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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

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

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

I think 4! x 3! x 2! x 3!/9 which is 24x2x2x2 which is 24x8 do the calculation yourself

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:

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

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

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

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

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

Yunseol can draw three cards with exactly two of the same color by following two scenarios:

Two cards from one color and one card from another color:

There are 4 choices of color for the two same-colored cards.

For each color choice, there are 7 options for the first card and 6 options for the second card (since two cards have the same color).

There are 4 choices of color for the different colored card.

Total ways for scenario 1 = 4 * (7 * 6) * 4 = 672

Three cards with one card from each of three different colors:

Yunseol can choose 3 colors out of 4 in 4C3 ways = 4.

For each color choice, there are 7 options for each card (since all three colors are different).

Total ways for scenario 2 = 4 * (7 * 7 * 7) = 1176

Adding the possibilities from both scenarios: 672 (scenario 1) + 1176 (scenario 2) = 1848

Therefore, Yunseol can draw 1848 three-card hands from Ms. Q's deck so that exactly two of the cards have the same color.

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