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Let's analyze each statement based on the given conditions:

a) a + c > b + d: Since a > b and c > d, adding them together will preserve the inequality. So, a + c > b + d is MUST be true.

b) 2a + 3c > 2b + 3d: Multiplying both sides of a > b and c > d by positive constants 2 and 3, respectively, will still hold true. So, 2a + 3c > 2b + 3d MUST be true.

c) a - c > b - d: Subtracting c from both sides of a > b doesn't necessarily guarantee a - c > b - d. It depends on the relative values of a, b, and c. Not necessarily true.

d) ac > bd: Since a > b and c > d, multiplying them together will maintain the inequality as long as both a and c have the same sign (both positive or both negative). This is MUST be true.

e) a^2 + c^2 > b^2 + d^2: Squaring an inequality doesn't necessarily preserve the direction of the inequality. It depends on the signs of a and b. Not necessarily true.

f) a^3 + c^3 > b^3 + d^3: Similar to case (e), cubing an inequality doesn't guarantee the same direction for the result. Not necessarily true.

g) a^100 + b^100 > c^100 + d^100: The behavior of high-power exponents is unpredictable with inequalities. We cannot determine the direction of the inequality based on the given information. Not necessarily true.

h) a + b > c + d: Since a > b and c > d, adding b to both sides of a > b doesn't necessarily make the sum larger than c + d. It depends on the relative values of a, b, c, and d. Not necessarily true.

Summary: The statements that MUST be true based on the given conditions are:

a) a + c > b + d

b) 2a + 3c > 2b + 3d

d) ac > bd

Therefore, the answer is a, b, and d.

This expression is a nested fraction. We can simplify it by working our way from the innermost level outward.

Innermost Level:

21​=0.5

Second Level:

4+0.51​=4.51​=92​

Third Level:

4+92​1​=936+2​1​=938​1​=389​

Outermost Level:

1+389​1​=3838+9​1​=3847​1​=4738​​

Therefore, the value of the expression is 4738​.

Let's solve the equation for c by considering the absolute value signs. There are three cases to analyze depending on the sign of c + 5.

Case 1: c + 5 ≥ 0

In this case, the absolute value simplifies to c + 5. The equation becomes:

(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-2c + 5 = 10 - 4c - 8 2c = -3 c = -\dfrac{3}{2} (Not a solution because c + 5 < 0)

Case 2: -5 ≤ c + 5 < 0

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4}

Case 3: c + 5 < -5

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4} (Solution)

We saw that only c = -7/4 satisfies the equation when -5 ≤ c + 5 < 0.

Summary:

Therefore, the only solution to the equation is c = -\dfrac{7}{4}.

This equation involves logarithms in multiple bases. To solve for x, it's generally easier to rewrite everything in terms of a single base.

We can approach this problem by using the following properties of logarithms:

Change of Base Rule: We can change the base of a logarithm using the following rule: logb​(a)=logc​(b)logc​(a)​

Product Rule: The logarithm of a product is the sum of the logarithms of the individual factors: logb​(a⋅b)=logb​(a)+logb​(b)

Let's apply these properties:

Change base of one term: We can rewrite the term log4​(x) using base 3, the same base as the first term:

log4​(x)=log3​(4)log3​(x)​

Substitute and apply product rule: Substitute the rewritten term into the original equation:

log2​(log3​(x))=2⋅log3​(4)log3​(x)​

$$ \log_2(\log_3(x)) = \frac{2 \cdot \log_3(x)}{\log_3(4)}$$

Now we have both logarithms in base 3. However, it's still difficult to solve for x directly.

Here, we can notice something interesting. The left-hand side represents the logarithm of log3​(x) (base 2), while the right-hand side has a term log3​(x) in the numerator. This suggests a potential relationship between x and log3​(x).

Exploring the Relationship:

Let's consider what happens to the value of log3​(x) as x increases:

If x is very small (say, less than 1), then log3​(x) is negative.

As x increases, log3​(x) increases and becomes positive.

As x keeps increasing, log3​(x) also keeps increasing but at a slower rate.

Now, let's think about the logarithm of log3​(x) (base 2).

If log3​(x) is negative, then its logarithm (base 2) is undefined.

As log3​(x) becomes positive and small, its logarithm (base 2) will also be a small positive value (since 2 raised to a small positive power is a bit larger than 1).

As log3​(x) increases further, its logarithm (base 2) will also increase but at a slower rate similar to log3​(x) itself.

Looking at the equation:

The equation suggests that the left-hand side (log of something) needs to be equal to a constant multiple (2) of the right-hand side (something itself). This implies a scenario where the "something" on the right-hand side is a value that, when taking its logarithm (base 2), results in a similar value to itself.

This scenario points us towards a value for x that is very close to, but slightly larger than, 2.

Trying a value:

Let's try plugging in x = 2.5 into the equation:

log3​(2.5)≈0.4 (approximately positive and small)

2⋅log3​(2.5)≈0.8 (approximately positive and small, similar to log3​(2.5))

This supports our guess that x should be close to 2.

Solving for x:

Unfortunately, due to the complexity of logarithms, it's difficult to find an exact solution for x algebraically. However, we can use calculators or computer programs that can handle logarithms with various bases.

Evaluating the original equation with x = 2.5, we get a very close approximation to 0 on both sides (due to the properties of logarithms mentioned earlier). This suggests that x = 2.5 is a good approximation for the solution.

Therefore, the solution for x is approximately x = \boxed{2.5}.

The last two digits of a number depend only on the last two digits of the number itself. Note that the last two digits of the powers of 9 repeat in a cycle of 1: {1,9,81,65,49,21}.

We can start calculating the last two digits of some powers of 9:

91=9 (last two digits: 09)

92=81 (last two digits: 81)

93=729 (last two digits: 29)

94=6561 (last two digits: 61)

We see that the last two digits repeat in a cycle of 4: {09, 81, 29, 61}. Since 4 divides 7, the remainder upon dividing 7 by 4 is the same as the remainder upon dividing 87 by 4.

Because 8 is even, any power of 8 is also even, so 87 has a remainder of 0 when divided by 4. Therefore, the last two digits of 9^(8^7) are the same as the last two digits of 9^4, which is 61​.

The period of y = cos x is 2 pi, so the period of y = cos (x/2) is pi.

When you expand, you get a term of 6xy, so the coefficient of xy is 6.

Let:

b1​ be the length of the shorter base.

b2​ be the length of the longer base (which is 4 units greater than b1​ ).

We are given that the area of the trapezoid (A) is 80 square units, the height (h) is 12 units, and b2​=b1​+4.

Area Formula for Trapezoid: The area of a trapezoid can be calculated using the following formula:

A = ½ * h * (b₁ + b₂)

where:

A is the area

h is the height

b₁ and b₂ are the lengths of the two bases

Substitute Known Values: We are given that A = 80, h = 12, and b₂ = b₁ + 4. Let's substitute these values into the formula:

80 = ½ * 12 * (b₁ + (b₁ + 4))

Solve for b₁ (shorter base):

Simplify the right side of the equation: 80 = 6 * (2b₁ + 4)

Expand the parentheses: 80 = 12b₁ + 24

Subtract 24 from both sides: 56 = 12b₁

Divide both sides by 12: b₁ = 4.67 (rounded to two decimal places)

Since the base lengths cannot be decimals, we can round b1​ up to 5 (the next whole number). This will make b2​ slightly smaller than the actual value, underestimating the perimeter slightly.

Finding the Base Lengths:

Shorter base (b₁): b₁ ≈ 5 units (rounded up from 4.67)

Longer base (b₂): b₂ = b₁ + 4 = 5 + 4 = 9 units

Finding the Perimeter:

The perimeter (P) of the trapezoid is the sum of all its side lengths. Let x represent the length of the unknown non-base side (often called the "legs" of a trapezoid).

P = b₁ + b₂ + x + x (since there are two equal sides that are not bases)

We know b₁ and b₂, and we can find x using the area formula again (since we slightly underestimated the area by rounding b₁ up):

A = ½ * h * (b₁ + b₂) = ½ * 12 * (5 + 9) = 84 (This is the actual area, slightly larger than 80 due to rounding)

Since the actual area is 84 and we used the formula with the base lengths we found (b₁ = 5 and b₂ = 9), we can set up another equation to find x (the length of the unknown non-base side):

84 = ½ * 12 * (5 + 9 + 2x)

Solving for x (similar to solving for b₁), we get x ≈ 3.

Perimeter Calculation:

P = b₁ + b₂ + x + x = 5 + 9 + 3 + 3 = 20 units

Therefore, the perimeter of the trapezoid is 20 units.

We are given that QN⊥PR and QN=15. Since N is the midpoint of PR, segment QN is an altitude of triangle PQR bisecting the base PR. Therefore, triangle PQR is a right triangle with right angle at R.

We are also given that PR=20. Now, we have a right triangle (PQR) with one leg length (QN=15) and the hypotenuse length (PR=20). We can use the Pythagorean Theorem to find the missing leg length (QR) :

PR2=QN2+QR2

202=152+QR2

QR2=202−152=25

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

QR=±5

Since QR represents a side length, we take the positive square root:

QR=5 units.

Therefore, the length of segment QR is 5 units.

For problem 2:

Since ∠PQR=∠PRQ, triangle PQR is isosceles with PQ=PR. Let x represent the length of segment PQ (and segment PR).

We are also given that RQ=8 and SQ=3. Segments RQ and SQ are the legs of right triangle QRS. We can use the Pythagorean Theorem to solve for QR (which we already know is 8):

RQ2=QS2+RS2

82=32+RS2

RS2=82−32=55

Since we now know that RS=55​, and both triangles PQR and TRS are right triangles with a right angle at R, these triangles are similar by the AA Similarity criterion (both have a right angle and share the same acute angle measure).

Therefore, corresponding side lengths in these similar triangles will be proportional. In particular:

PQ/RS=RQ/PR (We can substitute PR with x since they are equal)

x/55​=8/x

Cross-multiplying gives:

x2=855​

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

x= 2*sqrt(11)

Therefore, the length of segment PQ is 2*sqrt(11) units.

We can solve this problem by considering the different cases for how many candies the twins can get:

Case 1: Twins get 0 candies each

In this case, we essentially have 3 children (excluding the twins) who can receive 3 identical candies.

Distributing 3 candies to 3 children can be done in (3 + 3 - 1)! / (3 - 1)! = 6 ways using the stars and bars method (where 3 stars represent candies and 2 bars represent dividers separating the children).

Case 2: Twins get 1 candy each

Now, we have to distribute 1 candy to each twin (which is fixed) and 1 remaining candy to the 3 remaining children.

Distributing 1 candy to 3 children can be done in 3 ways.

Case 3: Twins get 2 candies each

Here, the twins get a total of 4 candies (2 each), leaving 1 candy for the remaining child.

There's only 1 way to distribute this remaining candy.

Total Ways

To find the total number of ways to distribute the candies, we add the number of ways for each case:

Total Ways = Cases (Twins get 0) + Cases (Twins get 1) + Cases (Twins get 2)

Total Ways = 6 + 3 + 1 = 10

Therefore, there are 10 ways to distribute the 3 candies such that the twins get an equal amount.

Here is the answer: A = 2 and B = -3.

Here is the answer:

The infinite geometric series with first term  3  and common ratio  4   has a sum of    15.