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kow [346]
3 years ago
7

Help me pls and thank you​

Mathematics
2 answers:
lana66690 [7]3 years ago
6 0

Answer:

I believe its D or B, sorry if im wrong.

Step-by-step explanation:

Alik [6]3 years ago
6 0

Answer:

B

Step-by-step explanation:

2.25-2.25 is zero, the others have nothing to do with zero. A is a "distractor", but it ends up being 20.

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I NEED HELP PLEASEE
xenn [34]

\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}

\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill

5 0
3 years ago
What is the solution to this equation?
solong [7]

Answer:

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8 0
2 years ago
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lawyer [7]

Answer:

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Step-by-step explanation:

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option B,C and D has negative exponents, so it is not a polynomial.

7 0
3 years ago
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