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Let O be the center of the inscribed circle. Let E be the midpoint of AB, and let F be the midpoint of CD. Draw segments AO, BO, CO, and DO. Since ABCD is an isosceles trapezoid, ∠ABE=∠CBE and ∠DCE=∠EDF, so triangles ABE, CBE, DCE, and EDF are all isosceles.

Since AE=BE, CE=DE, and AO=BO, triangles AEO, BEO, CEO, and DEO are all congruent. Therefore, OA=OB=OC=OD=r, where r is the radius of the inscribed circle.

Let E′ be the foot of the altitude from E to AB, and let F′ be the foot of the altitude from F to CD. Since triangles AEE′ and BEE′ are both right isosceles triangles, AE′=BE′=r, and similarly, CF′=DF′=r. Therefore, the sum of the lengths of the segments from E to D is: [ EE' + ED + DF' = 2r + x + 2r = x + 4r. ]

Similarly, the sum of the lengths of the segments from F to B is y+4r. Since ABCD is an isosceles trapezoid, the sum of the lengths of the segments from E to D is equal to the sum of the lengths of the segments from F to B. Therefore, x+4r=y+4r, so x=y. Hence, the radius of the inscribed circle is sqrt(xy)​​.

3)

Draw a tangent to the chord of P, forming a rectangle, labeling that point D.

QD = the length of the external tangent.

PD = 4

PQ = 20

\(QD^2=20^2-4^2\)

\(QD=8\sqrt{6}\).

1)

It's all ratios, so you can set BC = 3.

Then the radius of the semicircle is 1.

Label the point where the Semicircle is tangent to AB as D.

Label the center of the semicircle as O

Triangle ADO and triangle ACB are similar, and sset DA as x.

Using the properties of similar triangles:

\(\frac{x}{1}=\frac{\sqrt{x^2+1}+1}{3}\).

\(3x=\sqrt{x^2+1}+1 \)

\(x=\frac{3}{4}\).

\(AC = \frac{5}{4}+1=\frac{9}{4}\).

\(\frac{AC}{BC}=\frac{\frac{9}{4}}{3}=\frac{3}{4}\).

Since the arrangement is circular, any specific order will only be unique up to a rotation. So, choosing any person as the "starting point" and fixing them in a seat won't actually affect the number of possible arrangements.

Let's consider one adult, A. For A to be next to two students, there must be two students flanking A. There are 4 choices we can make for the first student to sit next to A.

Once we've chosen that student, there are 3 students left, and only one of them can sit on the other side of A. So, there are 4⋅1=4 ways to arrange the students such that A is flanked by two students.

Now, we have 3 adults left to seat. Since they each must be flanked by two students as well, we can repeat the same process for each adult, giving us 4⋅4⋅4=64​ total arrangements.

According to the Triangle Inequality, ABC is indeed a triangle, since 8 + 12 > 15.

Thus, the radius of the circle is the same as the circumradius of ABC, which can be found by:
\(R = \frac {abc} {4S}\)

Where S is the area of the triangle.

Using Heron's formula, we get:
\(S = \sqrt { \frac {35} {2} \left( \frac {35} {2} - 8 \right) \left( \frac {35} {2} - 15 \right) \left( \frac {35} {2} - 12 \right)}\)

\(S = \sqrt {\frac {35} {2} \left( \frac {19} {2} \right)\left( \frac {5} {2} \right)\left( \frac {11} {2} \right)}\)

\(S = \frac {5} {4} \sqrt {1463}\)

Therefore,

\(R = \frac {2520} {5\sqrt{1463}}\)

\(R = \frac {504\sqrt{1463}} {1463}\)

ratio of people: 5/6

ratio of work time: 6/5 = x/x-4

x = 24so 4 days - 25 more people

Both of these are inccorect, 

4t^2  + 4t + 14

This parabola opens  upward.....no max possible

Note  

3 + 4 + 5 =  3 *4

13 + 14 + 15 = 3 * 14......  etc.

3 [ 4  +  14 + 24  +  ....... + 994 ]  =

No. of terms =  100

Sum =  3 [  994 + 4 ] [ 100 / 2 ]   =  149,700

t^2  + 4t - 96

t that produces the smallest value  =  -4 / (2 * 1)  =   -2

-5s^2 - 16s +  39

This  is a parabola that opens  downward.....it has no minimum  value

Check alan's answer here

https://web2.0calc.com/questions/please-help_63707

In a rhombus, all the side lengths are $12,$ and one of the angles is equal to $20^\circ.$ Find the area of the rhombus.   

Visualize, or preferably draw, the rhombus, leaning sharply to the right.  

Drop a perpendicular from the upper left vertex to the bottom side. 

This has made a right triangle with a 20^o angle and the hypotenuse 12.  

Use trig to find that height..       sin(20^o)  =  height / 12  

                                                    0.3420  =  height / 12 

                                                      height  =  0.3420 • 12  =  4.104

                                            Area_rhombus  =  base • height 

                                                        Area  =  12 • 4.104 

                                                        Area  =  49.25  

Note that I rounded the decimal values that I wrote down.    

If you calculate it using those values, you'll get 49.248.   

I calculated it without rounding, that's where I got 49.25.  

_.

9 points

Laverne says  2  7   12   17

Sum

2  9  21

12 is the number

0   +  55d  =   1/5

55d = 1/5

d  =   1 / [ 5 * 55]  =  1 / 275

70th term is  0  + 60d  =   60 / 275   =12 / 55

S₇₀ =   [  12/55 + 0 ] [70 -10 + 1 ]  / 2   =

[12/55] * 61  / 2  =   

366 / 55

\($10 \sqrt{2 }\)

Let the sides be   a , b

2a + 2b =40

a + b =20      square both sides

a^2 + 2ab + b^2  = 400

a^2 + b^2 + 2ab  = 400

a^2 + b^2 = (10sqrt 2)^2

a^2 + b^2  = 200

200 + 2ab = 400

100 +ab  = 200

ab = 200 - 100

ab = 100  =  the area

https://web2.0calc.com/questions/coordinates_60311

300 =   10^2 * 3  =  2 * 2 * 5 * 5 * 3 * 1

No. of  different sequences  =  6!  / [ (no. of 2s )! * (no. of 5s )! ] =  6! / [ 2! * 2! ] =  720 / 4  =  180

y = log ₂ (2x)

y= log ₄ (16 + x)

Change of base theorem

log ( 2x) / log 2  =  log (16 + x) / log 4

log (2x) / log 2  = log (16 + x) / log 2^2

log (2x) / log 2  =  log (16 + x) / [ 2 log 2]            {multiply both sides by log 2 }

log (2x)  =  log (16 + x) / 2

2 log (2x)  = log (16 + x)

log (2x)^2  =  log (16 + x)     implies that

2x^2  = 16 + x

4x^2  - x  - 16  =  0        { x must be positive }

x = [ 1 + sqrt [ 1 + 256 ] ]  / 8  =   [ 1 + sqrt (257) ] /  8