bader

Since M is the midpoint of AM, we have that BM=AM=2AM​, so B is the midpoint of AM. Hence, median AM is also an altitude, so △ABC is right-angled at M. Since △ABC is isosceles, we have that ∠A=∠C, and ∠B=180∘−∠A−∠C=90∘.

Let BE=a and BM=b. Since D is the midpoint of AM, we have that MD=2AM​=2AB​=10. Since D is also the midpoint of BM, we have that BD=2BM​=b. Since ∠BME=∠BMD=45∘, triangles BME and BMD are congruent right triangles. Therefore, BE=BM=b, so E is also the midpoint of BM.

Since E is the midpoint of AB and M is the midpoint of AM, we have that AE=ME=2AB​=10. By the Pythagorean Theorem, we have that BE=AB2−AE2​=202−102​=103​. Since △BET is a 30-60-90 triangle with BE=103​, we have that ET=21​BE=53​. Therefore, the area of triangle BET is 21​⋅BE⋅ET=21​⋅103​⋅53​=75​ square units.

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)​​.

here are the steps to solve the problem:

1. Find the formula for the terms of the new sequence.

The terms of the new sequence are alternately equal to 9, 18, −27, 36, and so on. This can be written as an​=(−1)n(9+9(n−1)).

2. Find the sum of the first n terms of the new sequence.

The sum of the first n terms of the new sequence is given by 21​∑i=1n​ai​=21​∑i=1n​(−1)n(9+9(n−1)). This can be simplified to 21​∑i=1n​(−1)i(9+9(i−1)).

3. Set the sum of the first n terms equal to 108 and solve for n.

We have 21​∑i=1n​(−1)i(9+9(i−1))=108. Multiplying both sides by 2, we get ∑i=1n​(−1)i(9+9(i−1))=216.

We can evaluate the sum on the left-hand side by pairing up the terms:

\begin{align*} \sum_{i=1}^n (-1)^i (9 + 9(i-1)) &= (9 - 18) + (27 - 36) + (45 - 54) + \dots + (a_{n-1} - a_n) \ &= 9 + 9 + 9 + \dots + 9 \ &= 9n. \end{align*}

Therefore, 9n=216, so n=24.

Let's simplify the expression: 51​f(5x+5)−5

Steps to solve: 1. Combine multiplied terms into a single fraction: 51f(5x+5)​−5

2. Multiply by 1: 5f(5x+5)​−5

3. Distribute: 55fx+5f​−5

4. Simplify fraction: fx+f−5

Answer: fx+f−5

We can use the distance formula to find the distance between two points, and the equation of the line to find the points that lie on the line.

The distance formula tells us that the distance between two points (x1​,y1​) and (x2​,y2​) is:

sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

The equation of the line tells us that for any point (x,y) on the line, the following equation must hold:

y = 5x - 28

We can use these two equations to find all points (x,y) that are 5 units away from the point (2,7) and that lie on the line y=5x−28.

First, we can find the distance between the point (2,7) and any point (x,y) on the line. Using the distance formula, we get:

sqrt((x - 2)^2 + (y - 7)^2)

We can then substitute the equation of the line into this expression to get:

sqrt((x - 2)^2 + (5x - 35)^2)

We want this distance to be equal to 5, so we can set up the equation:

sqrt((x - 2)^2 + (5x - 35)^2) = 5

Squaring both sides of this equation, we get:

(x - 2)^2 + (5x - 35)^2 = 25

Expanding this equation, we get:

x^2 - 4x + 4 + 25x^2 - 70x + 1225 = 25

Combining like terms, we get:

26x^2 - 74x + 1249 = 0

Factoring this equation, we get:

(13x - 49)(2x - 25) = 0

Therefore, the possible values of x are 1349​ and 225​.

We can substitute these values of x back into the equation of the line to find the corresponding values of y. For x=1349​, we get:

y = 5 \cdot \frac{49}{13} - 28 = 7

For x=225​, we get:

y = 5 \cdot \frac{25}{2} - 28 = 7

Therefore, the two points that are 5 units away from the point (2,7) and that lie on the line y=5x−28 are (1349​,7) and (225​,7).

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Now let's substitute the values into the formula:

s = (-(-6000) ± √((-6000)^2 - 4*11*(-27500))) / (2*11)

Simplifying further:

s = (6000 ± √(36000000 + 1210000)) / 22
s = (6000 ± √(37210000)) / 22
s = (6000 ± 7200) / 22

We need s to be positive, so s = (6000 + 7200)/22 = 600.

The answer is 600.

By the Pythagorean Theorem, PQ2+PR2=QR2, so 52+82=112.

Thus △PQR is a right triangle where PQ is the hypotenuse, so by the Pythagorean Theorem, PM2=(2PR​)2=(28​)2=16.

Hence, PM=16​=4​.

By the Angle Bisector Theorem, [ \frac{AZ}{ZB} = \frac{AC}{BC} = \frac{CY}{BY} = \frac{8}{14-Y}. ] Since AY+Y=14, [\frac{AZ}{ZB} = \frac{8}{14-AY} = 8.] Let ZB=x. Then AZ=8x. Since AZ+ZB=14, x+8x=14, so x=2. Therefore, BZ=2.

Let CZ=y. Then CY=8y. Since CY+YZ=14, y+8y=214​=7. Therefore, [BC = CY + BZ = 8y + 2 = 8 \cdot 7 + 2 = \boxed{13.0909}.]

By the Angle Bisector Theorem, [ \frac{AZ}{ZB} = \frac{AC}{BC} = \frac{CY}{BY} = \frac{8}{14-Y}. ] Since AY+Y=14, [ \frac{AZ}{ZB} = \frac{8}{14-AY} = 8. ] Let ZB=x. Then AZ=8x. Since AZ+ZB=14, x+8x=14, so x=2.

Therefore, BZ=2​.

Let $x = AC$. Since $\overline{BC}\parallel\overline{DE}$, we have that $\angle BAD$ and $\angle BCD$ are corresponding angles, and thus congruent. Therefore, $\triangle ABD \sim \triangle DCB$ by AA similarity.

From the similarity, we have \[
\frac{AB}{BD}=\frac{BD}{CB}\quad\Rightarrow\quad \frac{x}{20}=\frac{20}{x+20}.
\] Cross-multiplying gives $x^2+20x = 20^2$ so $x^2+20x-400=0$. Factoring the quadratic gives $(x-20)(x+40) = 0$, so $x= \boxed{20}$.