# Tutor profile: Dennis W.

## Questions

### Subject: SAT II Mathematics Level 2

Example SAT II Math Level 2 Question: A sphere with diameter 50 cm intersects a plane 14 cm from the center of the sphere. What is the number of square centimeters in the area of the circle formed? (A) 49π (B) 196π (C) 429π (D) 576π (E) 2304π

1. Focus on the actual question: what is the area of the circle formed? Think: 1. Area of circle $$ = \pi r ^2$$, the only variable is $$r$$, so let's reduce our answers to the radius! The answers all have $$\pi$$ and some large numbers, let's convert them into just radius for easy comparison (remove $$\pi$$ and take rough square root): A) 7 B) 14 C) 20-ish D) 24 E) 40 to 50 2. This is multiple-choice, let's eliminate obvious wrong answers without requiring additional formulas: Radius of the biggest sphere cross-section $$ =50/2 = 25$$, so E) is out. A plane that is 14 cm off center is about half-way along the 25 cm sphere radius, so A) 7 would be too small, and D) 24 would be too large. B) 14 is too exact, is the obvious trick answer since "14" is part of the question, and also as you follow radius $$r_{sphere}$$ of the sphere out from the center towards the surface, the radius of the formed circle $$r_{circle}$$ at each point is always larger than $$r_{sphere}$$. So the answer is C)! 3. Of course, you can also answer the question/verify your answer using more exact calculations instead of elimination: Draw a quarter circle with radius 25 cm (representing the sphere). Make the straight edges the x- and y-axis. At the 14 cm mark on the x-axis, draw a vertical line up to meet the circle arc. This line is the $$r_{circle}$$ you need to find. Now draw a line from the origin to that intersection point. You are essentially solving for one leg of a triangle where the hypotenuse is $$r_{sphere} = 25$$ cm, and the other leg is 14 cm. Pythagorean theorem: $$r_{circle}^2 + 14^2 = 25^2$$ Again, looking at your answers, A) 7 and B) 14 are too small, D) 24 and E) 40-50 are too large. C) is your goldilocks. $$25^2-14^2 = 429$$ which is the original C) answer!

### Subject: Microsoft Suite

In Powerpoint, can I create easy animations that make my presentation pop without spending a lot of time?

Yes, absolutely! There are plenty of default functionalities available that can make presentations pop! 1. If you have a newer version of Powerpoint, look at the "Transitions" tab and select the "Morph" icon. This allows Powerpoint to figure out automatically how to animate similar objects between slides. A one-click solution! 2. Instead of animating, simply build your slides with one line of content at a time. This way, when you present it, you are creating "frames" with each line of content appearing sequentially. As a bonus, this kind of presentation can be saved as a PDF and you can just step through the PDF pages without losing the animation effects! 3. Powerpoint also has a "Zoom" tool in the "Insert" tab. This allows you to insert automatic links within your presentation to other slides! Using this, you can create a Summary slide where you can link to multiple other slides in your presentation to make it extremely professional.

### Subject: MATLAB

Create a simple graphical-user-interface (GUI) program to blur out specific areas of an image in Matlab!

%% Demo GUI to blur out selected areas in image % Xiaotian (Dennis) Wu close all clear %% Let user select image [image_file, image_folder] = uigetfile('*.jpg'); %% Load in this image im = imread([image_folder,image_file]); %% Display this image figure; imshow(im); %% Allow user to draw a region to blur title('Draw your region to blur!') h = drawfreehand; %% Get the area that user drew mask = h.createMask; mask = repmat(mask,[1 1 size(im,3)]); %duplicate mask however many color planes your image has! %% Blur a copy of the image blur_value = 100; %can change how much you want to blur! imcopy = imfilter(im, ones(blur_value)/blur_value^2); %% Replace the area of the original image with the blurred image part im(mask) = imcopy(mask); %% Display your results! figure; imshow(im); title('Blurred result!')