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Travka [436]
3 years ago
10

(13x+14)(6x-5)=0 Solve by factoring

Mathematics
1 answer:
NeTakaya3 years ago
5 0
(13x+14)(6x-5)=0
78x²-65x+84x-70=0
78x²+19x-70=0
It was equation.
______________
Factoring⇒  a×c=78×(-70)=-5460
84×(-65)=-5460
84+(-65)=19  which is b
__________________
78x²+84x-65x-70=0
6x(13x+14)-5(13x+14)=0
(13x+14)(6x-5)=0
13x+14=0             6x-5=0
13x=-14                6x=5
 x=-14/13                x=5/6
         Two solution X=-14/13;5/6
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First option is correct.

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malfutka [58]

Consider expression \log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).

1. Use property

\log_a\dfrac{b}{c}=\log_ab-\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.

2. Use property

\log_abc=\log_ab+\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.

3. Use property

\log_ab^k=k\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.

4. Use property

\log_{a^k}b=\dfrac{1}{k}\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.

Answer: correct option is B.

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3 years ago
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