hectictar

Also on the calculator you can click x^y . It is on the second row from the top, fourth column from the left.

I've been thinking about this some more.

\(3a^2 = b^2\)

The question is, "Are there any integer values of a and b that make this equation true?"

Zero works but I want to find something besides zero.

I can rewrite it like this:

\(b = \sqrt{3a^2} \\ b = (\sqrt{3})a\)

b needs to be an integer, so what integer for a can make b an integer?

Well since √3 is irrational, the only possible thing that, when multiplied by √3 comes out with something rational will be √3.

√3 times any integer will just be that integer times √3.

You could multiply by √(27), but all that is doing is multiplying by (√3√3√3). There must be an odd number of √3's in a to cancel out the first √3. And if there's an odd number of √3's in a, then a is irrational too and not an integer.

I think it is impossible with anything besides 0.

What is the value of sin(A) ?

\(\sqrt{3(0^2)} = 0\)

That probably doesnt help any ahaha...

The minimum selling price will be 100% of the original price plus 30% of the original price.

The minimum selling price will be 130% of the original price.

130% of $5.62 = (1.30)($5.62) = $7.306 ≈ $7.31 per book

\(\frac{x}{225}=\frac{13}{100} \\~\\ x = \frac{13(225)}{100} \\~\\ x = \frac{2925}{100} \\~\\ x = 29.25 \%\)

The correct answer should be 7.

Multiplication comes before addition.

A negative times a negative is a positive.

-2 * -2 + 3

4 + 3 = 7

I don't know what the other calculator is doing. Maybe it's thinking the negative signs are subtraction signs and it completely ignores the multiplication.

-2 - 2 + 3

-4 + 3 = -1

y - 10 = (4/3)(x-6)

Get it into y = mx + b form.

y - 10 = (4/3)x - (4/3)6

y - 10 = (4/3)x - 8

y = (4/3)x + 2

When it is in this form, the x coefficient is the slope.

4/3 is the slope.

Oh no! I wouldn't worry about that haha, there are lot of math things that I don't understand. :)

Test the points in each option to determine which one fits the data.

Start by testing the first point, x = 1 and y = 3.

A. 3 =? 2(1)

    3 ≠ 2 , so option A is out.

B. 3 =? 4(1²)

    3 ≠ 4 , so option B is out.

C. 3 =? 2(1²) + 1

     3 = 3 , so option C goes to the next round.

D . 3 =? 3(1²)

     3 = 3 , so option D goes to the next round.

Test the second point, x=2 and y = 12.

C. 12 =? 2(2²) + 1

     12 ≠ 9 , so option C is out.

D. 12 =? 3(2²)

     12 = 12 , option D goes to the next round.

Option D is the only remaining choice. Quickly test the other points for D:

27 =? 3(3²)

27 = 27

48 =? 3(4²)

48 = 48

y = 3x² fits all the data.

f(x) = 3x²