# Tutor profile: Natalia M.

## Questions

### Subject: Trigonometry

In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).

Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse. tan(A) = opposite side / adjacent side = a/b = 3/4 This implies that a = 3k and b = 4k , where k is a coefficient of proportionality. Now we can use Pythagora's Theorem to find h which gives us: h^2 = (3k)^2 + (5k)^2. We can now solve for h: h=5k. Now that we know the a, b and h values for this triangle we can plug these into the following formulas for sin and cos: sin(A) = a / h and cos(A) = 4k / 5k. This implies that sin(A)=3/5 and cos(A)=4/5.

### Subject: Calculus

Assume that y is a function of x . Find y' = dy/dx for x^3 + y^3 = 4 .

We will first want to differentiate both sides of the equation to get: 3(x^2) + (3y^2)y' = 0. In order to solve for y', we will need to first subtract 3(x^2) from both sides of the equation to get: (3y^2)y'= -3(x^2). We can finish solving for y' by dividing both sides by (3y^2) to get our final answer: y'= (-x^2)/y^2

### Subject: Algebra

For what value of the constant K does the equation 2*x + K*x = 3 have one solution?.

The first step is to solve the above equation for the variable x. In order to do this, we will first factor out the variable x from the left-hand side of the equation. We are left with the following equation: x*(2+K)=3. We now solve for x by dividing both sides of the equation by (2+K) to get: x=3/(2+K). This equation implies that the given equation has one solution for all real values of K which are not equal to -2.

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