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At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 8 mm, Theo's donut holes have radius 10 mm, and Akshaj's donut holes have radius 12 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?

Hello Guest!

\(\frac{N\cdot Parts}{8^2mm^2}+\frac{T\cdot Parts}{10^2mm^2}+\frac{A\cdot Parts}{12^2mm^2}\\ =\frac{N}{64}+\frac{T}{100}+\frac{A}{144}\\ =\frac{N}{16}+\frac{T}{25}+\frac{A}{36}\\ = \frac{900N+576T+400A}{14400}\cdot \frac{Parts}{mm^2}\)

In the period in which Akshaj 400 and Theo 576 coated donut holes,

Niraek coated 900 donut holes.

  !

Find all possible values of cos(x) if
sin(2x)=2tan(x) .

Hello Guest!

\(sin(2x)=2sin(x)cos(x)\\ \color{BrickRed}sin(2x)=2tan(x)\\ sin(x)cos(x)=tan(x)\\ tan(x)=\frac{sin(x)}{cos(x)}\\ sin(x)cos(x)=\frac{sin(x)}{cos(x)}\\ cos^2(x)=1 \)

\(cos(x)=\pm \sqrt{1}\)

\(cos_1(x)=1\\ \color{black}x_1=0\\ cos_2(x)=-1\\ \color{black}x_2=\pi\)

\(sin(2x)=2tan(x)\\ sin(2\cdot 0)=2\cdot tan(0)\\ 0=0\)

\(sin(2x)=2tan(x)\\ sin(2\cdot \pi)=2\cdot tan(\pi)\\ 0=0\)

  !

Sally has a cube of side length  units such that the number of square units in the surface area of the cube equals 1/10 of the number of cubic units in the volume. (condition1)

She also wants to make a square for which the number of square units in the area of the square is equal to the number of cubic units in the volume of the cube. (condition2)

What should the side length of the square be?

Hello Sally!

condition1

\(\color{blue}6a^2=\frac{1}{10}a^3\\ a^3-60a^2=0\\ a^2(a-60=0) \)

\(a=60\)      

The cube has a side length of 60 side length units.

condition2

\(x^2=y^3\)

\(2log(x)=3log(y)\\ log(y)=\frac{2}{3}log(x)\\ {\color{blue}y=10^{\frac{2}{3}log(x)}}\ |\ \{x,y\} \subset \mathbb N\)

x : 1, 8, 64, 1000 \( side\ length\ units\)     square

y : 1, 4, 16,   100 \( side\ length\ units\)     cube

The side length of the square is \(\{1, 8, 64, 1000\}\ side\ length\ units.\)

Condition1 and condition2 are not compatible.

The measure of one of the smaller base angles of an isosceles trapezoid is 60 degrees. The shorter base is 5 inches long and the altitude is 3 sqrt 3 inches long. What is the number of inches in the perimeter of the trapezoid?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The length of the longer base is    2(3√3 / √3) + 5 = 11

The length of the leg(s) is              sqrt[(3√3)² + 3³] = 6

The perimeter is                        P = 5 + 11 + 2*6 = 28 inches ☺

There are three cases:

1 ace 1 king: 2*1/13*4/51=8/663

1 ace 1 other: 2*1/13*47/51=94/221

1 king 1 other: 2*1/13*47/51=94/221

so, adding the probabilities up, we have 8/663+282/663+282/663=572/663=44/51.

.01 x  $ 275000 = $............

oh wait im an idiot. I completely forgot about that, thanks for your help

Melody hasn't done anything to you....

Hello GamemasterX40!  This is my attempt of solving this problem:

"Ms. Thayer bought 6 items for her physics class at a mean cost of 8.40 dollars per item. When she added a dial caliper for 46.50 dollars to her purchases, the new mean for all 7 items became about... Explain your thinking."

SOLUTION (and explanation):

We know that the mean is the average of a set of numbers. With this information in mind, we know that the total cost of the 6 items Ms Thayer bought for her class was: \[6 \times 8.40 = 50.40 \text{ dollars.}\] Now, Ms. Thayer bought a 7th item, the dial caliper, for 46.50 dollars. Adding this to our total, we get $50.40+46.50 = 96.90 \text{ dollars.}$ Again, we knot that the mean is the average of a set of numbers. That means the mean, or in the context of the problem, the average price, of the 7 objects is: \[\frac{96.90}{7}=13.8428571429... \approx 13.84 \text{ dollars.}\]

I hope this helped! 

No problem! Always happy to help :D

I like this one better. It's faster and easier to do. Thank you!

I usually prefer using the vertex method, where all you have to do is rearrange the equation.

But this is nice! This person will now know two methods to do these problems! 

You are correct....i do not know how I used 6 c 2 and got 20 !   it is 15

All right... Here is my attempt:

We need to rewrite this equation in vertex form. First, we isolate the $x$ to give $x = y^2 - 8y + 13$. Completing the square in $y$, we get \[x = (y - 4)^2 - 3.\]

The graph of this equation is a rightward-opening parabola, so the vertex is the leftmost point, not the top or bottom point as usual.


Since $(y-4)^2$ is nonnegative, the value of $x$ is smallest when $y=4$, which makes $x=-3$. Therefore, the vertex of the parabola is $\boxed{(-3,4)}$.

Hope this helped! 

x = y^2 -8y+13   is a sideways parabola

    the vertex y coordinate   is  - b/2a    where b = - 8 a = 1

       use this value in the equation to calculate the x coordinate of the vertex

6 C 2 =15 combinations as follows:

{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {3, 4} | {3, 5} | {3, 6} | {4, 5} | {4, 6} | {5, 6} (total: 15)

Probability = 8 / 15

x is the answer to your problem.....how many pounds of seeds  are required for the entire garden

standard form   f(x) =  a x^2 + b x   + c 

  you can solve it by the Quadratic Formula :

         \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

or if it is 'easy'   you can solve it by factoring

   example    x^2 + 6x +5 = 0

    factors to    (x+3)(x+2)  = 0       then x = -3 or -2  

20 possible selections

1 will divide into 5 selections

2 will divide into 2 selections (4 and 6)

3 will divide into 1 selection (6)

so   8 out of 20     8/20 = 2/5