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ratelena [41]
2 years ago
5

Pls answer correctly

Mathematics
1 answer:
Serga [27]2 years ago
7 0

Answer:

i think its a but im not sure

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Can someone help me with this question
Mashcka [7]
The correct answer here would be D. -1(3x+1)(x+5). You can find this answer by redistributing the problem.

-1(3x+1)(x+5)
-1(3x²+15x+1x+5)
-1(3x²+16x+5)
-3x²-16x-5

Using the math above, we can see that when we redistribute we get the original equation. That makes Choice D correct.
5 0
3 years ago
-extra points- please help with this. all you have to do is state which method (factoring, using square roots, complete the squa
MAXImum [283]

Answer:

Step-by-step explanation:

5 0
3 years ago
if a system of linear equations has infinitely many solutions, what does this mean about the two lines?​
emmainna [20.7K]

Answer:  If a system has infinitely many solutions, then the lines overlap at every point. In other words, they're the same exact line so this means that any point on the line is a solution to the system.

Step-by-step explanation: Hope this helps

8 0
3 years ago
What number is between 7 and 8
VashaNatasha [74]

Answer:

7 1/2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
We were never taught this in class, and I understand it, kind of... but i'm just confused with everything.
GaryK [48]
\bf \cfrac{x^3y}{xy^5}\cdot \cfrac{x^2y^9}{x^8}\implies \cfrac{x^3x^2y^1y^9}{x^1x^8y^5}
\\ \quad \\
\textit{now, using exponent rules, same base, different exponent}
\\ \quad \\
\cfrac{x^{3+2}y^{1+9}}{x^{1+8}y^5}\implies \cfrac{x^5y^{10}}{x^9y^5}\\
----------------------------\\\bf \textit{now, keep in mind that}
\\ \quad \\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
%  negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}
\\ \quad \\
thus\\
----------------------------\\
\bf \cfrac{x^5y^{10}}{x^9y^5}\implies \cfrac{x^5y^{10}}{1}\cdot \cfrac{1}{x^9}\cdot \cfrac{1}{y^5}\implies x^5y^{10}\cdot x^{\boxed{-9}}\cdot y^{\boxed{-5}}
\\ \quad \\
x^5x^{-9}y^{10}y^{-5}\implies x^{5-9}y^{10-5}\implies x^{-4}y^5
\\ \quad \\
\cfrac{1}{x^{\boxed{4}}}\cdot \cfrac{y^5}{1}\implies \cfrac{y^5}{x^4}
8 0
3 years ago
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