# Tutor profile: Ethan R.

## Questions

### Subject: Geometry

A right triangle has an angle $$ 30^{\circ}$$. The side opposite to this angle, the short leg of the triangle, is BC = 2. Solve for the hypotenuse, x.

Knowing the relationship of the sides in special case 30, 60, 90 right triangle we can determine the value of the hypotenuse. The hypotenuse in this type of triangle is equal to 2 times the short leg, the side opposing the $$ 30^{\circ}$$ angle. Thus the hypotonuse is simply 2x2=4.

### Subject: Basic Chemistry

Using the molecular weight of glucose, $$C_{6}H_{12}O_{6}$$ is 180.16 $$\frac{g}{mol}$$ How many moles of glucose are in 19.1 g of glucose. Use 3 significant figures.

Using the given information we know that the quantity of glucose, 19.1 g, is a fraction of the molecular weight. To solve simply multiply the 19.1 g of glucose by the inverse of the molecular weight. $$ (19.1g) (\frac{1mol}{180.16g})$$ This gives us the appropriate units and the answer of 0.106 mol.

### Subject: Algebra

Simplify the expression $$ \frac{x^{2}+10x+25}{x+ 5} $$

Start by factoring the top half of the equation. $$ {x^{2}+10x+25} $$ becomes $$ ({x+5})(x+5) $$ The problem then becomes $$ \frac{({x+5})(x+5)}{x+ 5} $$ Simplify by division to $$ x+5 $$

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