Ziggy

You too!!  

If a has 3 factors, then a will be the square of a prime number (it's factors are 1, a prime number, and the square of the prime number). The smallest square of a prime number is 2^2 = 4. Therefore, if b is divisible by 4 and has 4 factors, than the least possible value of b is 8.

Since we are told that the polynomial x^2+bx+c has exactly one real root, we know that the discriminant (b^2-4ac) is equal to 0. Thus, plugging in a = 1 and b = c-2 into the equation, we get c^2 - 8c + 4 = 0. I think this will give us c = 4 +- 2sqrt(3), so the product of all possible values of c is [4+2sqrt(3)][4-2sqrt(3)] = 16 - 12 = 4.

Here's my solution:

We know that the points that he labels with the number 25 has a distance of sqrt(25) = 5 from the origin. Thus, using the distance formula and calling the points that satisfy the condition (x, y), we have:

sqrt(x^2+y^2) = 5. Sqaring both sides gives us x^2+y^2 = 25. See if you can solve it from there! 

Hint: We have to find the number of points (x, y) that satisfy the equation x^2+y^2 = 25. We also have to remember that x and y can be negative numbers too, but they all have to be real integers. Listing the possibilities out, we get

(x, y) = (0, 5), (5, 0), (0, -5), (-5, 0), (3, 4), (-3, 4), (3, -4), (-3, -4), (4, 3), (-4, 3), (4, -3), and (-4, -3). How many pairs is that? 

12 times (I did this problem yesterday lol)

Sure, I can try! First, we can call the side length of the square "x". The area of this orginal square would be x^2. Next, if the length of the side of the square was doubled, then one side would have a length of 2*x = 2x. The area of this new sqaure would be (2x)^2 = 4x^2. Therefore, the ratio of the areas of the original sqaure and the area of the new sqaure is x^2 / 4x^2 = 1/4, or 1:4. I hope that helped! 

3.1536 × 10^6

First, we cross multiply for all three equations to get 

a^2 = b, b^2 = 2c, and c^2 = 4a. Thus, we know that a = sqrt(b) from the first equation and b = sqrt(2c) from the second. Next, using the last equation, we find that 

c^2 = 4a ---> c^2 = 4sqrt(b). We already found that b = sqrt(2c), so we can substitute it in to get c^2 = 4*sqrt(sqrt(2c)) = 4*∜2c. Further simplifying it, we get c^8 = 512c ---> c^7 = 512, so c equals the seventh root of 512. 

Lastly, we plug this value of c into the third equation to obtain a = about 1.48599429.

The expression (4k + 8)/4 can be written as 4k/4 + 8/4 = 1k + 2 = k + 2. Therefore, a = 1 and b = 2, so a/b = 1/2.  

The formula for the surface area of a sphere is given by 4pi*r^2, where r is the radius. We also know that the diameter of the sphere is 12 units, so its radius is 12/2 = 6 units. Thus, we can conclude that the surface area of that sphere is 4pi*36, which is about 452 units, or B. 

Here's a hint on how to do the problem: 

First, plug in the number 6 into f(6) and then use a similar technique to solve for f(10) and f(15) (Plug in 10 and 15 into f(x) to find f(10) and f(15)). Then, plug in the values of f(6), f(10), and f(15) into S and compute S. Lastly, divide S by 30 and you should get the answer.  

Note: I believe that you can probably do this problem with a little help, so I'm not going to write down the answer.

Here's another solution:

3^x = 5 can be written as log₃(5) = x and 5^y = 9 can be written as 2*log₅(3) = y. Thus, xy is just log₃(5)*2*log₅(3) = 2*1 = 2. 

The second graph is shifted three units to the right from the graph of y = f(x), so the equation of the second graph would be y = f(x - 3).

We can simplify the problem by doing 4*9*3*6 = 648, so the answer is 8.

I guess it's because marshlands will temporarily store water and then slowly release it so that there won't be too much water at once in a nearby town. Also, this isn't the kind of place to ask science questions.

Yay! I made an account!

Hi! I'm not sure about the answer, but I tried to do your problem.  Here's my solution:

First, take the square root of both sides to get +- x^2 = 2i. Then, you will get two cases:

Case #1; x^2 = 2i: We take the square root of both sides of the equation x^2 = 2i to get +- x = 1 + i. Thus, x can be either 1 + i or -1 - i in this case.

Case #2; -x^2 = 2i: We move the terms in the equation around to get x^2 = -2i. Then, taking the square root of both sides, we get

+-x = 1 - i. Therefore, this case yields the solutions x = 1 - i and x = -1 + i.

In the end, all of the complex numbers z such that z^4 = -4 are 1 + i, -1 -i, 1 - i, and -1 + i.