Member: bingboy

To find a linear inequality with the given solution set, we need to consider the shaded region. Let's denote the coordinates of the shaded region's vertices.

Looking at the image, we can see that the shaded region is below and to the left of the line passing through the points (4, 1) and (5, 3). We can use these two points to find the equation of the line in slope-intercept form (y = mx + b):

First, find the slope (m): m=5−43−1=2

Now, we can use the slope and one of the points (let's use (4, 1)) to find the y-intercept (b): 2⋅47b=−7

So, the equation of the line is y=2x−7.

Now, since we want the shaded region to be below and to the left of this line, we want the region where y<2x−7.

We can rewrite this inequality in standard form (ax + by + c > 0) by moving all terms to one side

−2x+y+7<0

So, the linear inequality that represents the shaded region in standard form is −2x+y+7<0.

bingboyNov 29, 2023

+743

To find the coordinates of the center of the circle, we can use the midpoint formula. The midpoint formula states that the midpoint of a line segment with endpoints (a,b) and (c,d) is the point (2a+c,2b+d).

Using the midpoint formula, we can find the midpoints of the following line segments:

The midpoint of the line segment connecting (22,15) and (−25,0) is (222+(−25),215+0)=(−1.5,7.5).

The midpoint of the line segment connecting$(-25,0)$ and (22,−29) is (2−25+22,20+(−29))=(−1.5,−14.5).

The midpoint of the line segment connecting (22,−29) and (22,15) is (222+22,2−29+15)=(22,−7).

Since the three midpoints coincide, we can conclude that the center of the circle is the point (−1.5,−7).

bingboyNov 26, 2023

+743

To solve this, let's first consider the case where there are no restrictions on the seating arrangements. In this case, there are 6! ways to arrange the 6 siblings in the 6 chairs.

Next, let's consider the number of arrangements where siblings are seated next to each other. For each pair of siblings, there are 2 ways to arrange them (either sibling can sit in the front or back chair). Since there are 3 pairs of siblings, there are 23=8 ways to arrange the siblings in pairs.

However, we have double-counted the arrangements where two pairs of siblings are seated next to each other. For each arrangement of two pairs of siblings, there are 2 ways to arrange the third pair. Therefore, we have overcounted the number of arrangements with two pairs of siblings by 8⋅2=16.

Finally, let's consider the number of arrangements where all three pairs of siblings are seated next to each other. There are 2 ways to arrange the three pairs of siblings (either all three pairs can be in the front row or all three pairs can be in the back row). Once the pairs are arranged, there are 2 ways to arrange each pair within their respective row. Therefore, there are 2⋅23=16 arrangements where all three pairs of siblings are seated next to each other.

Therefore, the total number of arrangements where siblings cannot be seated next to each other is 6!−8⋅2−16=552.

bingboyNov 26, 2023

+743

Since AB=AC, △ABC is isosceles, so ∠CAB=∠CBA, and the two medians CD and BE intersect at the incenter I. Also, since CD is a median, D is the midpoint of BC, so BD=14/2=7.

Since △ABC is isosceles, CD is also an altitude of △ABC. Since CD is an altitude, ∠ICD=90∘. Since ∠ICD=90∘, rectangle ICBD is formed, so CI=BD=7. Since IC=7, CDBD=21.

Now, since CDBD=21, △ICD is similar to △ABC. Since △ICD is similar to △ABC,

ABCI=BCCD.

Substituting the given values, we have

257=14CD.

Solving for CD, we get CD=2514⋅7=2598.

Since CD=2AC=2598, AC=2⋅CD=2598⋅2=25196.

Since ∠CAB=∠CBA, △ABC is 45-45-90 triangle. Since △ABC is a 45-45-90 triangle, BC=2⋅AC=2⋅25196=5249=5142.

Since BCCD=21,

51422598=21.

Simplifying both sides, we have

272=1425.

Multiplying both sides by 2142, we get

72=7175.

Squaring both sides, we have

49⋅2=4930625.

Solving for 2, we get 2=49625.

Then $BI = 7*sqrt(2)/5$.

bingboyNov 22, 2023

+743

Because the scale factor is negative, triangle A'B'C' is obtained by rotating 180 degrees triangle ABC about point G and then shrinking it. Hence, the overlap between triangles ABC and A'B'C' is the smaller triangle A'BC'. (The region consisting of the intersection of triangle ABC with dilated triangle A'B'C' is actually a kite.) We know that the side lengths of triangle A'B'C' are 3/5 the side lengths of triangle ABC. Since the area of a triangle is proportional to the square of its side lengths, the area of triangle A'B'C' is (3/5)2 times the area of triangle ABC. Therefore, the area of the overlap K is (1−(3/5)2) times the area of triangle ABC. Hence, \begin{align*} \frac{K}{ABC} &= 1-(3/5)^2 \ &= 1 - 9/25 \ &= \boxed{\frac{16}{25}}. \end{align*}

The answer is 16/25.

bingboyNov 13, 2023

+743

-1

Since Angle B = 2Angle D and they are both external angles of triangle ADC then Triangle BDC is isosceles. Let O be the circumcenter of triangle BDC. Since Triangle BDC is isosceles, BO = CO = r. Drop a perpendicular from B to AC at point E. Since BE = r, then BD = rsqrt(3). Also since Angle B = 60 then Triangle BDC is equilateral and therefore, Angle CBD = 120 degrees. Since Angle C = 90 - Angle B = 90 - 60 = 30 degrees then Angle ADC = 90 - Angle C = 60 degrees. Since Angle ADC = 60 degrees and BD = rsqrt(3), then AD = 2rsqrt(3). Now consider Triangle ADE. Since Angle ADE = 90 degrees and AE = 2r and AD = 2rsqrt(3), then DE = 2rsqrt(3). Also since DE = 2rsqrt(3), then CD = 2rsqrt(3). Since Angle C = 30 degrees and CD = 2rsqrt(3) and BD = rsqrt(3) then Angle BDC = 120 degrees. Therefore, Triangle BDC is equilateral and therefore, BD = r. Since BD = r and CD = 2r then by the Pythagorean Theorem, BC = rsqrt(3). Now consider Triangle ABC. Since Angle B = 60 degrees and BC = rsqrt(3) and AC = 4sqrt(3), then Angle A = 90 degrees. Therefore, Triangle ABC is a right triangle with legs AC = 4sqrt(3) and BC = rsqrt(3). Therefore, the circumradius of Triangle ABC is AC/2 = 2sqrt(3). Now consider Triangle ADC. Since Angle D = 30 degrees and CD = 2r and AC = 4sqrt(3), then Angle A = 60 degrees. Therefore, Triangle ADC is a right triangle with legs AC = 4sqrt(3) and CD = 2r. Therefore, the circumradius of Triangle ADC is AC/2 = 2sqrt(3). Therefore, the distance between the circumcenters of Triangle ABC and Triangle ADC are both 2sqrt(3) and therefore, the total distance between them is 22sqrt(3) = 4*sqrt(3). So the answer is 4*sqrt(3).

bingboyNov 8, 2023

+743

-1

There are (410)=210 ways to choose which 4 letters to use out of the 10 letters in MATHCOUNTS.

Once we have chosen our 4 letters, we can permute them in 4! ways.

Therefore, there are 210×4!=5040 distinct four-letter permutations that can be formed using four of the ten letters in MATHCOUNTS.

bingboyNov 7, 2023

+743

-1

The number 2^29 contains exactly 9 distinct digits, but it is missing the digit 1.

We can see this by considering the following:

The units digit of 2^29 must be 2, because 2^1 = 2 and 2^4 = 16.

The tens digit of 2^29 must be 4, because 2^2 = 4 and 2^8 = 256.

The hundreds digit of 2^29 must be 8, because 2^3 = 8 and 2^12 = 4096.

The thousands digit of 2^29 must be 6, because 2^4 = 16 and 2^16 = 65536.

The ten thousands digit of 2^29 must be 4, because 2^5 = 32 and 2^20 = 1048576.

The hundred thousands digit of 2^29 must be 2, because 2^6 = 64 and 2^24 = 16777216.

The millions digit of 2^29 must be 0, because 2^7 = 128 and 2^28 = 268435456.

This leaves us with only two digits left to fill in: the billions digit and the tens of billions digit. Since the number 2^29 must have 9 distinct digits, the missing digit must be 1.

Therefore, the number 2^29 contains all of the digits 0, 2, 3, 4, 5, 6, 7, 8, and 9, except for the digit 1.

bingboyOct 28, 2023

+743

-1

Let's start by plugging the coordinates ofAandBinto the distance formula:

d(A, B) = \sqrt{(2-a)^2 + (-1-2a-7)^2}

We know thatd(A,B)=5, so we can substitute this into the equation to get:

\sqrt{(2-a)^2 + (-1-2a-7)^2} = \sqrt{5}

Squaring both sides of the equation, we get:

(2-a)^2 + (-1-2a-7)^2 = 5

Expanding the parentheses, we get:

4-4a+a^2 + 1+4a+49 = 5

Simplifying, we get:

a^2 - 4a + 50 = 0

Factoring the quadratic, we get:

(a-10)(a-5) = 0

Therefore,a=10ora=5.

Since we are asked for the sum of all possible values ofa, the sum is 10+5=15.

bingboyOct 28, 2023

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