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Evaluate the infinite geometric series ​\(0.4 + 0.12 + 0.036 + 0.0108 + \dotsb\)

\(a_1=0.4\\ q=\dfrac{0.12}{0.40}=0.3\\ \dfrac{lim\ s_n}{n\to \infty}= \dfrac{a_1}{1-q}=\dfrac{0.4}{1-0.3}\\ \color{blue}\dfrac{lim\ s_n}{n\to \infty}=s= \dfrac{4}{7}=0.\overline{571428}\)

 !

The first term is 9 and the common ratio is 1/3, so the tenth term is 9*(1/3)^10 = 1/6561.

thats pretty challenging, but I got $12\sqrt2$

We can find the range of the function f(x)=2x2−8x+1 by completing the square and analyzing its extreme points.

Completing the Square:

We can rewrite the function as:

f(x)=2(x2−4x)+1

To complete the square, we take half of the coefficient of our x term, square it, and add it to both sides of the equation:

Half of the coefficient of our x term is -4 / 2 = -2.

Squaring -2 gives us 4.

Therefore:

f(x)=2(x2−4x+4)+1−8

f(x)=2(x−2)2−7

Finding the Vertex and Analyzing Extremes:

The expression inside the parentheses is squared, so it's always non-negative (0 or greater). Multiplying by 2 doesn't change this property. Therefore, the minimum value of f(x) is achieved when the squared term is 0, which happens when x=2. In this case, f(2)=2(2−2)2−7=−7.

Since the coefficient of the squared term (2) is positive, the parabola opens upwards. This means the function has a minimum point at x=2 and increases as we move away from it (either to the left or right on the interval).

Range of the Function:

The function is defined only on the interval [−1,2]. We saw that the minimum value within this interval is f(2)=−7. As x approaches either end of the interval, the function increases without bound because the squared term keeps getting larger. Therefore, the range of f(x) is the single interval:

(−7,∞)​

For f(x) to be defined, the inner term 1+x3​2​ must be defined. This inner term is undefined when its denominator, 1+x3​=0, so when x=−31​. However, even at x=−31​, the outer term 1+21​ is still defined, so f(x) is actually defined at x=−31​.

Therefore, the only restriction on the domain of f(x) comes from the denominator of the entire expression, 1+1+x3​2​. This denominator is undefined when its denominator, 1+x3​, is 0. So the function is undefined when x=−31​, as discussed above, and also when x=−1.

Finally, note that as x approaches positive or negative infinity, the entire expression approaches 11​, which is a defined value. Therefore, the only restrictions on the domain are x=−31​ and x=−1. The sum of these two numbers is −31​−1=−34​​.

We need to analyze when the expression inside the function f in g(x) becomes undefined. This happens when the denominator, (x+1), is zero. So, x=−1.

However, this restriction only applies if the numerator, (x2−2), is not also zero at x=−1. Since (−1)2−2=1=0, the function f((x2−2)/(x+1)) is defined even when x=−1.

Therefore, the only restriction on the domain of g(x) is that x=−1. This means g(x) is defined for all real numbers in the interval except for this single point.

The width of an interval that excludes a single point is 0​.

Analyze Triangle Properties:

We are given that triangle ABC is a right triangle with angle A = 30 degrees and angle C = 90 degrees. This makes angle B = 60 degrees (since the angles in a triangle sum to 180 degrees).

Since BD bisects angle ABC, we know that angles ABD and CBD are both equal to half of angle ABC, which is 30 degrees.

Utilize Isosceles Triangle Property:

Because BD bisects angle B and is perpendicular to side AC (since AC is the hypotenuse of right triangle ABC), segment BD also bisects side AC. This makes triangles ABD and BDC congruent.

Find Lengths in Triangle ABD:

Knowing AC = 12 and that BD bisects AC, we can determine that AD = DC = 12 / 2 = 6.

Apply Area Formula for Triangle ABD:

The area of a triangle can be calculated using the formula: Area = 1/2 * base * height.

In triangle ABD, we know the base (AB) is part of the hypotenuse of right triangle ABC. Since angle A = 30 degrees, we can use the 30-60-90 right triangle properties to find that AB = 2 * AC = 2 * 12 = 24.

The height of triangle ABD is the perpendicular distance from point D to side AB. Since BD is perpendicular to AC, the height is the same length as segment BD. However, we haven't yet determined the length of BD.

Find BD using Pythagorean Theorem:

Since triangle ABD is a right triangle with AB = 24 and AD = 6, we can use the Pythagorean Theorem to find BD (the hypotenuse):

BD^2 = AB^2 - AD^2 BD^2 = 24^2 - 6^2 BD^2 = 576 - 36 BD^2 = 540 BD = √540 (Since we want the positive side length)

However, working with the square root of 540 might be cumbersome. We can simplify this by noticing that 540 can be factored as 2^2 * 3^3 * 5. Taking the square root of both sides:

BD = √(2^2 * 3^3 * 5) = 2 * 3 * √5

Calculate Area of Triangle ABD:

Now that we know all side lengths, we can find the area of triangle ABD:

Area = 1/2 * base * height

Area = 1/2 * 24 * (2 * 3 * √5)

Area = 72 * √5

To solve this congruence, we need to find the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \).

First, let's observe that \( 5 \equiv 2 \pmod{3} \). Therefore, \( 5^n \equiv 2^n \pmod{3} \).

Now, let's calculate the values of \( 2^n \) modulo 3 for small values of \( n \):

- \( 2^1 \equiv 2 \pmod{3} \)

- \( 2^2 \equiv 1 \pmod{3} \)

- \( 2^3 \equiv 2 \pmod{3} \)

- \( 2^4 \equiv 1 \pmod{3} \)

From this, we can see a pattern emerge: the value of \( 2^n \) alternates between 1 and 2 modulo 3, with period 2.

Now, let's consider \( n^5 \). Since we're taking \( n^5 \) modulo 3, we can reduce \( n^5 \) to its residue modulo 3:

- \( 1^5 \equiv 1 \pmod{3} \)

- \( 2^5 \equiv 32 \equiv 2 \pmod{3} \)

- \( 3^5 \equiv 243 \equiv 0 \pmod{3} \)

The pattern for \( n^5 \) modulo 3 is not as obvious as \( 2^n \), but we can see that for \( n = 2 \), \( n^5 \) matches \( 2^n \) modulo 3.

So, the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \) is \( n = 2 \).

75%x35%=3/4x7/20=21/80 of it's total matches for the season won. you need 19/80, or 19/20 of 1/4, so the answer is 19/20

Sure, let's break it down:

We're given triangle \(STU\) with midpoints \(M\), \(N\), and \(P\) on sides \(TU\), \(US\), and \(ST\) respectively. \(UZ\) is the altitude of the triangle.

Given:

- \(\angle TSU = 62^\circ\)

- \(\angle STU = 29^\circ\)

1. We draw the altitude \(UZ\) from vertex \(U\) to side \(TS\), creating right triangles \(TUZ\) and \(SUZ\).

2. We find:

   - \(\angle SUT = 180^\circ - 29^\circ - 62^\circ = 89^\circ\)

   - \(NM\) is parallel to \(ST\), so \(\angle NZM = \angle TSU = 62^\circ\)

   - \(NP\) is parallel to \(UT\), so \(\angle NPM = \angle STU = 29^\circ\)

3. Then, we calculate the sum:

   \[ \angle NZM + \angle NPM = 62^\circ + 29^\circ = 91^\circ \]

So, the sum of angles \( \angle NZM + \angle NPM \) is \( 91^\circ \).

We want to find the probability that the coin will be tossed more than $10$ times. We can achieve this in two scenarios:

After $10$ tosses, we have not yet seen three consecutive heads or three consecutive tails.

We have seen either three consecutive heads or three consecutive tails, but not until the $10$th toss.

Let's focus on the first scenario:

In this scenario, we need to find the number of valid sequences of $10$ tosses where neither three consecutive heads nor three consecutive tails occur.

Valid sequences include any combination of heads and tails as long as there are no three consecutive heads or three consecutive tails. For example: $HHTHHTHHTT$ or $THTHTHTHTH$.

Invalid sequences include: $HHHHHHHHHH$ (three consecutive heads), $TTTTTTTTTT$ (three consecutive tails), $HHTHHHHHHH$ (three consecutive heads).

There are $2^{10}$ total possible sequences of $10$ tosses.

Out of these, we need to subtract the number of sequences that have three consecutive heads or three consecutive tails.

The number of sequences containing three consecutive heads or three consecutive tails is $2^8$.

So, the number of valid sequences is $2^{10} - 2^8$.

The probability of each valid sequence is $\left(\frac{1}{2}\right)^{10}$.

Thus, the total probability of getting more than $10$ tosses is:

Total probability = (2^10 - 2^8)/2^10 = 3/4.

So, the probability that the coin will be tossed more than $10$ times is $\frac{3}{4}$.

Since a rotation about point O maps A to B, B to C, and C to D, we know the following:

Angle AOD: This is given as 24 degrees.

Full Rotation: A full rotation around a point is 360 degrees.

Possible Rotations:

There are three possibilities for the rotation that maps A to B, B to C, and C to D, resulting in three possible measures for angle AOB:

Case 1: Single Rotation of 24 Degrees

In this case, the rotation about O maps A to B by 24 degrees clockwise, B to C by 24 degrees clockwise, and C to D by 24 degrees

clockwise, resulting in a total rotation of A to D of 24 + 24 + 24 = 72 degrees.

Since the full rotation is 360 degrees, the remaining angle for AOB must be 360 - 72 = 288 degrees.

However, angle measures are typically represented between 0 and 180 degrees. We can achieve this by subtracting a multiple of 180 (full rotations) from 288:

AOB = 288° - 180 = 108.

Case 2: Double Rotation (360 + 24 Degrees)

Here, the rotation about O maps A to B by a full rotation (360 degrees) followed by a 24-degree clockwise rotation. This effectively brings B back to its original position and then continues the rotation to C and D.

The total rotation for AOD in this case becomes 360 + 24 + 24 = 408 degrees.

Similar to case 1, we can adjust this to fit the 0-180 degree range:

AOB = 408° - 360° = 48°

Case 3: Triple Rotation (2 * 360 + 24 Degrees)

This case involves two full rotations followed by a 24-degree clockwise rotation from A to B. Again, the first two rotations effectively bring B back to its original position.

The total rotation for AOD becomes 2 * 360 + 24 = 744 degrees.

Adjusting for the 0-180 degree range:

AOB = 744° - 2 * 360° = 124°

Summary:

Therefore, the three possible degree measures for angle AOB, considering rotations between 0 and 180 degrees, are:

108° (Case 1)

48° (Case 2)

124° (Case 3)

Understanding the Transformations:

Tasha's Transformation: Moves the point sqrt(2) units to the right.

Richard's Transformation: Rotates the point clockwise about point O by 90 degrees.

Round 1:

A' is sqrt(2) units to the right of A (Tasha's translation).

A" is obtained by rotating A' by 90 degrees clockwise around O (Richard's rotation).

Round 2:

A1 is obtained by rotating A by 90 degrees clockwise around O (Richard's rotation).

A2 is sqrt(2) units to the right of A1 (Tasha's translation).

Finding A"A2:

To find the distance between A" and A2, we can consider the following:

A" and A1 are the same distance away from O (since both are obtained by a 90-degree rotation from A and they share the same original distance from O).

A'A2 forms a right triangle with A"A1 as the hypotenuse (due to the 90-degree rotations).

Pythagorean Theorem:

We can use the Pythagorean theorem to find A"A2:

a^2 + b^2 = c^2

where:

a = A'A2 (we know this is sqrt(2) from Tasha's translation)

b = A"A1 (distance between A" and A1, which is the same as the distance between A and O, which is sqrt(3))

c = A"A2 (what we want to find)

Substituting Values:

(sqrt(2))^2 + (sqrt(3))^2 = (A"A2)^2

2 + 3 = (A"A2)^2

5 = (A"A2)^2

Taking the Square Root (consider both positive and negative):

A"A2 = ±√5

Since distance cannot be negative, we take the positive square root:

A"A2 = √5

Therefore, the distance between A" and A2 is √5 units.

This is a series where the numerator follows the Fibonacci sequence and the denominator is a geometric sequence with common ratio 1/10. To find the sum of such a series, we can use a technique involving manipulation and summation of geometric series.

Here's how to solve it:

Step 1: Splitting the Series

We can represent the series as the sum of two series:

S = (1/10^2 + 1/10^3 + 1/10^4 + ...) + (2/10^4 + 3/10^5 + 5/10^6 + ...)

Step 2: Recognizing Geometric Series

The first series is a geometric series with first term (a = 1/10^2) and common ratio (r = 1/10). The second series is another geometric series with first term (a = 2/10^4) and common ratio (r = 1/10).

Step 3: Find the Sum of Each Series

The formula for the sum (Sn) of a finite geometric series is:

Sn = a(1 - r^n) / (1 - r)

where:

a is the first term

r is the common ratio

n is the number of terms

Step 4: Apply the Formula to Each Series

First Series:

S1 = (1/10^2) * (1 - (1/10)^n) / (1 - 1/10)

Second Series:

S2 = (2/10^4) * (1 - (1/10)^n) / (1 - 1/10)

Step 5: Note on Infinite Series

Since we're dealing with an infinite series (n tends to infinity), both (1/10)^n terms in the numerators approach zero. Therefore, we can simplify the expressions:

S1 ≈ (1/10^2) * (1 - 0) / (1 - 1/10) = 1/100

S2 ≈ (2/10^4) * (1 - 0) / (1 - 1/10) = 1/50

Step 6: Sum the Simplified Series

The total sum (S) of the original series is the sum of S1 and S2:

S = S1 + S2 ≈ 1/100 + 1/50 = 3/100

Answer:

Therefore, the sum of the series is 3/100.

The strategy here is to calculate the probability of the event's complement (i.e., the probability that the coin is tossed ten times or fewer) and subtract this from 1.

Favorable Outcomes:

There are two favorable outcomes:

HHH appears in exactly 10 flips: There are 3 choices for the position of the first heads (leaving 9 flips remaining), then 2 choices for the second heads (leaving 8 flips remaining), and 1 choice for the third heads (leaving 7 flips remaining).

So, there are 3⋅2⋅1=6 successful outcomes where HHH appears in exactly 10 flips.

TTT appears in exactly 10 flips: This follows the same logic as scenario 1, so there are also 6 successful outcomes.

Total Favorable Outcomes:

There are a total of 6 + 6 = 12 successful outcomes where the coin is tossed exactly 10 times.

Total Outcomes:

Since the coin is fair, there are 210=1024 total possible outcomes for 10 flips (heads or tails for each flip).

Probability of Favorable Outcomes:

The probability of needing exactly 10 flips is the number of successful outcomes divided by the total number of outcomes:

P(exactly 10 flips) = 12 / 1024

Complement's Probability:

We want the probability of needing more than 10 flips. This is the complement of the event where the coin is tossed ten times or fewer.

P(more than 10 flips) = 1 - P(exactly 10 flips)

Final Answer:

Substitute the probability of needing exactly 10 flips:

P(more than 10 flips) = 1 - (12 / 1024)

Simplify:

P(more than 10 flips) = (1024 - 12) / 1024 = 1012 / 1024

Both the numerator and denominator have a common divisor of 4, so we can simplify:

P(more than 10 flips) = 253 / 256

Therefore, the probability of needing more than 10 flips is 253/256.

For question 2:

Absolutely, we can find the positive values of n for which both n and 3n are four-digit base-9 integers.

Understanding the Problem:

Base-9 Integers: Numbers are represented in base-9 using digits 0 to 8, just like base-10 uses 0 to 9.

Four-digit Integers: We're interested in n such that both n and 3n have four digits in base-9.

Conditions for Four-digit Numbers:

For a base-b number to be four-digit, it must satisfy:

b^3 ≤ n < b^4

In this case, the base is b = 9. Therefore:

9^3 ≤ n < 9^4

729 ≤ n < 6561

Constraint for 3n:

We also need 3n to be a four-digit base-9 integer. This translates to:

9^3 ≤ 3n < 9^4

243 ≤ 3n < 6561

Dividing the second inequality by 3, we get:

81 ≤ n < 2187

Combining Conditions:

To satisfy both conditions, n must lie between 729 and 2187 (inclusive):

729 ≤ n ≤ 2187

Counting Possible Values of n:

Within this range, n must be a multiple of 3 (since 3n is a multiple of 3 and needs to be a four-digit integer).

The first multiple of 3 within the range is 732, and the last multiple of 3 is 2187.

To count the number of multiples of 3 in this range, we can divide the last multiple by the first multiple and subtract 1 (since we only consider integers from 732 to 2187, inclusive):

(2187 / 732) - 1 = 3

Answer:

Therefore, there are 3 positive values of n (732, 735, and 738) for which both n and 3n are four-digit base-9 integers.

For question 1:

Here's how to find the number of positive five-digit integers containing the digit grouping "24" at least once:

Total Possible Numbers:

There are 9 choices (excluding 0) for each digit in a five-digit number (1, 2, 3, 4, 5, 6, 7, 8, 9).

So, the total number of possible five-digit positive integers is 9 * 9 * 9 * 9 * 9 = 59,049.

Unwanted Scenarios (Numbers without "24"):

No "24" group: There are 8 choices for the digit in the place where "24" could be (all digits except 2 and 4). For the remaining 4 digits, there are 9 choices each. So, there are 8 * 9 * 9 * 9 = 5832 cases where "24" doesn't appear in any of the five digits.

"24" only appears once, but not next to each other:

Choose the digit that will be between "2" and "4": There are 7 choices (all digits except 0, 2, and 4).

Choose the positions for "2" and "4" (not next to each other): There are 4 choices for the first position (remaining slots after fixing one digit). There are 3 choices for the second position (remaining slots after fixing the first and the middle digit). So, there are 4 * 3 = 12 ways to arrange "2" and "4" with another digit in between.

Choose the remaining 2 digits: There are 9 choices each.

Total unwanted cases with "24" not next to each other: 7 * 12 * 9 * 9 = 5832.

Total Numbers with "24" at least once:

We want to find the number with "24" at least once. So, subtract the unwanted cases from the total:

Total numbers - Unwanted cases = Numbers with "24"

59,049 - (5832 + 5832) = 47,385

Therefore, there are 47,385 positive five-digit integers containing the digit grouping "24" at least once.

We can solve this problem using the properties of isosceles triangles and the Pythagorean theorem.

Isosceles Triangle Properties:

Since AB = AC in an isosceles triangle ABC, angles B and C are congruent.

Dividing the Base:

The height from A to BC is the perpendicular bisector of the base (divides it into two segments of equal length) in an isosceles triangle.

Segment Lengths:

Given that BC = 8 cm and BD = 3 cm, we know DC = BC - BD = 8 cm - 3 cm = 5 cm.

Finding the Height:

We can use the Pythagorean theorem on right triangle ABD (since the height is perpendicular to the base).

Pythagorean Theorem:

a^2 + b^2 = c^2 (where a and b are the lengths of the legs and c is the length of the hypotenuse)

In this case:

a (length of leg BD) = 3 cm

c (length of hypotenuse AB) = unknown (represents the height of the triangle)

b (length of leg AD) = unknown

Solving for Height (c):

b^2 = c^2 - a^2 (rearranging the equation)

b^2 = c^2 - 3^2 (substitute known values)

Since AD is half of the base (property of isosceles triangle):

AD = 1/2 * BC = 1/2 * 8 cm = 4 cm

Substitute the value of AD (b) in the equation:

4^2 = c^2 - 3^2

Simplify:

16 = c^2 - 9

Add 9 to both sides:

25 = c^2

Take the square root of both sides (remember to consider both positive and negative since squaring either results in 25):

c = ± 5

Positive Height:

Since the height represents a distance, we take the positive value:

c (height of the triangle) = 5 cm

Therefore:

Length of segment DC = 5 cm

Height of the triangle = 5 cm