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Hello.

If: A+B=C, then we can make A=C-B or vice versa. Now:
(C-B)C = C^2-BC

That is your answer ;)

Hello.

A circle has no end, therefore this problem would be indeterminate. Biologically, the chicken would have developed first. This is because Gallus gallus domesticus evolved from a dinosaur family known as theropods, and the rest of genus Gallus gallus evolved from other close ancestors. Therefore, by genetic mutation, a chicken was developed and reproduced to have more favorable outcomes in offspring. Through natural selection alongside artificial selection by humans of these first chickens, the species evolved to what we know today. Therefore, biologically, the chicken developed first.

Hello. 

The answer is below.

Smart!

Checking cases, there is only one solution: x = -8:

For x = -8, |x| = 8

-1/2*x + 4 = (-1/2)*(-8) + 4 = 4 + 4 = 8

Therefore, x = -8 is the only solution.

Another point on the graph is (3,-7).

Hmmm... IDK.

It seems there was an error in the previous response where the solution was not complete. I apologize for that. Here is the complete solution:

Let (a, a^2) be the point of tangency shared by the two circles and the parabola. The first circle is tangent to the parabola at this point and has its center on the line y = a^2 + r. The second circle is tangent to the x-axis at the point (a - r, 0) and has its center on the line y = -r.

By distance formula, we know that the distance between the centers of the two circles is 2r, and the distance from each circle's center to the point of tangency is also r. Therefore, we have the following equation:

(2r)^2 = (a - (a^2 + r))^2 + (-r - a^2)^2

Solving for a yields:

a = (2r + 1)/2

Also, the equation of the tangent line to the parabola at (a, a^2) is y = -(x - a)^2 + a^2. Since this is a circle of radius r and center (a, a^2 + r), we can write:

(x - a)^2 + (y - a^2 - r)^2 = r^2

Substituting the expression for a we found earlier and solving for r, we get:

r = 1/2

Therefore, the radius of the circles is 1/2.

Thank you Melody...you really are there for me every time!!!...Stay blessed

Thanks, just solved it on my own

How would I prove that they are the only ones that exist? the theme of the week is being able to prove rigously and also to be able to prove it to an algebra 1 student.

This is easy.  Try starting with linear polynomials.

Using the areas of two triangles works.

Draw a right-angled triangle, vertical leg v, hypotenuse 1; the length of the base will be sqrt(1 - v^2) and the base angle will be arcsin(v).

Add the mirror image of that to form an isosceles triangle with sides 1, 1, 2v and say that the area of the second triangle (1/2)1.1.sin(2arcsin(v)) is double the area of the earlier triangle.

Having been told that you are not allowed to use a double angle theorem.

Somewhat quicker, at the third line of text,

\(1=a^{2}+ab+b^{2}, \\ 1-3ab=a^{2}-2ab+b^{2}=(a-b)^{2}=1,\\ 3ab=0, \\a=0 \text{ or }b=0,\text{ etc.}\)

(a) To find the possible values of ab, we can use the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2) and substitute a - b = 1 and a^3 - b^3 = 1 to get:

1 = (a - b)(a^2 + ab + b^2) = (1)(a^2 + ab + b^2) = a^2 + ab + b^2

Now, we can use the identity (a + b)^2 = a^2 + 2ab + b^2 and substitute a - b = 1 to get:

(a + b)^2 = (a - b)^2 + 4ab = 1 + 4ab

So, ab = ((a + b)^2 - 1)/4. Therefore, ab can take any value greater than or equal to -1/4.

(b) To find the possible values of a+b, we can add a - b = 1 to a^3 - b^3 = 1 to get:

a^3 + a - b^3 - b = 2a

Now, we can use the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2) and substitute a - b = 1 to get:

a^2 + ab + b^2 = 1

So, a+b = 2a - 1 = 2b + 1. Therefore, a+b can take any value.

the meaning of your question is not clear.

If I understand the question correctly

10C5  *  0.7^5 * 0.3^5 = approx 10.29 %

Post one question at a time !  I will do (a) only and you do the rest so that you can learn:

We can use a combinatorial argument to prove this identity.

Suppose we want to choose a committee of k people from a group of n individuals. We can count the number of ways to do this in two different ways.

First, we can choose one person to be the leader of the committee, and then choose the remaining k - 1 members from the n - 1 remaining individuals. There are C(n - 1, k - 1) ways to do this.

Second, we can simply choose k members from the n individuals without designating one as the leader. There are C(n, k) ways to do this.

Since both methods count the same thing (the number of ways to choose a committee of k people from a group of n individuals), we have:

C(n - 1, k - 1) + C(n - 1, k) = C(n, k)

Therefore, we have proven the identity C(n - 1, k - 1) + C(n - 1, k) = C(n, k) using a combinatorial argument.

It is easiest if it is of the form

\(f(x)=Acos[b(x-p)]+M\)

A is the amplitude (if neg then it is upside down)

2pi/b  is the period

y=M is the midline

p is the phase schift

Now I think of it in bits, sketch it as you go. So that you build up the graph step by step

consider

\(1) y=cos(x)\\ 2) y=-cos(x)\\ 3) \text{make the numbers on the x axis fit with a period of } 2\pi/b\\ 4) \text{drop or raise the graph so the centre line is }y=M\\ \\5) \text{Change the amplitude by making the bottom of the wave M-A and the top M+A }\\ 6) \text{Shift the phase by p units to the right}\), 

f(x)=-4cos(2/3x+pi/3)-3 

We can solve this problem using the binomial distribution. Let X be the number of new players who stick with the new gaming mechanic, and let p be the probability that a new player sticks with the new gaming mechanic. Since 70% of all other gaming developers report that the new gaming mechanic will bring in a new surge of players, we have:

p = 0.7

We want to find the probability that five new players stick with the new gaming mechanic, so we have:

X ~ Binomial(n = 10, p = 0.7)

P(X = 5) = (10 choose 5) * (0.7)^5 * (1 - 0.7)^(10 - 5)

Using the formula for the binomial coefficient, we can simplify the expression as:

P(X = 5) = (10! / (5! * (10 - 5)!)) * (0.7)^5 * (0.3)^5

Simplifying further, we get:

P(X = 5) = 252 * 0.16807 * 0.00243

P(X = 5) = 0.102

Therefore, the probability that five new players will stick with the new gaming mechanic is approximately 0.102 or 10.2%.

Let's start with the expression sin(2arcsin[v]). We know that:

sin(2θ) = 2sin(θ)cos(θ)

So, we can rewrite our expression as:

sin(2arcsin[v]) = 2sin(arcsin[v])cos(arcsin[v])

Using the fact that sin(arcsin[x]) = x and cos(arcsin[x]) = sqrt(1 - x^2), we can simplify further:

sin(2arcsin[v]) = 2v*sqrt(1 - v^2)

Therefore, sin(2arcsin[v]) simplifies to 2v*sqrt(1 - v^2) in terms of v.

Let the base be x.  Then each leg is 1.5x, so 1.5*x + 1.5*x + x = 63.

This becomes 4x = 63, so x = 63/4.

The length of the base is 63/4.