All Answers 358187 Answers

Hi CPhill,

thank you for replying, but that is not it....the line they draw is actually a very narrow rectangular, and it stretches from the x axis upwards to where it meets the graph. I know it is not anything to do with symmetry since it is plotted anywhere (of course calculated). 

the book explains how in the past before Integration was developed, small rectangles were used to get an approximate area size. So the area would be filled with rectangles as narrow as possible (x values), and then were the areas of each rectangle calculated and then added together. The area of each rectangle is lb, so since the rectangles are positioned vertically, the length would be y and the breadth would be \(\Delta x\). The greater this value, the less accurate the final answer. (Total Area)

Then they explain that integration makes the \(\Delta x\) value to be zero, thus we have millions of very thin rectangles (with lengths (y) and zero breadths) of which the areas need to be calculated and added together, hence the introduction to integration. Now, one of these "strips" are called "a representative strip"....please bear in mind I translated directly from Afrikaans...so this may not even be the true name for it. 

The question is: Show the "representative strip" you are going to use to calculate the area. Then the second question is to use integration to actually calculate the area size. 

If nothing makes sense, I will understand....Thanx for trying though...

I'm not sure, but I think this might be something known as the "average value"

For instance......let us find the area of the curve between the function  1 - x^2   and the x axis

The curve will intercept the x axis at   x = - 1    and x  = 1

So.....using symmetry, the area is

    1                                                      1

2* ∫  1 - x^2  dx     =    2  [ x - (1/3)x^3 ]       =   2 [  1 - 1/3]  =  2 [ 2/3] =  4/3  units^2

    0                                                      0

Note that we can construct a rectangle with the same area if we let the width range   from x  = - 1 to x  = 1  =  2 units = the interval width

So....the height of this rectangle  must be      (4/3) / 2  =  4/6  = 2/3

See this image :

Note that the rectangle ABCD  has the same area as the definite integral

To   always find this average value  we can calculate this :

[ area  of integration ]

_________________    =      height of rectangle  =  average value

  interval width

Zan has created this iterative rule for generating sequences of whole numbers:

1) If a number is 25 or less, double the number.
2) If a number is greater than 25, subtract 12 from it.

Let F be the first number in a sequence generated by the rule above.

F is a "sweet number" if 16 is not a term in the sequence that starts with F.
How many of the whole numbers 1 through 50 are "sweet numbers"?

 1 -> 2 -> 4 -> 8 -> 16
 2 -> 4 -> 8 -> 16
 3 -> 6 -> 12 -> 24 -> 48 -> 36 -> 24 sweet number
 4 -> 8 -> 16
 5 -> 10 -> 20 -> 40 -> 28 -> 16
 6 -> 12 -> 24 -> 48 -> 36 -> 24 sweet number
 7 -> 14 -> 28 -> 16
 8 -> 16
 9 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 10 -> 20 -> 40 -> 28 -> 16
 11 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 12 -> 24 -> 48 -> 36 -> 24 sweet number
 13 -> 26 -> 14 -> 28 -> 16
 14 -> 28 -> 16
 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 16 -> 32 -> 20 -> 40 -> 28 -> 16
 17 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 18 -> 36 -> 24 -> 48 -> 36 sweet number
 19 -> 38 -> 26 -> 14 -> 28 -> 16
 20 -> 40 -> 28 -> 16
 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 23 -> 46 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 24 -> 48 -> 36 -> 24 sweet number
 25 -> 50 -> 38 -> 26 -> 14 -> 28 -> 16
 26 -> 14 -> 28 -> 16
 27 -> 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 28 -> 16
 29 -> 17 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 31 -> 19 -> 38 -> 26 -> 14 -> 28 -> 16
 32 -> 20 -> 40 -> 28 -> 16
 33 -> 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 35 -> 23 -> 46 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 36 -> 24 -> 48 -> 36 sweet number
 37 -> 25 -> 50 -> 38 -> 26 -> 14 -> 28 -> 16
 38 -> 26 -> 14 -> 28 -> 16
 39 -> 27 -> 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 40 -> 28 -> 16
 41 -> 29 -> 17 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 43 -> 31 -> 19 -> 38 -> 26 -> 14 -> 28 -> 16
 44 -> 32 -> 20 -> 40 -> 28 -> 16
 45 -> 33 -> 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 46 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 47 -> 35 -> 23 -> 46 -> 34 -> 22 -> 44 -> 32 -> 20 -> 40 -> 28 -> 16
 48 -> 36 -> 24 -> 48 sweet number
 49 -> 37 -> 25 -> 50 -> 38 -> 26 -> 14 -> 28 -> 16
 50 -> 38 -> 26 -> 14 -> 28 -> 16

There are 16 sweet number:
  3 -> 6 -> 12 -> 24 -> 48 -> 36 -> 24 sweet number
  6 -> 12 -> 24 -> 48 -> 36 -> 24 sweet number
  9 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 12 -> 24 -> 48 -> 36 -> 24 sweet number
 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 18 -> 36 -> 24 -> 48 -> 36 sweet number
 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 24 -> 48 -> 36 -> 24 sweet number
 27 -> 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 33 -> 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 36 -> 24 -> 48 -> 36 sweet number
 39 -> 27 -> 15 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 45 -> 33 -> 21 -> 42 -> 30 -> 18 -> 36 -> 24 -> 48 -> 36 sweet number
 48 -> 36 -> 24 -> 48 sweet number

Let     f(x)  =  x * (3x + 1)  =  3x² + x

Let's see what  f(x)   is  when  x  is at the endpoints of the interval.

f(-7/3)  =  3(-7/3)² + (-7/3)  =  14

f(2)  =  3(2)² + 2  =  14

Aha! they are the same, just as I suspected! 🕵️‍♀️

Let's see what  f(x)  is when  x  is in the interval.

f(0)  =  3(0)² + 0  =  0

And it is true that  0 < 14

Since  f(x)  is a parabola, we can be sure that  f(x) < 14  if and only if  x  is in the interval  (-7/3, 2)

"Modulo \(m\) graph paper" consists of a grid of \(m^2\) points, representing all pairs of integer residues \((x,y)\) where \((0\le x)\) .
To graph a congruence on modulo m graph paper, we mark every point \((x,y)\) that satisfies the congruence.
For example, a graph of \(y\equiv x^2\pmod 5\)  would consist of the points \((0,0),\ (1,1),\ (2,4),\ (3,4),\ \text{and}\ (4,1)\).

The graph of
\(3x\equiv 4y-1 \pmod{35}\)
has a single x-intercept \((x_0,0)\)  and a single y-intercept \((0, y_0)\), where \(0\le x_0,y_0<35\).

What is the value of \(x_0+y_0\)?

\(\begin{array}{|lrclrcl|} \hline P(x_0,\ 0): & 3x &\equiv& 4y-1 \pmod{35} \\ & 3x_0 &\equiv& 4\cdot 0 -1 \pmod{35} \\ & 3x_0 &\equiv& -1 \pmod{35} \\\\ & 3x_0 &=& -1 +35n \quad n\in \mathbb{Z} \\ & x_0 &=& \dfrac{-1 +35n} {3} \\ & x_0 &=& \dfrac{-1 +36n-n} {3} \\ & x_0 &=& 12n-\underbrace{\dfrac{1+n} {3}}_{=a}\quad a\in \mathbb{Z} \\ & x_0 &=& 12n-a & a&=& \dfrac{1+n} {3} \\ & & & & 3a&=& 1+n \\ & & & &\mathbf{ n}&=&\mathbf{ 3a-1} \\ & x_0 &=& 12(3a-1)-a \\ & x_0 &=& 36a-12-a \\ & x_0 &=& -12+35a \\\\ & x_0&\equiv& -12 \pmod{35} \\ & x_0&\equiv& 35 -12 \pmod{35} \\ & x_0&\equiv& 23 \pmod{35} \\ & \mathbf{x_0}&=& \mathbf{23} \qquad 0\leq x_0 < 35\\ \hline \end{array}\)

\(\begin{array}{|lrclrcl|} \hline P(0,\ y_0): & 3x &\equiv& 4y-1 \pmod{35} \\ & 0 &\equiv& 4y_0 -1 \pmod{35} \quad | \quad \cdot (-1) \\ & 0 &\equiv& -4y_0 +1 \pmod{35} \\ \\ & 0 &=& -4y_0 +1 +35n \quad n\in \mathbb{Z} \quad | \quad +4y_0 \\ & 4y_0 &=& 1 +35n \quad | \quad : 4 \\ & y_0 &=& \dfrac{1 +35n} {4} \\ & y_0 &=& \dfrac{1 +36n-n} {4} \\ & y_0 &=& 9n+\underbrace{\dfrac{1-n} {4}}_{=a}\quad a\in \mathbb{Z} \\ & y_0 &=& 9n+a & a&=& \dfrac{1-n} {4} \\ & & & & 4a&=& 1-n \\ & & & &\mathbf{ n}&=&\mathbf{ 1-4a} \\ & y_0 &=& 9(1-4a)+a \\ & y_0 &=& 9-36a+a \\ & y_0 &=& 9-35a \\\\ & y_0&\equiv& 9 \pmod{35} \\ & \mathbf{y_0}&=& \mathbf{9} \qquad 0\leq y_0 < 35\\ \hline \end{array}\)

\(\mathbf{x_0 + y_0} = 23+9 \mathbf{=32}\)

...or  2/7   if you look at the web2.0calc display

0.9*(3/2.25- 1/8* (1+ 2/3 + 1/9 ))- 4/7 /0.8 = 0.2857142857142857

Remember, the radius of a circle is the distance between the center and a point on the circle.

The center of the circle is  (2, 3)

The point  (-2, 0)  is on the circle.

Let's use the distance formula to find the radius.

radius  =  distance between  (2, 3)  and  (-2, 0)

radius  =  \(\sqrt{(0-3)^2+(-2-2)^2}\)

radius  =  \(\sqrt{(-3)^2+(-4)^2}\)

radius  =  \(\sqrt{9+16}\)

radius  =  \(\sqrt{25}\)

radius  =  5

Let the number of unripe oranges  =  U

Let the number of ripe oranges  =  R

There were 18 more ripe oranges than unripe oranges. In other words,

the number of ripe oranges  =  the number of unripe oranges + 18

R  =  U + 18

The number of rotten oranges  =  24

84 – 24  =  60

R + U  =  60

Since  R  =  U + 18  we can substitute  U + 18  in for  R:

R + U  =  60

U + 18 + U  =  60

U + U  =  60 – 18

2U  =  42

U  =  21

Now that we know the value of  U  we can find the value of  R

R  =  U + 18

R  =  21 + 18

R  =  39

the number of ripe oranges  =  39

the number of rotten oranges  =  24

the number of unripe oranges  =  21

num of ripe oranges : num of rotten oranges : num of unripe oranges   =   39 : 24 : 21   =   13 : 8 : 7

84 oranges - 24 rotten= 60 oranges left

39 are ripe and 21 are unripe        ripe : rotten : unripe  = 39:24:21     simplifies to   13:8:7

Let Mai's height  =  m

Let Jon's height  =  j

We want to find  \(\frac{m}{j}\)

\(\frac14\cdot m\ =\ \frac25\cdot j\)

                             Multiply both sides of the equation by  4

\(m\ =\ \frac85\cdot j\)

                             Divide both sides of the equation by  j

\(\frac{m}{j}\ =\ \frac85\)

The ratio of  m  to  j  is  \(\frac85\)

Draw a height from the base to create two congruent 30-60-90 triangles.

the side across from the  60°  angle  =  10

the side across from the  30°  angle  =  10/√3

area of isosceles triangle  =  ( 1/2 )( base )( height )

area of isosceles triangle  =  ( 1/2 )( 20 ft )( 10/√3 ft )

area of isosceles triangle  =  100 / √3     square feet

area of isosceles triangle  =  100√3 / 3   square feet

Here's equilateral triangle ABC and a circle with radius 1 centered on point A:

The intersecion of the triangle and the circle is the highlighted sector.

probability that a randomly chosen point lands in highlighted sector  =  area of sector / area of triangle

So we just have to find the area of the sector and the area of the triangle.

Let's find the area of the sector:

\(\frac{\text{area of sector}}{\text{area of circle}}\ =\ \frac{\text{measure of central angle}}{360^\circ} \\~\\ \frac{\text{area of sector}}{\pi\cdot1^2}\ =\ \frac{60^\circ}{360^\circ} \\~\\ \text{area of sector}\ =\ \frac{60^\circ}{360^\circ}\cdot\pi\cdot1^2 \\~\\ \text{area of sector}\ =\ \frac{\pi}{6} \)

Now let's find the area of the triangle:

\(\text{area of triangle}\ =\ \frac12\cdot\text{base}\cdot\text{height}\\~\\ \text{area of triangle}\ =\ \frac12\cdot3\cdot\frac{3\sqrt3}{2}\\~\\ \text{area of triangle}\ =\ \frac{9\sqrt3}{4}\)

Now we can find the probability in question.

\(\text{probability} = \frac{\text{area of sector} }{ \text{area of triangle}}\\~\\ \text{probability} = \text{area of sector} \div \text{area of triangle} \\~\\ \text{probability} = \frac{\pi}{6} \div \frac{9\sqrt3}{4} \\~\\ \text{probability} = \frac{\pi}{6} \cdot \frac{4}{9\sqrt3} \\~\\ \text{probability} = \frac{4\pi}{54\sqrt3} \\~\\ \text{probability} = \frac{4\sqrt3\pi}{162} \\~\\ \text{probability} = \frac{2\sqrt3\pi}{81} \)_

i wrote it in the instructions

how do you figure that key

It is 120. :D

Answered here.

https://web2.0calc.com/questions/help_3820

Plenty of answers to choose from :D

-\(\pi\) KeyLimePi

it seems i am the only one on so ill try but i do not know what the vertex angle is can you infom me on that?