Question

Helppp plssss I’ll mark brainliest

Answer 1

Answer:

1) $b^{6}$

3) $\frac{a^2}{c^{3} }$

Step-by-step explanation:

I'm only going to give you the answer for number 1 and 3, because this is a very simple concept.

When dividing real numbers with exponents, variables or not, the exponents can be subtracted from one another as long as the base is the same.

So the answer to number one is $b^6$, because in $\frac{b^9}{b^3}$, the base is the same $b$, so we can just subtract the exponent in the denominator from the exponent in the numerator. $9-3=6$

In number 3's case, we have 2 expressions divided by each other. But notice, there are similarities in the bases. There is an $a$ to some power in the numerator and the denominator, and the same with $c$. The exponents can be subtracted yet again.

Let's look at $a$ first. $4-2=2$, so the simplified expression with $a$ will be to the 2nd power. Now let's look at $c$. $3-6=-3$. So the final, simplified expression witn $c$ will be to the negative 3rd power. So we have $a^2c^{-3}$. Since we have a negative exponent, you can apply the rule $x^{-n}=\frac{1}{x^{n} }$.

Your final answer is $a^2*\frac{1}{c^{3} } =\frac{a^2}{c^{3} }$

Try using this for the rest of the problems. If you have any questions, please comment and I'll try to help.

Answer 2

1. b^ 7-3 = b^4
2. b ^6-7 = b^ -1 = 1/b^1
3. a^ 4-2 c ^ 3-6 = a^2 c ^-3 = a^2 * 1:c^3
4. 1/4^3-1 = 1/4 ^ 2
5. 6y ^3 z^3/ 2y ^5 = 3y^3z^2/y^5 = 3y ^3-5 z^ 5 = 3y^-2 z^5= 1/3y^2 z^5
6. F^3-1 g^2-5 = f^3 * 1/g^3