bader

We can solve this problem using the Pythagorean Theorem applied to both triangles ABC and ACD since we are given a side length and looking for the hypotenuse of each right triangle.

For triangle ABC, we are given that AC=4 and solving for AB (the hypotenuse) using the Pythagorean Theorem: AB2=AC2+BC2=42+x2=42+452, so AB=42+452​.

For triangle ACD, we are given that AC=4 and solving for AD (the hypotenuse) using the Pythagorean Theorem: AD2=AC2+CD2=42+y2=42+632, so AD=42+632​.

To find the difference in length between AB and AD, we subtract the length of AC from each and find the difference: AD−AB=42+632​−42+452​.

We can evaluate these square roots on a calculator and round to the nearest hundredth to find that AD − AB ≈ 8.79 − 6.65 = 2.14 units.

Let the roots of Babette's monic quadratic be r and s. Since the leading coefficient is 1, the quadratic can be written as:

p(x)=(x−r)(x−s)=x2−(r+s)x+rs.

We are given that squaring the roots gives another monic quadratic with those squares as its roots. In other words:

(x−r2)(x−s2)=x2−(r2+s2)x+r2s2

Expanding the left side gives:

x2−(r2+2rs+s2)x+r2s2

Equating the coefficients of the corresponding terms in both quadratics, we get the system of equations:

r+s=r2+s2

r2s2=rs (notice this simplifies to r2s2−rs=0)

From the first equation, we can rearrange to get r2+s2−r−s=0. Factoring this gives (r−1)(s−1)=0. This means either r=1 or s=1.

We can consider two cases:

Case 1: r=1

Substituting r=1 into the second equation gives s2−s=0, which factors as s(s−1)=0. This means either s=0 or s=1. However, if s=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, s=1 and the quadratic is simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Case 2: s=1

Substituting s=1 into the second equation gives r2−r=0, which factors as r(r−1)=0. This means either r=0 or r=1. However, if r=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, r=1 and the quadratic is again simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Since both cases lead to only one possible quadratic, Babette could have been thinking of at most 2​ different monic quadratics.

Find the intersection points:

The line x = 2 intersects the y-axis at (2, 0).

To find the intersection of y = -x - 1 and y = 2, we set the y values equal: -x - 1 = 2 --> x = -3. Substituting this x value back into y = -x - 1 gives the point (-3, -2).

Find the midpoint of each side:

Midpoint of side connecting (2, 0) and (-3, -2):

x-coordinate: ((2) + (-3)) / 2 = -1/2

y-coordinate: ((0) + (-2)) / 2 = -1

Midpoint is (-1/2, -1).

Since the line x = 2 is a vertical line, any point on this line with a y-coordinate of -1 will be the midpoint of the side connecting (2, 0) and the intersection point on this line (which doesn't matter since it's a vertical line). Therefore, the midpoint is (2, -1).

Since all three vertices lie on the line x = 2, the center of the circle that passes through all three vertices must also lie on this line. Therefore, the center of the circle has an x-coordinate of 2.

Since the midpoint of one side is also the midpoint of another side that lies on the same line, the center of the circle must also coincide with the midpoint we found: (2, -1).

Therefore, the center of the circle is at (2,−1)​.

The radius of the circle is 22/9.

Since I is the incenter, then ∠BIM=∠CIN=90∘. Also, since MN is parallel to BC, then ∠AMN=∠ABC and ∠ANM=∠ACB.

By the Law of Cosines on triangle ABC, [\cos C = \frac{AB^2 + AC^2 - BC^2}{2AB \cdot AC} = \frac{5^2 + 5^2 - 8^2}{2 \cdot 5 \cdot 5} = -\frac{3}{25}.]Then sinC=1−cos2C​=2546​​.

pen_spark

Let x=BM and y=CN. Then by the Law of Sines on triangle ABM, [\frac{BM}{\sin C} = \frac{AB}{\sin \angle BAM} = \frac{AB}{\sin 90^\circ} = AB,]so BM=5⋅sinC=546​​.

Similarly, by the Law of Sines on triangle ACN, [\frac{CN}{\sin C} = \frac{AC}{\sin \angle CAN} = \frac{AC}{\sin 90^\circ} = AC,]so CN=5⋅sinC=546​​.

Then triangle AMN is a right triangle with legs of length 546​​, so its area is [\frac{1}{2} \cdot \frac{4 \sqrt{6}}{5} \cdot \frac{4 \sqrt{6}}{5} = \boxed{\frac{32}{5}}.]

We can solve the inequality by distributing the terms, combining terms on each side, and isolating x. Here's how:

Distribute the terms on the left side: -4(x + 4) = -4x - 16

Combine terms on both sides: -4x - 16 > x + 7 + 8x + 21

Combine like terms: -4x - 16 > 9x + 28

Isolate x by getting all the x terms on one side: -13x - 16 > 28

Add 16 to both sides to isolate x further: -13x > 44

Divide both sides by -13 to isolate x (remembering to flip the inequality sign since we're dividing by a negative number): x < - \dfrac{44}{13}

Simplifying the answer: x < - \dfrac{33}{9}

Therefore, the solution to the inequality is the set of all real numbers less than - \dfrac{33}{9}. We can express this mathematically as: x ∈ {real numbers | x < - \dfrac{33}{9}}.

Let x be the maximum weight each of my neighbor's small trucks can carry. Then my truck can carry at most x+500 pounds.

The total amount her four trucks can carry is 4x pounds. Since my truck's maximum load is at most 41​ of the total her trucks can carry, we know x+500≤41​⋅4x=x.

Simplifying the inequality on the right gives x+500≤x, which means 500≤0. Since this is not possible, there must be no value of x that satisfies both of our original inequalities. Therefore, there is no upper bound on the amount my truck can carry, so the greatest load it can carry is infinite​ pounds. However this makes no sense in the real world, so we can determine that there is an error in the original problem statement.

Simplifying the right side of the inequality gives: 2x - 5 <= -8x + 30

Combining terms with x on the left side: 10x - 5 <= 30

Adding 5 to both sides: 10x <= 35

Dividing both sides by 10 (and flipping the inequality since we're dividing by a positive number): x <= 3.5

Therefore, the solution is the set of all real numbers less than or equal to 3.5. We can express this mathematically as: x ∈ {real numbers | x ≤ 3.5}.

By the distance formula,

\begin{align*} PA^2 &= (x - 4)^2 + (y + 4)^2, \ PB^2 &= (x - 3)^2 + (y - 5)^2, \ PC^2 &= (x + 1)^2 + (y - 2)^2. \end{align*}

Hence, the given equation becomes \begin{align*} (x - 4)^2 + (y + 4)^2 + (x - 3)^2 + (y - 5)^2 + (x + 1)^2 + (y - 2)^2 &= 3((x - Q)^2 + (y - Q)^2) + k \ \Rightarrow \quad 3x^2 + 3y^2 - 14x - 14y + 12 &= 3x^2 - 6Qx + 3Q^2 + 3y^2 - 6Qy + 3Q^2 + k \ \Rightarrow \quad -14x - 14y &= -6Qx - 6Qy + k - 12 \ \Rightarrow \quad 7x + 7y &= 3Q(x + y) + \frac{k - 12}{7}. \end{align*}

Since this equation holds for all points P, the coefficients of x and y must be zero.

Hence, Q=−7/3 and k−12=0, so k = 12​.

To estimate the density of air \( \rho \), we can rearrange the terminal velocity formula:

\[ |v_{\text{terminal}}| = \sqrt{\frac{2mg}{C_D \rho A}} \]

We can solve this equation for \( \rho \):

\[ \rho = \frac{2mg}{C_D A |v_{\text{terminal}}|^2} \]

Given that \( g \approx 9.8 \, \text{m/s}^2 \) and \( C_D \approx 1 \), we can use our measurements for \( m \), \( A \), and \( |v_{\text{terminal}}| \) to estimate \( \rho \).

\[ \rho = \frac{2 \times 0.1 \times 9.8}{1 \times 0.01 \times 5^2} \]

\[ \rho \approx \frac{1.96}{0.25} \]

\[ \rho \approx 7.84 \, \text{kg/m}^3 \]

So, based on these measurements and estimates, the density of air \( \rho \) is approximately \( 7.84 \, \text{kg/m}^3 \).