# Tutor profile: Emma H.

## Questions

### Subject: Linear Algebra

Show that the vectors $$(1, 0)$$ and $$(1, 1)$$ form a basis of $$R^2$$

To show that vectors form a basis of $$R^2$$, we have to show that they span $$R^2$$ and that they are linearly independent. First, let's show that they're linearly independent. We want to show for any numbers a and b, if $$a(1,0) + b(1,1) = (0,0)$$ then both a and b are equal to 0. $$a(1,0)+b(1,1) = (a,0) + (b, b) = (a+b, b)$$, so if $$(a+b, b) = (0,0)$$ then $$a+b = 0$$ and $$b = 0$$. Now we already have that $$b = 0$$, let's substitute that into $$a+b = 0$$ So $$a + 0 = 0$$, so $$a = 0$$. This shows that $$(1, 0)$$ and $$(1,1)$$ are linearly independent. Now we want to show that $$(1,0)$$ and $$(1,1)$$ span $$R^2$$. This means that we want to be able to write any vector (x, y) in $$R^2$$ as a sum of multiples of $$(1, 0)$$ and $$(1,1)$$. Let's write $$a(1,0) + b(1,1) = (x, y)$$. We'd like to show that there is a way to find a and b for any x and y we choose. $$(a,0) + (b, b) = (a+b, b) = (x, y)$$. This means that $$b = y$$ and $$a+b = a+y = x$$. Moving terms around we get $$b = y$$ and $$a = x-y$$. Now we have a way to write any vector $$(x, y)$$ as a sum of multiples of $$(1,0)$$ and $$(1,1)$$ so we have shown that they span $$R^2$$. We've shown that they span, and that they're linearly independent, so this shows that they're a basis.

### Subject: Pre-Calculus

Find the roots of the function $$f(x) = 2x^2+7x+6$$

We want to find what values of x will make $$f(x) = 0$$. We'll do this by factoring the equation. Because there's a number in front of the $$x^2$$ term we can't look for product and sum like we normally would. Instead, we'll try to split up the middle term, 7x, in a clever way: $$2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6$$. We can add brackets to make it easier to see that we've split up the equation and because it's all addition it doesn't change anything. $$(2x^2+4x) + (3x + 6)$$ Now we want to look and see if there's anything we can factor out of these two brackets. In the first bracket, we can factor out an x, or a 2, or both. We want to make what's inside the brackets as simple as possible, so let's factor out both. $$(2x^2+4x) = 2x(x + 2)$$ In the second bracket, the only thing we can factor out is 3. $$(3x+6) = 3(x+2)$$ So now we have $$2x(x+2) + 3(x+2)$$. This is great, because now we have the same thing in both of the sets of brackets. That means we can factor (x+2) out of the whole expression! This leaves us with: $$(x+2)(2x+3)$$, and our equation is factored. Now that our equation is factored, we can set each of the factors to be equal to zero to solve for the values of x. $$x+2 = 0$$ gives us $$x = -2$$ and $$3x+2 = 0$$ gives us $$x = -\frac{2}{3}$$. These are the roots of $$f(x)$$

### Subject: Pre-Algebra

Simplify the following expression: $$(-4)*(-12) - (3)*\frac{-1}{2}$$

First, let's simplify the two terms in the expression. $$(-4)*(-12) = 4 * 12$$ because multiplying two negatives gives a positive, and $$4 * 12 = 48$$ $$(3)*\frac{-1}{2} = \frac{3}{1} * \frac{-1}{2} = \frac{-3}{2}$$ Now we can substitute these simplified terms into our expression: $$48 - (\frac{-3}{2})$$ -3/2 is negative, and subtracting a negative number is the same as adding the positive number, so let's change that: $$48 + \frac{3}{2}$$ Now we have a much simple addition question :) We can write 48 as a fraction with 2 as the denominator to add them: $$\frac{2*48}{2} + \frac{3}{2} = \frac{96}{2} + \frac{3}{2} = \frac{99}{2}$$ And now we have just one number so we're done.

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