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3 years ago

Solve for z 7-√3z+2=4 What is the root? If there is no root, choose none. z = 7/3 z = 11/3 none

Mathematics

1 answer:

cestrela7 [59]3 years ago

6 0

Answer:

Desmos is Function Calculator is a great tool for this, the correct answer is z= 7/3

7/3= 2.333~ and so √(3*2.333~) + 2= 2.9999999833333~

Now 7- 2.9999999833333~= 4.0000000166667~ but I don't know the true infinate sign for this so thats why my numbers look a little weird, just plug (7/3) in for z and you will get 4, trust me

Step-by-step explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZz

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Que is on pic.i can't able to type in text.

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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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umka2103 [35]

Answer:

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Step-by-step explanation:

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