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enot [183]
2 years ago
9

PLEASE HELP!!!!!!!!!!!!

Mathematics
2 answers:
tresset_1 [31]2 years ago
8 0
Answer= 21x+14
7(3x)+ 7(2)= 21x+14
Igoryamba2 years ago
3 0
<h3>Answer:  21x+14  (choice A)</h3>

Explanation:

Use the distribution rule to multiply the outer 7 by each term inside

  • 7 times 3x = 21x
  • 7 times 2 = 14

We go from 7(3x+2) to 21x+14

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Solve. (K+1)(k-5)=0<br> Explin how to solve ???
Anna71 [15]
Alright remember, if any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0. 
k+1=0
k-5=0
Set the first factor equal to 0 and solve
k=-1
Set the next factor equal to 0 and solve
k=5
The final solution is all the values that make (k+1)(k-5)=0 true. 
k=-1, 5

Hope this helped you out :)
5 0
3 years ago
Read 2 more answers
Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)
Leni [432]
Alrighty

remember some rules
(ab)^c=(a^c)(b^c)
and
x^{-m}=\frac{1}{x^m}
and
x^0=1 for all real values of x
and
(a^b)^c=a^{bc}
and
(\frac{a}{b})^c=\frac{a^c}{b^c}
and
(a/b)/(c/d)=(ad)/(bc)
and
(a^b)(a^c)=a^{b+c}
and
\frac{a^m}{a^n}=a^{m-n}
and don't forget pemdas
example: -x^m=-1(x^m) but (-x)^m=(-1)^m(x^m)

so

4.
(\frac{c^{-2}}{2})^{-2}=

\frac{(c^{-2})^{-2}}{2^{-2}}=

\frac{c^4}{\frac{1}{2^2}}=

\frac{c^4}{\frac{1}{4}}=

4c^4



5.

\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=

\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=

\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=

\frac{(a^4bc^5)b^3}{(-1)(a^2)}=

\frac{-a^4b^4c^5}{a^2}=

-a^2b^4c^5
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Gwar [14]

Answer:

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Step-by-step explanation:

you multiply.

7 0
3 years ago
How do you sketch the graph of y=(x 3)^2 6 and describe the transformation?
mr_godi [17]
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