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stealth61 [152]
3 years ago
12

I need to solve and show work for 3x over4 divided by 5-1 over 8

Mathematics
2 answers:
likoan [24]3 years ago
5 0
3or4vxe
     4  



Is the answer 
Margaret [11]3 years ago
4 0
Sounds as tho' you want to evaluate

3x
---
  4

over

5x-1
------
    8

To accomplish this, invert the 2nd fraction and then multiply:


3x     5x-1
---   * ------  = 15x^2-3x   divided by 32.
  4       8
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Que is on pic.i can't able to type in text.
ad-work [718]
It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
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Step-by-step explanation:

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Eric practiced piano and guitar for a total of 8 hours this week. he practiced the piano for 1/4 or that time. how many hours di
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