All Answers 358110 Answers

Since triangle ABC is isosceles with AB=AC=20, segment AM is a median and an altitude (since the base angles are congruent). Therefore, △ABM is a right triangle with hypotenuse AB=20 and leg BM=21​⋅AB=10

Because D is the midpoint of AM, MD=21​⋅AM=21​⋅AB=10. Then △BDM is a 3-4-5 right triangle, so BD=4⋅10=40.

Similarly, since E is the midpoint of AB, AE=21​⋅AB=10. Then △AEM is also a 3-4-5 right triangle, so ME=4⋅10=40.

Triangle BET shares altitude EM with △AEM, and the ratio of their bases is AEBE​=21​. Therefore, the area of △BET is half the area of △AEM, which is 21​⋅21​⋅AE⋅EM=21​⋅21​⋅10⋅40=100​.

Let's denote the side length of the small square as "s".

Equilateral Triangles:

Since the triangles are equilateral, each side length is equal to "s". The height of an equilateral triangle with side length "s" is given by 23​s​.

Large Square:

The large square consists of the small square in the center and four right triangles formed by the sides of the equilateral triangles and the sides of the large square.

Finding the Side Length of the Large Square:

The base of each right triangle is "s" (side of the small square).

The height of each right triangle is the height of the equilateral triangle, which is 23​s​.

By the Pythagorean Theorem, the hypotenuse (which is the side of the large square) can be found as:

Side of Large Square^2 = s^2 + (sqrt(3)s/2)^2 Side of Large Square^2 = s^2 + (3s^2)/4 Side of Large Square^2 = (7s^2)/4 Side of Large Square = sqrt((7s^2)/4) Side of Large Square = (s * sqrt(7))/2

Area of the Large Square:

The area of a square is calculated by squaring the side length. Therefore, the area of the large square is:

Area of Large Square = (Side of Large Square)^2 Area of Large Square = ((s * sqrt(7))/2)^2 Area of Large Square = (s^2 * 7) / 4

Therefore, the area of the large square is a function of the small square's side length "s" and is proportional to 7/4 times the area of the small square.

Since △ABC is isosceles, let AB=AC=x. Then the area of △ABC is [[ABC] = \frac{1}{2} \cdot x \cdot h,]where h is the altitude drawn from A to BC.

Setting the area equal to 150, we get [\frac{1}{2} \cdot x \cdot h = 150.]

Then xh=300. However, by the Pythagorean Theorem, [x^2 = (h + 8)^2 = h^2 + 16h + 64,]so x2−h2=16h+64.

Substituting 300 for xh, we get [x^2 - 300 = 16h + 64.]Then x2−16h−364=0.

By Vieta's formulas, the sum of the roots of this quadratic is 16, so x=20.

Hence, the perimeter is x+x+8=48​.

Because no two consecutive squares can have the same color, to have at least three yellow squares, they must be placed in squares 1, 3, and 5.

The remaining squares, 2 and 4, must then be colored red and blue in some order. There are 2!=2 ways to order these remaining colors, so there are 2​ total colorings.

There are two main cases to consider for how the siblings can be arranged:

Case 1: Two pairs sit together in one row

There are 3 ways to choose which row will have two pairs of siblings sitting together. For the chosen row, there are 2 ways to arrange the two siblings within each pair (brother first or sister first). Once we've placed these four children, the remaining two children can fill the remaining seats in the other row in 2!=2 ways. This gives a total of 3⋅2⋅2=12​ arrangements for this case.

Case 2: All pairs are split between the rows

Here, there is only one way to arrange the siblings: each sibling sits in a different row from their brother or sister. Once we've placed one child from each pair in the first row, the remaining three children can fill the second row in 3!=6 ways. So, there are 1⋅6=6​ arrangements for this case.

Adding the number of arrangements for these two cases, there is a total of 12+6=18​ ways to seat the siblings.

Let the sphere have radius r and consider a right triangle through the center of the sphere and the base of the cone, with one leg being the radius of the base, 4, and the hypotenuse being the slant height of the cone.

By the Pythagorean Theorem, the other leg (which is the height of the cone minus the radius of the sphere) has length 32−r2​.

Since this right triangle is similar to a right triangle with legs 4 and 3 (i.e. a 4-3-5 triangle), we have [\frac{3-r}{3} = \frac{r}{4},]so r=43​.

The volume of the cone is 31​π(42)(3)=16π, and the volume of the sphere is 34​π(43​)3=83​π, so the ratio of the volume of the sphere to the volume of the cone is 3/128.

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.

Find the radius of the circle.

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

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

  !

This is actually quite an easy question!

At first, it might look like we have to distribute out and actually go through and raise 3x^2 + 11 to the power of 12, which is insanely time consuming and honestly, just not worth it. 

However, there is a simple trick we could use to solve this problem. Notice that the problem is asking for the degree of polynomial, which is the highest power in this polynomial. 

This means we wouldn't have to actually find the entire polynomial. 

Now, let's note something. We see the orinal polynomial has degree 2, since it's leading term has a power of 2. This means that not matter what, it is not possible to get a higher degree than whatever this orginal degree is raised to the power of. What I mean is, when we raise 3x^2 to the power of 12, the other terms in the polynomial will not have a higher power than that! 

So the highest degree is just (x^2)^12, which is just x^24. So the new polynomial has degree 24!

I hope you understood my explenation!

Thanks!

This problem looks really complicated and a nightmare to solve, but it actually is quite simple!

First, we have to factor in the 4m-2 into the parenthesis, getting us \(\frac{{1}^{4m-2}}{{49}^{4m-2}} = {7}^{-10m-2}\). 

This still looks far off from our goal of finding m, but notice that 49 is just 7^2 and the numerator is just 1. This means that we just have \(\frac{1}{{7}^{2(4m-2)}}\). 

Alright, now we're closer to comparable levels. However, we have once term in the denominator and one in the numerator. What should we do about that?

Well, the answer lies in the fact that \({a}^{-1} = \frac{1}{a}\). If we factor out -1 from the power in the right side of the equation, we have \({7}^{-(10m+2)}\), meaning that we technically just have the final equation \(\frac{1}{{7}^{8m-4}} = \frac{1}{{7}^{10m+2}}\). In order for the equation to be satisfied, we must have \({7}^{8m-4} = {7}^{10m+2}\), meaning that \(8m - 4 =10m + 2\). 

Combining like terms gets us \(-2m = 6\) so \(m = -3\)!

The key to this problem is to realize that we have to kind of force our way to a similar expression on both sides so that we could write an equation. I hope I answered your question correctly and clearly!

Thanks!

To find the area of triangle ABC, we can first determine the height of the triangle from point A to side BC. Let's denote the height as h.

1. Let's use the formula for the area of a triangle: Area = 1/2 x base x height. 

2. Since we know AX = 10, BC = 10, and CX = 5, we can find the length of BX by subtracting CX from BC: BX=BC-CX = 10-5=5.

3. Now, let's consider triangle ABX and triangle ACX. Given that the circumcircles of these triangles have the same radius, we can deduce that triangles ABX and ACX are similar triangles. 4. By the property of similar triangles, the ratio of the sides of the two triangles is equal: 5. Since AX = 10, CX = 5, and BC = 10, we can find AB using the ratio: AX/AB = CX/AC = BX/BC Solving for AB gives: AB=2xAX=2x10=20

6. Now that we have AB = 20 and BC = 10, we can use these side lengths to find the height of triangle ABC.
 

7. To find the height h, we can use the Pythagorean theorem: Solving for h gives:\(h=\sqrt{400-25}=\sqrt{375}=5\sqrt3\)

8. Finally, we can calculate the area of triangle ABC using the formula we mentioned earlier: Therefore, the area of triangle ABC is Area= 1/2 x BC x h= 1/2x10x\(5\sqrt3=25\sqrt3\)

Let's denote the side length of the tetrahedron as a.

Volume of Tetrahedron ABCD: The formula for the volume of a regular tetrahedron is V = (a^3 * √2) / (12). We are given that V = 18, so:

(a^3 * √2) / (12) = 18 a^3 * √2 = 216 a^3 = 216 / √2 (rationalize the denominator by multiplying top and bottom by √2) a^3 = 216√2 / 2 a^3 = 108√2

Volume of Pyramid EFGH: EFGH is a smaller tetrahedron similar to ABCD. The ratio of side lengths between similar figures is the same as the ratio of their corresponding edge lengths.

In this case, the edges of EFGH are half the length of the edges of ABCD because E, F, G, and H are the centers of the faces. Therefore, the side length of EFGH is a/2.

We can now find the volume of EFGH using the same formula for a tetrahedron, but with the side length a/2:

V_EFGH = [(a/2)^3 * √2] / (12) V_EFGH = (a^3 * √2) / (48)

Relating the Volumes: We know a^3 = 108√2 from part 1. Substitute this into the formula for V_EFGH:

V_EFGH = [(108√2) * √2] / (48)

V_EFGH = (108 * (√2)^2) / (48)

V_EFGH = (108 * 2) / (48)

V_EFGH = 216 / 48 = 9/2

Therefore, the volume of pyramid EFGH is 9/2.

Alright, not too bad of a problem!

So, we just have to set up the slope equation to find what a+b is equal to because spoiler alert, this is a really easy problem once we write out the equation. 

The slope formula is where the slope of a line is equal to \(\frac{x2-x1}{y2 - y1}\) where (x2, y2) and (x1, y1) are points on that line. 

In this case, we have point 1 as (a,a^2) and point 2 as (b,b^2). 

Plugging these numbers into the formula, we have \(\frac{ b^2 - a^2}{b - a} = 2\). Now, there's on trick that helps us solve thus problem immediately. That is, Mr. Good Old Difference of Squares. 

We can write the numerator as \((b-a)(b+a)\), meaning we can just cancel out the b - a. 

\(\frac{(b -a) (b + a)}{b -a} = b + a = 2\)

Boom, this means b + a = 2, so our final answer is simply 2! Simple enough!

Thanks!

Alright! I geuss I can try to solve this question!

First, let's note that the only points where the lines intersect is when we have \(f(x) = g(x)\). 

This means that we only have to solve the equation \(x^3 + x^2 -3x + 5 = x^3 + 2x^2 \) to find the points where the lines intersect. 

Combining like terms and moving all the terms to one side, we get the equation \(x^2 + 3x - 5 = 0\).

We can't factor this polynomial with integers, so the next best thing to do is to complete the square and find x that way.

We set up the equation by moving 5 to the other side, getting us \(x^2 + 3x = 5\). Now we complete the square by adding 9/4 to both sides. 

\(x^2 + 3x + 9/4 = 5 + 9/4 \)

\((x + 3/2)^2 = 29/4\)

Now we just have to square root both sides to get us \(x+3/2= \pm \sqrt{29/4}\)

This means that \(x = \frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}\).

Subsituting these x values back into the g(x) and f(x) functions, we get \(y = \frac{-\sqrt{29}-3} 2, \frac{\sqrt{29}-3}{2}\)

So for the final answer, we have \((\frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}), (-\frac{\sqrt{29}-3}{2}, \frac{\sqrt{29}-3}{2})\)

Thanks! :) :) :)

We know that f(3) = 4, so we want the inside of f(2x/3) to equal 3. We notice that when we have x=9/2, then we have \(y=\frac{5+f(3)}{11} = \frac{5+4}{11}\). 

This is easily solvable, so we get y=9/11. 

This means that the point (9/2, 9/11) has to be on the line. 

Thanks!

Find PM.

\(f_Q(x)=\sqrt{5^2-x^2}\\ f_R(x)=\sqrt{8^2-(x-11)^2}\\ 25-x_P^2=64-x_P^2+22x_P-121\\ x_P=\dfrac{121+25-64}{22}\\ x_P=3.\overline{72}\\ y_P=3.333\)

\(P(3.\overline{72},\ 3.333)\\ M(5.5,\ 0)\\ \overline{PM}=\sqrt{(x_M-x_P)^2+y_P^2}=\sqrt{(5.5-3.727)^2+3.333^2}\\ \color{blue}\overline{PM}=3.775\)

 !

Let's first notice that both triangle BPQ and triangle CQP are both right triangles. 

In order for the sin of angle PAQ, we need to find the length of side PQ and PA and then use the ratio for the sin.

Using the pythagorean thereom, we can find that the length of PQ is \(PQ^2 = BP^2 + PC^2 PQ^2 = 4^2 + 1^2 PQ^2 = 17 PQ = \sqrt{17} \)

Use the Pythagorean Theorem, we can find the length of PA is \(PA^2 = CQ^2 + QP^2 PA^2 = 4^2 + \sqrt{17}^2 PA^2 = 33 PA = \sqrt{33}\)

Now, we simply take the ratio of the two numbers getting us \(\frac{\sqrt{17}}{\sqrt{33}} = 17/33\)

So the sin of PAQ is simply \(\sqrt{\frac{17}{33}}\)!

Thanks!

Alrgiht! Not too bad of a problem!

First of all, if the polygon has 3 sides then there are 3 angles, pretty simply in that term. It also means that the sides add up to 180!

Another thing to point out is the fact that because the 3 angles make an arithmatic sequence, the difference between the smallest and the middle angle is the same as the difference between the biggest and the middle angles. Since the total is 56, we just do 56/2 = 28, so the common difference for angles ig 28.

Now, let's write an equation. 

Let s be the smallest angle. We have: \(d +(d+28) + (d+56) = 180\)

Combining like terms gets us \(3d + 84 = 180\), meaning \(3d = 96\) so 

\(d = 96/3 = 32\)

So the smallest angle is 32 

Thanks!

You can't really tell the exact sidelength or an integer for this problem, but you can tell what the sudelength is in terms of the area or perimeter. 

First off, we know that the perimeter of a regular hexagon is 6 times the sidelength, meaning the area is \(3/2 * 6 = 9 \) times the area. 

So the sidelenghth of the hexagon is 1/9 the area and 1/6 the perimeter!

I hope I answered your question!

Thanks!

Here's my take:

A baking club wants to form an executive committee.  There are $18$ people in the baking club, including Mark.  In how many ways can the baking club form an executive committee with $2$ people?

It's just \(\binom{18}{2}\)because you just have to choose \(2\) people out of the 18 candidates.

\(\binom{18}{2}\)= \(\boxed{153}\).

A baking club wants to form an executive committee.  There are $18$ people in the baking club, including Mark.  In how many ways can the baking club form an executive committee with $2$ people, including Mark?

Well, one of the spots is taken (by Mark), and there are 17 candidates left, with 1 spot left, so it's just \(\binom{17}{1}\), or \(\boxed{17}\).

A baking club wants to form an executive committee.  There are $18$ people in the baking club, including Mark.  In how many ways can the baking club form an executive committee with $2$ people, not including Mark?

For this one, because Mark is no longer a candidate that can be put on the committee, that would leave 17 candidates, and 2 need to be chosen, so it would just be \(\binom{17}{2}\), or \(\boxed{136}\)

What do you notice about your answers to all three parts?

The answers of the second and third parts add up to the answer of the first part. This makes perfect sense because the committee either has Mark on it, or it doesn't.

Hope this helps!