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# Tutor profile: Dylan F.

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Dylan F.
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## Questions

### Subject:Computer Science (General)

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Question:

Write a method called allLess that accepts two arrays of integers and returns true if each element in the first array is less than the element at the same index in the second array. Your method should return false if the arrays are not the same length. For example, if the two arrays passed are {45, 20, 300} and {50, 41, 600}, your method should return true. If the arrays are not the same length, you should always return false.

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Dylan F.

//needs to be static because it's not acting on an object, //return a boolean (true/false), and accept two integers as arguments. public static boolean allLess(int[] arr1, int[] arr2){     if(arr1.length != arr2.length){ //first check to see that the lengths match up         return false; //if they don't, we can just return false and exit the method     }     boolean isLess = true; //otherwise, default to true     for(int i = 0; i<arr1.length;i++){ //loop through every index         if(arr1[i]>arr2[i]) //if the arr2 value is ever less,             isLess = false; //override that true to be false     }     return isLess; //if it made it all the way through as true, it'll return true, //but if arr2 value was ever too small, it'll return false }

### Subject:Calculus

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Question:

Find the volume of the solid formed by rotating the area of the region in the 1st quadrant between y=x² and x=2 around the y-axis.

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Dylan F.

One can consider the rotated segment as a stacking of infinitely thin cylinders on top of each other. To set up an integral for this, consider the formula for cylinder volume, πr²h. In this example, rotating about the y-axis, our radius is the distance from the y-axis and the height will be dy, an infinitesimal vertical distance. In total, that vertical distance will range from 0, the bottom of Quadrant I, to y=4, the intersection of the two lines. To find the total volume, without the gap that y=x² will leave, the integral would be π∫(2²) dy, with 2 being the radius and dy being the height in the cylinder equation. To factor in the missing area, we need to subtract the volume that y=x² would have. Plugging into our cylinder equation as it is yields π∫(x²)² dy—but wait, the integral needs to be in terms of y! We need to rearrange y=x² to √y=x for it to work, giving π∫(√y)²dy (again, √y is our radius and dy is our height), or just π∫y dy. Putting it all together, what we need to solve is π∫(2²) dy - π∫y dy, or π∫(4-y) dy after factoring out the π and consolidating integrals. Keep in mind that this is all still on the interval of 0 to 4. The antiderivative is fairly simple, just 4y-.5y², and using the fundamental theorem of calculus, we get 4(4)-.5(4)² - 4(0)-.5(0)², which simplifies to 16-8, or 8. Multiplying by the π gives us the final answer of 8π.

### Subject:ACT

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Question:

What is the perimeter of an isosceles right triangle whose hypotenuse is 8√2 inches long?

Inactive
Dylan F.

An isosceles triangle's base angles are, by the base angle theorem, congruent. As a right triangle, the angle across from the hypotenuse must be 90º. Since all triangles' angles must sum to 180º, the other two base angles have to form the remaining 90º (the total 180º - the right angle's 90º = 90º remaining). Since the two base angles are congruent, both of them must be 45º (90/2). Now that we know all the angles: 45, 45, and 90, we can find the side lengths. One property of 45-45-90 triangles is that the hypotenuse is √2 times the leg length, so given that the hypotenuse is 8√2, the legs are both 8 inches long, giving a perimeter of 16+8√2. If the √2 property of 45-45-90 triangles is hard to remember, though, we can work past it with the pythagorean theorem, a²+b²=c². In this example, with an isosceles triangle, a=b; the legs are of equal length. For simplicity's sake, I'll replace them both with "x," changing our equation to 2x²=c². c, the hypotenuse, is 8√2, so substituting that in gives us 2x²=(8√2)². From here, we can go ahead and work out (8√2)², which is the same as (8)²(√2)², or 64*2. Now, to isolate x, divide both sides by 2, giving x²=64. Finally, square root both sides to find that the legs' length is 8 inches. Two legs plus the hypotenuse gives us the same answer as above, 16+8√2.

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