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\((1+r)^2=(\overline{AB}-r)^2+1^2\\(1+r)^2=(2-r)^2+1\\ 1+2r+r^2=4-4r+r^2+1\\ 6r=4\\\color{blue}r=\frac{2}{3}\)

\(\color{blue}The\ radius\ of\ the\ circle\ is\ \frac{2}{3}.\)

 !

To find the probability that Sam chooses a 2, we need to determine how many times the digit 2 appears in the numbers from 1 to 315.

First, let's count the number of times the digit 2 appears in the units place. From 1 to 315, there are 31 numbers ending in 2 (2, 12, 22, ..., 312).

Next, let's count the number of times the digit 2 appears in the tens place. From 1 to 315, there are 10 numbers in each ten block that contain the digit 2, except for the numbers from 20 to 29 where there are 11 numbers. So, there are \(10 \times 3 + 11 = 31\) numbers with 2 in the tens place.

Finally, let's count the number of times the digit 2 appears in the hundreds place. From 1 to 315, there is only one number with 2 in the hundreds place, which is 200.

Adding up all the occurrences of the digit 2:

\[31 + 31 + 1 = 63\]

So, there are 63 occurrences of the digit 2 in the numbers from 1 to 315.

Now, since Sam chooses one of the digits written down at random, the total number of possible outcomes is the total number of digits written down, which is 315.

Therefore, the probability that Sam chooses a 2 is:

\[ P(\text{choosing a 2}) = \frac{\text{Number of occurrences of 2}}{\text{Total number of digits}} = \frac{63}{315} = \frac{1}{5} \]

So, the probability that Sam chooses a 2 is \( \frac{1}{5} \) or 20%.

There are two cases to consider:

Case 1: Siblings of one pair sit in the same row

There are 3 ways to choose which pair of siblings will have siblings sitting in the same row (either the first row, the second row, or both).

Once a pair is chosen, there are 2 ways to arrange the siblings within that pair (who sits in the first seat and who sits in the second).

For the remaining two pairs, there are only 4 choices for the first seat in the row they occupy (since two seats are already taken by siblings from the chosen pair). Then there are 3 choices for the second seat (since one child has already been seated), for a total of 4 * 3 = 12 ways to arrange the children in that row.

Therefore, the total number of arrangements for Case 1 is 3 (ways to choose the pair) * 2 (arrangements within the pair) * 12 (arrangements in the other row) = 72.

Case 2: Siblings of each pair sit in different rows

There are 6 choices for the child in the first seat of the first row. It doesn't matter which sibling takes it, so suppose child A sits there (denoted by A).

Then there are 4 choices for the second seat (child B, child C, child D, or child E). The last seat in the first row cannot be filled by a sibling of child A, so it must be filled by one of the remaining two children (child D or child E).

This leaves a pair of siblings (let's say child B and child C) and one person (child D or child E) for the second row. So the only order that will work is child B - empty seat - child C. There are 2 choices for which sibling (B or C) sits in the first seat of the second row.

Therefore, the total number of arrangements for Case 2 is 6 (choices for child in first seat of first row) * 4 (choices for second seat) * 2 (choices for who sits first in the second row) = 48.

Adding the ways from both cases, there are 72 (Case 1) + 48 (Case 2) = 120 ways to seat the siblings.

To solve this problem, let's break it down step by step:

1. **Arrange the pairs of siblings in one row**: There are 3 pairs of siblings, so we need to arrange them in one row. This can be done in \(3!\) ways (3 factorial) because there are 3 pairs and the order matters.

2. **Arrange the pairs in the first row**: We have three pairs of siblings, and they can't sit next to each other in the same row. We can start by placing the first pair in one of the three seats, then place the second pair in one of the remaining two seats, and finally, place the third pair in the last remaining seat.

This can be done in \(3 \times 2 \times 1 = 3!\) ways.

3. **Arrange the pairs in the second row**: After arranging the pairs in the first row, there are three pairs left to be seated in the second row. Since no siblings can sit next to each other, the first pair can be seated in any of the three available seats.

Then, the second pair can be seated in one of the remaining two seats, and the last pair will take the remaining seat. This can be done in \(3 \times 2 \times 1 = 3!\) ways.

Now, to find the total number of arrangements, we multiply the number of arrangements for each step:

Total arrangements = (Arrangements of pairs in one row) * (Arrangements of pairs in the first row) * (Arrangements of pairs in the second row)

So, the total number of arrangements is:

\[3! \times 3! \times 3! = 6 \times 6 \times 6 = 216\]

Therefore, there are 216 ways to seat three pairs of siblings from different families in two rows of three chairs, if siblings may not sit next to each other in the same row.

Let's denote the length of the rectangle as "l" and the width as "w". We are given that the perimeter (total length of all sides) is 40 units and the diagonal is 16 units.

Perimeter: Perimeter of a rectangle is calculated by adding the lengths of all its sides. In this case, Perimeter = 2l + 2w = 40.

Diagonal: The diagonal of a rectangle divides it into two congruent right triangles. We can use the Pythagorean theorem to relate the lengths of the sides of the rectangle and the diagonal.

The Pythagorean theorem states: a^2 + b^2 = c^2, where a and b are the lengths of the shorter sides (length and width in this case), and c is the length of the hypotenuse (diagonal). In this case, we are given c (diagonal) as 16.

We can approach this problem in two ways:

Method 1: Solve for l and w directly

From the perimeter equation, we can rewrite it to isolate "l": l = (40 - 2w) / 2.

Substitute this expression for "l" in the Pythagorean theorem equation: [(40 - 2w) / 2]^2 + w^2 = 16^2.

Solve this equation for "w" (tedious but solvable). Once you find "w", you can plug it back into the equation for "l" to find the length.

Method 2: Exploit the relationship between l and w using the diagonal

We know the diagonal is 16, and a rectangle's diagonals cut each other in half, creating right triangles with legs equal to half the length and half the width.

Therefore, in each right triangle formed by the diagonal, one leg (half the length) is 8 (half of 16).

Since a rectangle has opposite sides equal, the other leg of the right triangle (half the width) must also be 8.

Now we know that half the width (w/2) is 8. Solve for the actual width (w) by multiplying by 2: w = 16.

Finding the Area

Once you have the width (w = 16 from Method 2), you can find the length (l) using the perimeter equation (l = (40 - 2 * 16) / 2 = 4).

Now that you know both the length (l = 4) and width (w = 16), the area of the rectangle can be calculated using the formula: Area = l * w = 4 * 16 = 64 square units.

Therefore, the area of the rectangle is 64 square units.

We know that a quadratic function with roots at x=2 and x=4 can be written in the factored form as:

a(x−2)(x−4)

We are also given that the function takes the value 6 when x=3. This translates to the equation: a(3−2)(3−4)=6 which simplifies to −a=6

Since we already know that a cannot be 0 (the function wouldn't be quadratic), we can divide both sides by −1 to get a=−6

Plugging this value of a back into the factored form, we get the quadratic function: −6(x−2)(x−4)

Expanding this expression, we get the answer in the desired form: −6x^2 + 36x + 48

So the answer is -6x^2 + 36x + 48.

By the British Flag Theorem, if \(X\) is a point in the interior of a rectangle \(ABCD\), then $$AX^2+CX^2=BX^2+DX^2.$$

It follows that $$9+25=16+DX^2 \implies DX^2=18 \implies DX = \boxed{3\sqrt{2}}.$$

In triangle $PQR$, we are given that $M$ is the midpoint of $\overline{PQ}$, $N$ is the midpoint of $\overline{PR}$, and $O$ is the intersection of $\overline{QN}$ and $\overline{RM}$. Additionally, we are told that $\overline{QN}\perp\overline{PR}$, $QN = 15$, and $PR = 20$.

Since $M$ is the midpoint of $\overline{PQ}$ and $N$ is the midpoint of $\overline{PR}$, we can conclude that $MO$ is the median of triangle $PQR$. Therefore, $MO$ passes through the midpoint of $\overline{QR}$.

Since $\overline{QN}\perp\overline{PR}$, triangle $PQN$ is a right triangle. According to the Pythagorean theorem, we have:

𝑃𝑄2=𝑄𝑁2+𝑃𝑅2PQ2=QN2+PR2 𝑃𝑄2=152+202PQ2=152+202 𝑃𝑄2=225+400PQ2=225+400 𝑃𝑄2=625PQ2=625 𝑃𝑄=25PQ=25

Since $M$ is the midpoint of $\overline{PQ}$, we have $PM = MQ = \frac{1}{2}PQ = \frac{1}{2} \times 25 = 12.5$.

Now, since $MO$ is the median of triangle $PQR$, we have $MO = \frac{1}{2}QR$. Therefore, $QR = 2 \times MO$.

To find $MO$, we can use the fact that $MO$ is perpendicular to $QN$, so triangles $QMO$ and $QNO$ are similar triangles. Therefore:

𝑀𝑂𝑄𝑁=𝑄𝑁𝑄𝑅QNMO​=QRQN​ 𝑀𝑂15=15𝑄𝑅15MO​=QR15​ 𝑀𝑂=152𝑄𝑅MO=QR152​ 𝑀𝑂=225𝑄𝑅MO=QR225​

Since $MO = \frac{1}{2}QR$, we can set up the following equation:

12𝑄𝑅=225𝑄𝑅21​QR=QR225​

Solving for $QR$, we get:

𝑄𝑅2=450QR2=450

𝑄𝑅=450QR=450​

𝑄𝑅=1510QR=1510​

Therefore, the length of $QR$ is $15\sqrt{10}$.

In triangle ABC,  the angle bisector of  BAC meets  BC at D.

If   BAC=60°, CAD=45°, and  AD=24, then find the area of triangle ABC.

\(\color{blue}If\ D\ lies\ on\ the\ bisector\ of\ \angle BAC=60°, \angle CAD \ cannot\ be\ equal\ to\ 45°.\)

 !

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.

  2x^2 + 3x + 8x - x^2 + 4x + k has a double root    

  combine like terms                          (2x² – x²) + (3x +8x + 4x) + k  

                                                           x² + 11x + k    

  to have a double root,    

  i.e., both roots are the same,   

  the discriminate must equal zero        b² – 4ac = 0    

                                                            (11)² – (4)(1)(k)  =  0    

                                                             121 – 4k  =  0    

                                                                     –4k  =  –121    

                                                                          k  =  121 / 4    

_.

1.   At the vertex of a polygon, the sum of the exterior angle and the interior angle is 180^o.   

      Since the interior angle is 12 times the exterior angle, the exterior angle has to be 15^o.  

      The sum of all exterior angles of a polygon is 360^o, irrespective of the number of sides.  

       Therefore, the polygon has 360 / 15 = 24 vertices.   

       We shall presuppose that a polygon has the same number of sides as it has vertices,    

       Therefore, this polygon has 24 sides.    

_.

                                       y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25    

group like terms              y  =  (–2x² – 3x²) + (8x – 14x) + (–15 + 25)    

combine like terms           y  =  –5x² – 6x + 10    

first derivitive                    ^dy/_dx  =  –10x – 6    

set equal to zero                    0  =  –10x – 6    

                                           10x  =  –6   

                                               x  =  – 0.6    this is the x-coordinate of the vertex

sub x into original                   y  =  [ –5 • (–0.6)² ]  –  [6 • (–0.6) ]  +  10    

                                               y  =  (–5 • 0.36)  –  (–3.6)  +  10    

                                               y  =  –1.8 + 3.6 + 10    

                                               y  =  11.8      this is the y-coordinate of the vertex

the vertex is at                        (–0.6 , 11.8)    

_.

The only constraint for the domain here, is the fact that the inside of the square root also known as the radicand must be greater or equal or 0. 

Essentially, we get \(6-x-x^2-2x^2 \geq 0\)

Then, we simply solve this inequality and there's your domain. Now, I'm not going to write the entire process down for this inequality, so try solving it yourself, and give it a try. It's pretty simple. Combine the like terms to get \(-3x^2 -x + 6 \geq 0\). Plug this into the quadratic formula, and you achieve the asnwer of 

In inequality form, it would be \(-\frac{1+\sqrt{73}}{6} \leq x \leq -\frac{1-\sqrt{73}}{6}\)

In Interval Notation, it would be \([-\frac{1+\sqrt{73}}{6}, -\frac{1-\sqrt{73}}{6}]\)

Thanks!

Find the area of the strip.

Let a be the distance of the base 10 touched by the strip.

Let alpha be the slope angle of the strip.

\(tan(\alpha)=\frac{8}{10-a}\\ sin(\alpha )=\frac{1}{a}\\ \alpha =atan(\frac{8}{10-a})=asin(\frac{1}{a})\\ atan(\frac{8}{10-a})-asin(\frac{1}{a})=0\)

\(\color{green}WolframAlpha\\ \color{green}a=\frac{8\sqrt{163}}{63}-\frac{10}{63}\\\)

\(\color{blue}a=1.4625\\ A=8\cdot 1.4625\\ \color{blue}A=11.7\)

\(\color{blue}The\ area\ of\ the\ strip\ is\ 11.7\ .\)

                                                 x^2-mx+24 = 10   

subtract 10 from both sides      x² – mx + 14  =  0   

integers that multiply to 14        (+7)(+2)   

                                                  (–7)(–2)   

                                                  (+1)(+14)   

                                                  (–1)(–14)  

possible values of m                   –9   

                                                   +9   

                                                  –15   

                                                  +15   

total number of

possible values of m                      4    

_.

                                        5x^2 - 2x + 8 - 2x^2 + 6x + 2 - 2x^2 + 4x - 1   

can be regarded as        (5x² – 2x² – 2x²) + (–2x + 6x + 4x) + (8 + 2 – 1)   

combine like terms            x² + 8x + 9   

discriminant is                   8² – (4)(1)(9)  =  64 – 36  =  28   

The discriminant is the expression under the radical   

in the quadratic formula.   

When a quadratic is stated in the form ax² + bx + c  

the discriminant is b² – 4ac.  

_.

We can divide this problem into two cases:

Case 1: Each row has exactly one child from each family.

Choose one child from each family for the first row: there are 3 choices

.

Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.

Arrange the second row similarly: 3⋅2=6​ ways total.

Case 2: One row has two children from the same family.

There are two subcases depending on the arrangement of siblings in the rows:

Subcase 2a: The first child in each row is from the same family.

Choose one pair of siblings: 3 choices

.

Arrange the siblings within the pair: 2 ways.

Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.

Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.

Total: 2!⋅2!3⋅2⋅4!​=36 ways.

Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.

Choose one pair to have their children occupy the third seats in each row: 3 choices.

Arrange the remaining 4 children in the first row: 4! ways.

Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.

Total: 2!⋅2!3⋅4!​=36 ways.

Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:

6 + 36 + 36 = 78​.

This gives us a final answer of 78.