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lord [1]
3 years ago
6

Is the answer B? Wanted to check my answer..

Mathematics
2 answers:
Lelechka [254]3 years ago
8 0
Yes! B is the correct answer
Yanka [14]3 years ago
6 0
1/2=3/6
2/3=4/6

The answer is B.

I hope this helped!!!:)
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How to solve 3+p=8 step by step
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5 0
3 years ago
Read 2 more answers
If f(x) = 2x + 3 and g(x) =4x - 1, find f(4).
fomenos

Solve the "f" function with substitute 4 and solve the "g" function with what we get for the "f" function.

f(4) = 2(8) + 3

f(4) = 16 + 3

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Best of Luck!

6 0
3 years ago
So would the answer be 5 eighths?
adelina 88 [10]
No, the answer is 3/8 because if 3/4 of the cars are Sudan's, and half of that three quarters are white, you need to divide three quarters into two to get the answer. I did this question by using the equivalent fraction 6/8 and dividing both numbers into two.
3 0
3 years ago
Please help I am so lost!!!
ASHA 777 [7]
\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\
-----------------------------\\\\
tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\

\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\

\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}
\end{cases}\\\\

\bf -----------------------------\\\\
\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}
\\\\\\
\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies 
\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)
4 0
3 years ago
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