Question

More math that I suck at!

Answer 1

You can apply the same rule we used last time.

\[3x^0 = 3 \times 1\]

Answer 2

3

Answer 3

Yes, now comes the more tricky part.

We have:
\[2y^{\frac{ -4 }{ 3}} \times y^3\]

Answer 4

This is like terms since both of these numbers just have the variable y in them. We can combine them. Since these exponents are being multiplied by each other, we need to remember the rules:

\[x^a \times x^b = x^{a + b}\]

Answer 5

i still don't understand.

Answer 6

Basically focus on the exponents. We're multiplying these two y numbers right? They have exponents, we must add the two exponents together, so we get:
\[2y^{ - \frac{ 4}{3 }+ 3}\]

Answer 7

Does that make sense?

Answer 8

kind of

Answer 9

\[x^a \times y^b = xy^{a + b}\]

When adding with frictions, you need to convert the number you're adding with into a fraction that has the same denominator as the fraction. So we get:

\[\frac{ -4 }{3 }+ \frac{ 9 }{3 } = ?\]

Since:

\[\frac{ 9 }{ 3 } = 3 \]

Answer 10

so it would be D

Answer 11

yes