All Answers 356344 Answers

G

2       8

F           E

FE  = sqrt [ 8^2 - 2^2]  = sqrt 60  =  2 sqrt 15

sin G =  FE / EG =    2sqrt (15) / 8  =  sqrt (15)  / 4

sec G  =  GE / FG  =  8 / 2  =   4

tan E  =  FG / FE  =  2  /  [ 2sqrt 15]  =  1 /sqrt 15 =  sqrt (15)  / 15

A

N

C      M       B

AC^2  +  (BC/2)^2  = 1

(AC/2)^2 + BC^2 = 1           simplify this system

AC^2 + BC^2 / 4  =  1

AC^2/4 + BC^2 = 1

4AC^2 + BC^2  = 4

-AC^2/4 - BC^2  =  -1       add these

(15/4)AC^2  =  3

AC^2  = 4/5

And

AC^2 + (BC/2)^2  = 1

4/5 + BC^2 / 4  =  1

BC^2 / 4  = 1/5

BC^2  = 4/5

Hypotenuse length =  sqrt [ AC^2 + BC^2 ] =   sqrt [ 4/5 + 4/5 ] =  sqrt (8/5) = 2sqrt (2) / sqrt (5)  =

2 sqrt (10) /  5

British Flag Theorem

AP^2  + CP^2 = BP^2 + DP^2

3^2  + 3^2  = 3^2  +  DP^2

9  = DP^2

DP  =  3

https://web2.0calc.com/questions/help-with-triangle_26

DE = EC

EA  =  sqrt [ AC^2 -  EC^2] =  sqrt [5^2 - 4^2] = sqrt [ 9 ]=   3

2 * area ADC  = area of parallelogram

2 * (1/2) DC * EA   =

DC * EA  =

8 * 3  =  

24

Let adjacent  sides be x, y

So.....

2x + 2y  =  40

x + y  = 20

y = 20 - x

By the Pythagorean Theorem

x^2  + y^2  = (10 sqrt 2) ^2

x^2 + ( 20 - x)^2  =  200

x^2 + x^2  - 40x + 400 =  200

2x^2  - 40x + 200  = 0

x^2 - 20x + 100  = 0

(x -10)^2  = 0           take positive root

x  - 10  =  0

x = 10

y = 20 - 10  =  10

Area = xy =  10 * 10  =   100

Area =  2 * (1/2) * 12^2 * sin (20°)  =  144 * sin (20°)  ≈  49.25

The ray will strike X  and then , because angle A = angle B  it will travel to  vertex D

The  total path length =   2 sqrt [ .5^2  + 1^2 ] =  2 sqrt [ 5/4]  = sqrt 5

Semi-perimeter  =  [ 15 + 9 + 10 ] /  2  =  17

Area  =   sqrt [ 17 * 2 * 7 * 8 ]  = 4 sqrt 119

Shortest altitude is  drawn  to the longest side

4 sart 119 =  (1/2) AB * Altityude

4 sqrt 119 = (1/2) 15 * Altitude

[ 4 sqrt 119  ]  /  [ 1/2 * 15 ]  =   [8 sqrt 119] / 15 =  Altitude  ≈  5.82

Without explanation , side of  square =

[  product of leg lengths  ]    [ sum of leg lengths ] = 

[ 6 * 6 ] / [ 6 + 6 ] = 

36  /  12     =

3

https://web2.0calc.com/questions/hexagon_12

Let:

b1​ be the length of the shorter base.

b2​ be the length of the longer base (which is 4 units greater than b1​ ).

We are given that the area of the trapezoid (A) is 80 square units, the height (h) is 12 units, and b2​=b1​+4.

Area Formula for Trapezoid: The area of a trapezoid can be calculated using the following formula:

A = ½ * h * (b₁ + b₂)

where:

A is the area

h is the height

b₁ and b₂ are the lengths of the two bases

Substitute Known Values: We are given that A = 80, h = 12, and b₂ = b₁ + 4. Let's substitute these values into the formula:

80 = ½ * 12 * (b₁ + (b₁ + 4))

Solve for b₁ (shorter base):

Simplify the right side of the equation: 80 = 6 * (2b₁ + 4)

Expand the parentheses: 80 = 12b₁ + 24

Subtract 24 from both sides: 56 = 12b₁

Divide both sides by 12: b₁ = 4.67 (rounded to two decimal places)

Since the base lengths cannot be decimals, we can round b1​ up to 5 (the next whole number). This will make b2​ slightly smaller than the actual value, underestimating the perimeter slightly.

Finding the Base Lengths:

Shorter base (b₁): b₁ ≈ 5 units (rounded up from 4.67)

Longer base (b₂): b₂ = b₁ + 4 = 5 + 4 = 9 units

Finding the Perimeter:

The perimeter (P) of the trapezoid is the sum of all its side lengths. Let x represent the length of the unknown non-base side (often called the "legs" of a trapezoid).

P = b₁ + b₂ + x + x (since there are two equal sides that are not bases)

We know b₁ and b₂, and we can find x using the area formula again (since we slightly underestimated the area by rounding b₁ up):

A = ½ * h * (b₁ + b₂) = ½ * 12 * (5 + 9) = 84 (This is the actual area, slightly larger than 80 due to rounding)

Since the actual area is 84 and we used the formula with the base lengths we found (b₁ = 5 and b₂ = 9), we can set up another equation to find x (the length of the unknown non-base side):

84 = ½ * 12 * (5 + 9 + 2x)

Solving for x (similar to solving for b₁), we get x ≈ 3.

Perimeter Calculation:

P = b₁ + b₂ + x + x = 5 + 9 + 3 + 3 = 20 units

Therefore, the perimeter of the trapezoid is 20 units.

We are given that QN⊥PR and QN=15. Since N is the midpoint of PR, segment QN is an altitude of triangle PQR bisecting the base PR. Therefore, triangle PQR is a right triangle with right angle at R.

We are also given that PR=20. Now, we have a right triangle (PQR) with one leg length (QN=15) and the hypotenuse length (PR=20). We can use the Pythagorean Theorem to find the missing leg length (QR) :

PR2=QN2+QR2

202=152+QR2

QR2=202−152=25

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

QR=±5

Since QR represents a side length, we take the positive square root:

QR=5 units.

Therefore, the length of segment QR is 5 units.

For problem 2:

Since ∠PQR=∠PRQ, triangle PQR is isosceles with PQ=PR. Let x represent the length of segment PQ (and segment PR).

We are also given that RQ=8 and SQ=3. Segments RQ and SQ are the legs of right triangle QRS. We can use the Pythagorean Theorem to solve for QR (which we already know is 8):

RQ2=QS2+RS2

82=32+RS2

RS2=82−32=55

Since we now know that RS=55​, and both triangles PQR and TRS are right triangles with a right angle at R, these triangles are similar by the AA Similarity criterion (both have a right angle and share the same acute angle measure).

Therefore, corresponding side lengths in these similar triangles will be proportional. In particular:

PQ/RS=RQ/PR (We can substitute PR with x since they are equal)

x/55​=8/x

Cross-multiplying gives:

x2=855​

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

x= 2*sqrt(11)

Therefore, the length of segment PQ is 2*sqrt(11) units.

AM =  sqrt [ 20^2 - 15^2]  =  sqrt [175 ]  =  5sqrt (7)

DM = AM / 2  =   2.5 sqrt (7)

Draw ED

Triangle AED ≈ Triangle ABM

AD / AM  = ED / BM

(1/2)AM / AM  = ED / BM

1/2  = ED / BM

ED = (1/2) BM

Triangle EDT ≈ BMT

Since ED = 1/2 BM

Then the height of triangle EDT  is 1/2 the height of triangle BMT

So the height of triangle BMT =  [ 2 /( 1 + 2)  ] * DM  = (2/3)DM

And the  height of triangle BME = DM

So  the area of triangle BME = (1/2)(BM)((DM)

And  the area  of triangle BMT = (1/2)(BM)(2/3)(DM)  = (1/3) (BM)(DM)

So  the area of triangle BET = [ BME ]  - [ BMT ]   = 

(1/2)(BM)(DM)   -  (1/3) (BM)(DM) =  (1/6) (BM) (DM)  =

(1/6) (15) ( 2.5 sqrt (7) )  = 

6.25 sqrt 7 =

25sqrt (7)  / 4

Ooops!!!....I found the area of CBD !!!

See my corrected answer....

What is UX?

\(\color{blue}S(0;0)\\\color{blue} T(13,0)\\ \color{blue}M(6.5, 0)\\f_S(x)=\pm \sqrt{14^2-x^2}\\ f_T(x)=\pm \sqrt{4^2-(x-13)^2}\\ x^2-14^2=(x-13)^2-4^2\\ x^2-14^2=x^2-26x+13^2-4^2\\ 26x=169-16+196\\ x_U=\dfrac{349}{26}\\ x_U=13.4\overline{230769}\\ y_U=3.9776\)

\(\color{blue}U(13.423,\ 3.978)\\ m_{TU}=\dfrac{y_U}{x_U-x_T}=\dfrac{3.9776}{13.42308-13}=9.4015\\ m_{NS}=-\dfrac{1}{m_{TU}}=-0.10637\)

\({\color{blue}f_{SN}(x)=}m_{NS}(x-x_S)+y_S=\color{blue}-0.10637x\)

\(m_{UM}=\dfrac{y_U}{x_U-x_M}=\dfrac{3.9776}{13.423-6.5}\\ m_{UM}=0.5387\)

\(f_{UM}(x)=m_{UM}(x-x_M)+y_M\\ f_{UM}(x)=0.5387x-0.5387\cdot 6.5\\ f_{UM}(x)=0.5387x-3.502\)

\(f_{UM}(x)=f_{SN}(x)\\ 0.5387x-3.5020=-0.10637x\\ x_X=\dfrac{3.5020}{0.5387+0.10637}\\ x_X=5.4285\\y_X=-0.57743\\ \color{blue}X(5.4285, -0.57743)\)

\(\overline{UX}=\sqrt{(y_U-y_X)^2+(x_U-x_X)^2}\\ \overline{UX}=\sqrt{(3.9776-(-0.57743))^2+(13.4231-5.4285)^2}\\ \color{blue}\overline{UX}=9.2012\)

  !

Using  partial fraction decomposition :

1/ [n * (n + 3)] = A / n  + B / (n +3)             multiply through by n *(n + 3)

1 = A (n + 3)  + Bn

1 = (A + B)n  + 3A

3A  = 1

A = 1/3

A+B  = 0

B = -1/3

So    

1 / [1x 4]  =   (1/3)/1 - (1/3)/4 =   1/3 - 1/12

1/ [ 4 x 7] =   (1/3)/4 - (1/3)/7  =  1/12 - 1/21

1/[7 * 10] =   (1/3)/7 - (1/3)/10 =  1/21 - 1/30

......

1/(34 *37)  = ((1/3)/34 - (1/3)/37  = 1/ 102 - 1/111

1 / (37 * 40) =  (1/3)/(37) - (1/3)/40  =  1/111  - 1/120

All the intermediate terms  "cancel"     and  we  are left with

1/3  -  1/120   =

[ 120 - 3 ] / 360   =

117 / 360    =

13 /  40

That is wrong...

Diagonal^2  =  20'2 + 20^2 -2 ( 20 *20) * cos (144°)

Diagonal^2  =  800 - 800 *(-1 - sqrt 5) / 4

Diagonal  =   sqrt [ 800 + 200 ( 1 + sqrt 5) ] =  10 sqrt [ 10 + 2sqrt 5 ] = 10 [ 2 (5 + sqrt 5)] ≈ 38.04

https://web2.0calc.com/questions/altitudes_39

From A,B  draw  AE  and BF perpendicular to BC

Let FC = x

By the PythagoreanTheorem

sqrt [24^2 - x^2 ]  =  height of  trapezoid

And DE = 104 - (78 + x)  =  26  - x

And, again,

sqrt [ 10^2  - ( 26 - x)^2 ] = height of trapezoid

So

sqrt [ 24^2 - x^2 ] = sqrt [ 10^2 - (26-x)^2 ]    square both sides

24^2 - x^2  = 10^2 - (26 - x)^2

576 - x^2  = 100  - x^2 + 52x - 676

52x  =  1152

x =  1152 / 52   = 288/13

height of trapezoid = sqrt [ 24^2  - (288/13)^2 ]    =  120 /  13

Area =  (1/2) (120 / 13) ( 104 + 78)  =  (1/2) (120/13) ( 182)   =  (91) (120 / 13)  =   840

Without  explanation.....the  side of the square = [ product of leg lengths ]  / [ sum of the leg lengths ]  =

 [6 * 6 ]  /  [ 6 + 6 ] = 36 / 12  =  3

I can show you how I got this  if you want me to.......

J / S =  3 / 2

J = (3/2)S

[J + 12 ] / [ S - 1 ]  =  2 / 1

[ (3/2)S + 12 ]  / [ S - 1 ] = 2 / 1

(3/2)S + 12  = 2  [S -1]

(3/2)S + 12 = 2S  - 2

2S - (3/2)S = 12 + 2

(1/2)S  = 14

S = 28

J = (3/2)S = (3/2) (28)  = 42

After the changes  there are  54 J   and  27 S  =    81  students total