Hello there!

I’m not sure if they want me to add the answers together or just do them separately.

Exponents
(-2)^2=(-2)(-2)=4
(7/6)^2=7/6(7/6)=49/36

I hope this helps!
Best wishes.
-HuronGirl

3 0

3 years ago

HEEELLLPPP!!!!!!!!! PLEEAASEEE!!!!!!!!

Ksivusya [100]

Answer:

Step-by-step explanation:

$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$

4 0

2 years ago

Consider the following system:

Goryan [66]

You did not include the choices. However, I answered one that just included them. I've included the possible answers below and then the correct answers.

<span>A multiple of Equation 1.
B. The sum of Equation 1 and Equation 2
C. An equation that replaces only the coefficient of x with the sum of the coefficients of x in Equation 1 and Equation 2.
D. An equation that replaces only the coefficient of y with the sum of the coefficients of y in Equation 1 and Equation 2.
E. The sum of a multiple of Equation 1 and Equation 2.

</span>A, B and E.

Adding and multiplying the terms allow them to keep working. However, you must make sure that each variable is changed each time. Not just one as in C and D.

4 0

3 years ago

Read 2 more answers

If 2, x,18 and 54 are in proportion then what is the value of x.​

If 2, x,18 and 54 are in proportion then what is the value of x.