Questions 271
Answers 18


We can divide the problem into two triangles: △ABC and △ACE. Since we are given three side lengths for each triangle, we can use the [Law of Cosines] to solve for DE.


For △ABC, we have

\begin{align*} BC^2 &= AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos ( \angle A) \ 8^2 &= 3^2 + 7^2 - 2 \cdot 3 \cdot 7 \cos ( \angle A) \ \cos ( \angle A) &= \frac{38}{42} = \dfrac{19}{21}. \end{align*}


Similarly, for △ACE, we obtain

\begin{align*} AC^2 &= AE^2 + CE^2 - 2 \cdot AE \cdot CE \cos ( \angle A) \ 7^2 &= (AE + DE)^2 + 2^2 - 2 \cdot (AE + DE) \cdot 2 \cos ( \angle A) \ 7^2 &= AE^2 + DE^2 + 4AE + 4 - 4(AE + DE) \cdot \dfrac{19}{21} \ 0 &= AE^2 + DE^2 - \dfrac{32}{7} AE - \dfrac{32}{7} DE + 20. \end{align*}


We can complete the square on both the AE2 and DE2 terms. To do this, we take half of the coefficient of our AE term, square it, and add it to both sides of the equation. Similarly, we do this for the DE term.


This gives us

\begin{align*} 0 &= (AE^2 - \dfrac{16}{7} AE + \dfrac{64}{49}) + (DE^2 - \dfrac{16}{7} DE + \dfrac{64}{49}) + 20 - \dfrac{64}{7} (AE + DE) \ &= (AE - \dfrac{8}{7})^2 + (DE - \dfrac{8}{7})^2 + 20 - \dfrac{64}{7} (AE + DE). \end{align*}


Let x=AE−78​ and y=DE−78​. Then AE=x+78​ and DE=y+78​, so

\begin{align*} 0 &= (x)^2 + (y)^2 + 20 - \dfrac{64}{7} ((x + \dfrac{8}{7}) + (y + \dfrac{8}{7})) \ 0 &= x^2 + y^2 + 20 - \dfrac{64}{7} x - \dfrac{64}{7} y - 16. \ 0 &= x^2 + y^2 - \dfrac{64}{7} x - \dfrac{64}{7} y + 4. \end{align*}


Now we can complete the square again, this time on the entire right side of the equation. Taking half of the coefficient of our x term, squaring it, and adding it to both sides gives


\begin{align*} 0 &= (x^2 - \dfrac{64}{7} x + \dfrac{16^2}{49}) + (y^2 - \dfrac{64}{7} y + \dfrac{16^2}{49}) + 4 - \dfrac{16^2}{49}. \ &= (x - \dfrac{32}{7})^2 + (y - \dfrac{32}{7})^2 - \dfrac{225}{49}. \end{align*}


Since the square of a real number is never negative, the only way for the right side of this equation to equal zero is if both (x-32/7​)2 and (y−32/7​)2 are zero. Thus, x = 32/7 and y = 32/7​, which means AE=8 and DE=8. 


Therefore, DE=8​.

May 21, 2024

Absolutely, I’ve been improving my problem-solving abilities in maximizing functions under constraints. Let's maximize the expression: xy

subject to the constraint: 30x2​+305y2​=1


We can solve this problem using Lagrange multipliers. Here's how we can approach this problem:


Introduce Lagrange Multiplier: We'll introduce a Lagrange multiplier, λ, to incorporate the constraint into the objective function. The new function to maximize will be:




Take Partial Derivatives: We will take partial derivatives of L with respect to x, y, and λ, and set them equal to zero. This will give us a system of equations to solve for x, y, and λ.








Solve the System of Equations: From the first equation, we get:




Substitute this into the second equation:








Since x is a nonnegative real number, we can ignore the case where x = 0. Therefore:






This equation has no real solutions for λ. However, we can notice that the constraint equation is already satisfied regardless of the value of x and y. This means that any values of x and y that satisfy the constraint will also satisfy the condition where the partial derivatives are zero.


Find Maximum Value: Since the constraint forces x2+5y2=30, the largest possible value of xy will occur when x and y are as close in value as possible, subject to the constraint. In a 2D plane, this would correspond to a point on the ellipse where x and y have the same absolute value.


Looking at the constraint equation, we see that when x2=18 and y2=12 (or vice versa), the condition is satisfied. So, one possible solution is:


x= sqrt(18​) = 3 sqrt(2)​

y= sqrt(12) = 2 sqrt(3)


Therefore, the maximum value of xy is:


xy = 3*sqrt(2)*2*sqrt(3) = 6*sqrt(6)


In conclusion, the maximum value of xy subject to the constraint is 6*sqrt(6).

May 4, 2024

We can solve this system of equations by adding and subtracting the equations strategically to eliminate one of the variables.

Adding the Equations:

Adding the two given equations eliminates the term ab^2:

(a^3 + 3ab^2) + (3a^3 - ab^2) = 679 + 615

Combine like terms:

4a^3 = 1294

Solving for a^3:

Divide both sides by 4:

a^3 = 1294 / 4 = 323.5

Subtracting the Equations:

Subtracting the second equation from the first equation eliminates the term 3a^3:

(a^3 + 3ab^2) - (3a^3 - ab^2) = 679 - 615

Combine like terms:

4ab^2 = 64

Solving for ab^2:

Divide both sides by 4:

ab^2 = 16

Relating a and b:

Since we know both a^3 and ab^2, we can try to express one variable in terms of the other.

From the equation for a^3, we can write a =∛(323.5).

Relating a - b:

We want to find a - b. Since we don't have a direct equation for b, we can try to manipulate the equation for ab^2.

Rewrite the equation for ab^2:

b^2 = a(ab^2) = a * 16

Substitute a = ∛(323.5):

b^2 = ∛(323.5) * 16

Now, we can express b in terms of a:

b = ± 4 * ∛(323.5)

Finding a - b:

Since a and b are real numbers, we can consider both positive and negative values of b. However, we only care about the difference a - b.

There are two cases:

Case 1: b = 4 * ∛(323.5)

a - b = ∛(323.5) - (4 * ∛(323.5))

Factor out ∛(323.5):

a - b = ∛(323.5) (1 - 4)

a - b = -3 * ∛(323.5)

Case 2: b = -4 * ∛(323.5)

a - b = ∛(323.5) - (-4 * ∛(323.5))

Factor out ∛(323.5):

a - b = ∛(323.5) (1 + 4)

a - b = 5 * ∛(323.5)


Since we don't know the signs of a and b beforehand, both cases are valid. Therefore, a - b can be either -3 * ∛(323.5) or 5 * ∛(323.5).

Both answers are negative and positive multiples of the same cube root, so they essentially represent the same value with opposite signs. The absolute value of a - b is:

|a - b| = |(-3) * ∛(323.5)| = |5 * ∛(323.5)| = 5 * ∛(323.5)

Apr 13, 2024

We can solve this problem by recognizing that subdividing a triangle into smaller similar triangles reduces the area by a factor based on the ratio of side lengths.


Similarity of Triangles:


Since M, N, and O are midpoints of sides, segments NO, OM, and MN are parallel to sides KL, LJ, and JK respectively (corresponding sides theorem).


Due to alternate interior angles, angles in triangles JKL, JMN, JNO, etc. are congruent (alternate interior angles theorem).


Therefore, triangles JKL, JMN, JNO, etc. are similar by Angle-Angle (AA) Similarity.


Ratio of Side Lengths:


Since M, N, and O are midpoints, segments NO, OM, MN, PN, QM, and PR are all half the length of their corresponding sides in the larger triangle (JKL).


Area Ratio of Similar Triangles:


The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding side lengths.


Let k be the scaling factor between triangles JKL and PQR (i.e., the ratio of side lengths between corresponding sides).


The area of triangle PQR is 12.


Relating Areas of JKL and PQR:


Area(PQR) / Area(JKL) = k^2 (ratio of areas of similar triangles)


Substitute the given value:


12 / Area(JKL) = k^2


Relating Side Lengths:


Since PN, QM, and PR are half the length of their corresponding sides in JKL, k = 1/2 (ratio of side lengths).


Finding Area(JKL):


Substitute k = 1/2 in the equation from step 4:


12 / Area(JKL) = (1/2)^2


Multiply both sides by Area(JKL):


12 = Area(JKL) / 4


Solve for Area(JKL):


Area(JKL) = 12 * 4 = 48


Therefore, the area of triangle JKL is 48 square units.

Apr 13, 2024

Angles in Triangle STU:


We are given that angle TSU = 62 degrees and angle STU = 29 degrees. Since the angles in a triangle add up to 180 degrees, we can find the third angle (angle SUT):


angle SUT = 180 degrees - angle TSU - angle STU = 180 degrees - 62 degrees - 29 degrees = 89 degrees.


Angles in Triangles MNU and MNP:


Since M, N, and P are midpoints of sides, segments MN, NP, and PM are parallel to TU, US, and ST respectively (corresponding sides theorem).


Due to alternate interior angles, angle MNU = angle TSU = 62 degrees and angle MNP = angle STU = 29 degrees (alternate interior angles theorem).


Angle NZM and Angle NPM:


Triangle NUZ is a right triangle (angle NUZ = 90 degrees) since UZ is an altitude.


Since M is the midpoint of TU, segment NZ is half of segment TU. Similarly, segment NP is half of segment US.


Therefore, triangles NUZ and MNP are similar (AA Similarity).


Corresponding angles in similar triangles are congruent. Thus, angle NZM = angle NMP (corresponding angles).


Finding Angle NZM + Angle NPM:


We know angle MNU = 62 degrees and angle NMP = angle NZM (from step 3). Since angles in triangle MNU add up to 180 degrees:


angle NZM + angle NMP + angle MNU = 180 degrees


Substitute the known values:


angle NZM + angle NZM + 62 degrees = 180 degrees


Combine like terms:


2 * angle NZM = 180 degrees - 62 degrees


Solve for angle NZM:


angle NZM = 118 degrees / 2 = 59 degrees


Therefore, angle NZM + angle NPM = angle NZM + angle NZM (since they are congruent) = 59 degrees + 59 degrees = 118 degrees.

Apr 13, 2024

To calculate the total cost of the Chromebook with the payment plan, we can leverage the concept of compound interest. Here's how:


Monthly Interest Rate:


The annual interest rate is 6%. Since the payments are compounded monthly, we need to convert this to a monthly rate.


Monthly interest rate = Annual interest rate / Number of compounding periods per year


Monthly interest rate = 6% / 12 months = 0.5% per month (converted to a decimal)


Number of Payments:


The payment plan involves 18 equal monthly payments.


Cost of Chromebook:


The Chromebook's cost is $499.99.


Formula for Monthly Payment:


We can use the following formula to calculate the monthly payment amount:


Monthly payment = Loan amount * (Monthly interest rate / (1 - (1 + Monthly interest rate)^(-Number of payments)))




Plug in the values:


Loan amount = $499.99


Monthly interest rate = 0.005 (converted from 0.5%)


Number of payments = 18


Calculate the monthly payment:


Monthly payment = $499.99 * (0.005 / (1 - (1 + 0.005)^(-18)))


Monthly payment = $28.61 (rounded to two decimal places)


Total Cost:


Since there are 18 equal payments of $28.61 each, the total cost will be:


Total cost = Number of payments * Monthly payment


Total cost = 18 payments * $28.61/payment


Total cost = $514.98 (rounded to two decimal places)


Therefore, by taking the payment plan with monthly compounding interest, you will end up paying a total of $514.98 for the $499.99 Chromebook.

Mar 30, 2024

Since children must be next to two adults, let's consider the children as a single unit. There are now 5 "units" to arrange around the table (4 adults + 1 unit of 4 children).


Here's how to solve the problem:


Total Arrangements without Restriction:


Initially, ignoring the child requirement, we can arrange 5 units (adults and children) around the circular table.


For circular arrangements of n distinct objects, there are (n-1)! ways.


In this case, there are (5-1)! = 4! = 24 ways to arrange the units.


Overcounting due to Rotation:


However, in a circular arrangement, rotating the entire configuration doesn't create a new seating order.


So, we've overcounted the arrangements by the number of ways to rotate 5 objects in a circle.


There are 5 positions, and rotating one slot to the right fills all the positions eventually.


Therefore, we've overcounted by a factor of 5.


Correcting for Overcounting:


To get the actual number of unique arrangements, we need to divide the initial arrangements by the overcounting factor:


Unique Arrangements = Total Arrangements / Overcounting Factor


Unique Arrangements = 24 arrangements / 5 rotations


Unique Arrangements = 4.8 (Since the answer deals with arrangements, we can't have a fraction of a seating)


Accounting for Child Arrangement:


So far, we've treated the children as a single unit. But within that unit, the 4 children can be arranged in 4! ways.


Final Answer:


To get the total number of arrangements where each child sits next to two adults, we multiply the number of unique arrangements for the units (adults and children) by the number of ways to arrange the children within their unit:


Total Arrangements = Unique Unit Arrangements * Child Arrangements


Total Arrangements = 4 * 4! = 4 * 24 = 96


Therefore, there are 96 ways to seat the children and adults around the table such that each child sits next to two adults.

Mar 30, 2024