 questions correctly and answered 10 items incorrectly what was his total points?  

correct answers .....    30 correct    @   3 pts per correct      =    90 points  

incorrect answers ...   10 incorrect  @ –2 pts per incorrect   =  –20 points  

                                                                                                 ————  

added together the total points received is                                  70 points  

_.  

I still have not worked out how to do the general case, however, a square is a special case of a parallelogram.

The question implies that the answer will always be the same.

so...

If you look at this specific case (A square of side length 2)

distance HF is easy to work out

then work out the coordinates of Q

then distance OQ can be found.

Now you can finish it.

Sketch a regular pentagon and one of its diagonals

--  that diagonal, along with two sides forms an isosceles triangle

You know two of the sides of that triangle; they are each 1

You can find the size of the vertex angle of the pentagon by using this formula:  (n - 2)·180^o / n

--  since  n = 5:  each vertex angle  =  (5 - 2)·180^o / 5  =  108^o.

To find the length of the diagonal (the base of the isosceles triangle), use the Law of Cosines:

--  c²  =  a² + b² - 2·a·b·cos(C)

--  c²  =  1² + 1² - 2·1·1·cos(108^o)

The value of c will be the length of a diagonal of a regular pentagon of unit side length ...

I would use Heron's formula for the area of a triangle:

--  if a, b, and c are the sides of a triangle:

--  first find the semiperimeter, s, by using this formula:  s  =  (a + b + c) / 2

--  and then find the area by using this formula:  area  =  sqrt( s · (s - a) · (s - b) · (s - c) )

To find a, b, and c  --  find the square roots of the areas of the squares to find their side lengths:

--  a  =  sqrt(116)

--  b  =  sqrt(370)

--  c  =  sqrt(74)

Although there may be a way to do this without a calculator, I would suggest using one ... preferably one that allows

you to store and recall values.

Interestingly, the result is a whole number.

Thanks Alan :)

One yard is equivalent to 91.44 centimeters.

You can calculate how many "full" 10s go into 91.44

If you don't understand anything feel free to ask.

The circumference of a circular garden is 8.7 m. Find the radius of the garden.

\(\begin{array}{|rcll|} \hline 2\pi r &=& 8.7\ \text{m} \quad | \quad : 2\pi \\ r &=& \dfrac{8.7}{2\pi} \\ \mathbf{r} &=& \mathbf{1.38464800490\ \text{m}} \\ \hline \end{array}\)

The radius of the garden \(\mathbf{1.38\ \text{m}}\)

The perimeter of a rectangle is 60 cm. Its length is twice its width. Find its length.

w = width

l = length

p = perimeter

\(\begin{array}{|rcll|} \hline p=60 &=& 2(w+l) \quad | \quad l = 2w \\ 60 &=& 2(w+2w) \quad | \quad : 2 \\ 30 &=& w+2w \\ 30 &=& 3w \quad | \quad : 3 \\ 10 &=& w \\ \mathbf{w} &=& \mathbf{10} \\\\ l &=& 2w \quad | \quad w=10 \\ l &=& 2*10 \\ \mathbf{l} &=& \mathbf{20} \\ \hline \end{array}\)

The length is 20

The link works Melody.

Regular pentagon ABCDE and regular hexagon AEFGHI are drawn on opposite sides of line segment AE such that they are coplanar.

What is the degree measure of exterior angle DEF?

Here is the picture

\(\begin{array}{|rcll|} \hline \angle DEA &=& \dfrac{(5-2)*180^\circ}{5} \\ \angle DEA &=& \dfrac{3*180^\circ}{5} \\ \angle DEA &=& 3*36^\circ \\ \mathbf{\angle DEA} &=& \mathbf{108^\circ} \\\\ \angle AEF &=& \dfrac{(6-2)*180^\circ}{6} \\ \angle AEF &=& \dfrac{4*180^\circ}{6} \\ \angle AEF &=& 4*30^\circ \\ \mathbf{\angle AEF} &=& \mathbf{120^\circ} \\\\ \angle DEF &=& 360^\circ - \angle DEA - \angle AEF \\ \angle DEF &=& 360^\circ - 108^\circ - 120^\circ \\ \angle DEF &=& 360^\circ - 108^\circ - 120^\circ \\ \angle DEF &=& 360^\circ - 228^\circ \\ \mathbf{\angle DEF} &=& \mathbf{132^\circ} \\ \hline \end{array}\)

Hint: Remember that if a and b are distinct integers, then P(a) - P(b) is divisible by a-b.

In parallelogram EFGH, let M be the midpoint of side EF, and let N be the midpoint of side EH. Line segments FH and GM intersect at P, and line segments FH and GN intersect at Q. Find PQ/FH. You don't need to give an answer if you don't have one. A hint or a diagram would be just as helpful. Thanks!

Here is a link to an adjustable graph. I do not remember sharing a link to a GeoGebra graph before so I will be interested in whether it works.  I suppose it will....

The ratio of A and B is 3:7 and the sum of the LCM and HCF of A and B is 352.
Find A+B.

\(\begin{array}{|rcll|} \hline \dfrac{A}{B} &=& \dfrac{3}{7} \\\\ \dfrac{A}{B} &=& \dfrac{3n}{7n} \\ \hline A &=& 3n \\ B &=& 7n \\ \mathbf{A+B} &=& \mathbf{10n} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline LCM(3n,\ 7n) &=& 3*7*n \\ \mathbf{LCM(3n,\ 7n)} &=& \mathbf{21n} \\ \mathbf{HCF(3n,\ 7n)} &=& \mathbf{n} \\ \hline LCM(3n,\ 7n)+ HCF(3n,\ 7n) &=& 21n+n \\ LCM(3n,\ 7n)+ HCF(3n,\ 7n) &=& 22n \quad | \quad LCM(3n,\ 7n)+ HCF(3n,\ 7n) = 352 \\ 352 &=& 22n \\ n &=& \dfrac{352}{22} \\ \mathbf{n} &=& \mathbf{16} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{A+B} &=& \mathbf{10n} \quad | \quad n=16 \\ A+B &=& 10*16 \\ \mathbf{A+B} &=& \mathbf{160} \\ \hline \end{array}\)