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dyslexics of the world, untie!

20000 = 13000 (1 + .035)^t       divide both sides by 13000

20/13 =  (1 + .035)^t  

20/13 = (1.035)^t    take the log of both sides

log (20/13)  = log  (1.035)^t    and we can write

log (20/13)  = t * log  (1.035)        divide both sides by log (1.035)

log (20/13) /  log  (1.035) = t  ≈ 12.52  years

I assume that we have this : 

Tha area of AEDCB = area of square ABCD + area of triangle AED =

10^2  + (1/2) (8) (6)  =

100 + 24  =  124  sq units

However....if E is in the interior of ABCD the area of AEDCB 100 - 24  = 76 sq units

The inradius is given by :

2A / P    where A is the area and P is the perimeter

The perimeter is easy =  16 + 2*17  =  16 + 34  = 50

The triangle is isosceles.....let the base =16

And the height can be found using the Pythagorean Theorem as

√ [17^2 - 8^2]  = √ [289 - 64]  = √ 225  = 15

So....the area  =  (1/2)B*H = (1/2) 16 * 15   =  120

So.......the inradius  is    2 * 120 / 50  =  240 / 50   =  24/5  = 4.8

Here's a pic.....AB is the inradius

(1 . x)^20 = 3

20 ln (1 .x) = ln 3

ln (1 .x) = (ln 3)/20  =  0.0549306

( 1 .x) = e^0.549306 = 1.0564

x = 0.0564.

Answer is  5. 64 %

FV = PV x [1 + R]^N

20,000 = 13,000 x [1 + 0.035]^N   divide both sides by 13,000

1.53846..= 1.035^N    Take the log of both sides

N =Log(1.53846...) / Log(1.035)

N=12.52 ~ 13 Years.

THAT IS NO QUESTION!

FV=PV[1+R]^N

$3=$1[1+ R]^20 take the 20th root of both sides

3^(1/20) = 1 + R

1.056467 = 1 + R  subtract 1 from both sides.

R =0.0565 x 100

R=5.65% -  annual interest rate required to triple your money in 20 years.

This is a "combinations" problem  --  choosing an unordered group of 2 from a group of 5.

C(5, 2)  =  5! / ( 2! x 3! )  =  10

I agree with hectiar

Well.....it kinda' looks like Alan, heureka and I have beaten this horse to death !!!!

Here's yet another approach:

.

ARe the post boxes and letter boxes all different?

Connect MN

Now......MB = 3 and BN = 4...and angle ABC is right....so....triangle MBN is a 3-4-5 Pythagorean Triple right triangle with  MN = 5  and the area = (MB * BN)/2 =  (3*4)/ 2 = 6

And triangle ABC is a 6-8-10 Pythagorean Triple right triangle with AC  = 10

Now....MN divides the sides of triangle ABC proportionally so MN is parallel to AC

And AN is a transversal between two parallels so that angles CAN and MNP  form equal alternate interior angles....and since they are vertical angles, angle APC = angle  MPN.....

So....by angle- angle congruency,  triangle APC is similar to  triangle NPM

And AC can be considered the base of triangle APC and NM can be considered the base of triangle NPM.......but AC = 2NM.....so, by the square of the scale factor, 2....the area of triangle APC = 4 times the area of triangle NPM

Now....... triangle MBC  has an area of (MB * BC) / 2 =  (3 * 8) / 2 =  24/2 = 12

And triangle ANC has an area of  (AB * NC) / 2 =  ( 6 * 4) / 2 = 12

And MBC = NPMB + PNC

And ANC = APC + PNC

So....since MBC = ANC, then

NPMB + PNC = APC + PNC .....subtract PNC from both sides

NPMB  = APC

But NPMB  = area of triangle MBN + area of triangle NPM

So......substituting

6 + NPM  = APC

6 + NPM  = 4 (NPM)

6 = 3 (NPM)

2 = NPM

And APC is 4 times this  =  8  (square units)

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

another approach

Let \(BC = 8\)
Let \(AB = 6\)

\(\text{Let } A = \text{area}_{ABC} = \frac{AB\cdot BC}{2} = \frac{6\cdot 8}{2} = 24\)

\(\text{Let } A_1 = \text{area}_{BPN} = \text{area}_{NPC} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_2 = area_{BCP} = area_{CAP} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_3 = area_{APC} = \ ? \\ \text{Let area}_{BAN} = \frac{ AB\cdot \frac{BC}{2} } {2} = \frac{ AB\cdot BC } {4}\\ \text{Let area}_{BMC} = \frac{ BC\cdot \frac{AB}{2} } {2} = \frac{ AB\cdot BC } {4}\)

\(\begin{array}{|rcll|} \hline \text{area}_{BAN} &=& \text{area}_{BMC} = \frac{ AB\cdot BC } {4} \quad & | \quad \text{area}_{BAN} = A_1 +2A_2 \qquad \text{area}_{BMC} = A_2 +2A_1 \\ A_1 +2A_2 &=& A_2 +2A_1 \\ 2A_2-A_2 &=& 2A_1-A_1 \\ A_2 &=& A_1 \\\\ A_1 +2A_2 &=& \frac{ AB\cdot BC }{4} \quad & | \quad A_2 = A_1 \\ 3A_1 &=& \frac{ AB\cdot BC }{4} \\ A_1 &=& \frac{ AB\cdot BC }{12} \quad & | \quad AB = 6 \qquad BC = 8 \\ A_1 &=& \frac{ 6\cdot 8 }{12} \\ \mathbf{A_1} & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A &=& 2A_1+2A_2+A_3 \quad & | \quad A_2 = A_1 \\ A &=& 2A_1+2A_1+A_3 \\ A &=& 4A_1+A_3 \\ A_3 &=& A-4A_1 \quad & | \quad A = 24 \qquad A_1 = 4\\ A_3 &=& 24-4\cdot 4 \\ A_3 &=& 24-16 \\ \mathbf{A_3 } & \mathbf{=} & \mathbf{8} \\ \hline \end{array} \)

The area of triangle APC is 8 square units

Have a look at https://en.wikipedia.org/wiki/Newton%27s_method for a derivation of the Newton-Raphson method.

Applying the technique tp f(x) = x² - 5 we have:

f(x) = x² - 5

f`(x) = 2x

x_n+1 = x_n - (x_n² - 5)/(2x_n)

x_n+1 = (2x_n² - (x_n² - 5))/(2x_n) 

x_n+1 = (2x_n² - x_n² + 5))/(2x_n) 

x_n+1 = (x_n² + 5))/(2x_n) 

x_n+1 = (x_n + 5/x_n)/2 

.

Thank you. THat make sense.

Ex:

The number of ways that 5 children can be chosen from 8 is     8 chose 5     or 8C5

So it is used in counting problems.

Also note that every time 5 children are chosen, 3 are not chosen.

So 8C5=8C3

Or more generally

nCr  = nC(n-r)

If that is not quite what you want to know then say so and someone, maybe  me, will try to help further. :)

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

Centroid of a Triangle: The point where the three medians of the triangle intersect.

So P is the centroid of the triangle ABC.

\(\begin{array}{lcll} \text{Let } AB = 6 \\ \text{Let } BC = 8 \\\\ \text{Let } \vec{B} = \binom {0}{0} \\ \text{Let } \vec{A} = \binom {0}{AB} \\ \text{Let } \vec{C} = \binom {BC}{0} \\ \end{array} \)

\(\vec{P} \text{ the centroid of the triangle ABC}\ =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 (\vec{A}+\vec{B}+\vec{C}) \\ \vec{P} &=& \frac13 \Big( \binom {0}{AB}+\binom {0}{0}+\binom {BC}{0} \Big) \\ \vec{P} &=& \frac13 \binom {BC}{AB} \quad & | \quad AB = 6 \qquad BC = 8 \\ \vec{P} &=& \frac13 \binom {8}{6} \\ \vec{P} &=& \binom {\frac{8}{3}}{2} \\ \hline \end{array} \)

\(\text{Area}_{\text{APC}} = \ ? \)

\(\begin{array}{|rcll|} \hline \text{Area}_{\text{APC}} &=& \frac12 | (\vec{C}-\vec{P}) \times (\vec{A}-\vec{P}) | \\ &=& \frac12 | \Big(\binom {BC}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{AB}- \binom {\frac{8}{3}}{2} \Big) | \quad & | \quad AB = 6 \qquad BC = 8 \\ &=& \frac12 | \Big(\binom {8}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{6}- \binom {\frac{8}{3}}{2} \Big) | \\ &=& \frac12 | \binom {8-\frac{8}{3}}{0-2} \times \binom {0-\frac{8}{3}}{6-2} | \\ &=& \frac12 | \binom {\frac{16}{3}}{-2} \times \binom {-\frac{8}{3}}{4} | \\ &=& \frac12 \cdot \Big( \frac{16}{3} \cdot 4 - (-2)\cdot (-\frac{8}{3}) \Big) \\ &=& \frac12 \cdot \Big( \frac{64}{3} - \frac{16}{3} \Big) \\ &=& \frac12 \cdot \frac{48}{3} \\ &=& \frac{48}{6} \\ \text{Area}_{\text{APC}} &=& 8 \text{ square units } \\ \hline \end{array}\)

Use the  " / "  symbol for the fraction bar.

For example, to do:     \(\frac25\)

Type:                           2/5

If the numerator or denominator contains another operation, you might need to put parenthesees around it. Make sure the fraction displays how you want at the top.  

Let's call the number  " n " .

The problem tells us that...

n * 2n  =  72

2n²  =  72              Divide both sides of this equation by  2  .

n²  =  36                Take the  ±  square root of both sides.

n  =  ±√36

n  =  ± 6

So..your number can be either  6  or  -6  .

Half of   6  is    3  .

Half of  -6  is  -3  .

    (8x³)   *   (-27)

=  (8 * x³) * (-27)

=  8   *  x³  *  -27

=  8  *  -27  *  x³

=  -216  *  x³

=  -216x³

???????