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2 years ago

Math question 1. 3-8=?, 2. 6-(-7)=?, 3. -10+(-9)=?, 4. -5-(-4)

Mathematics

1 answer:

kiruha [24]2 years ago

5 0

3-8=-5
6-(-7)=13
-10+(-9)=-19
-5-(-4)=-1

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Answer:

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Step-by-step explanation:

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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:

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

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

Now,

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

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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Delicious77 [7]

Answer:

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Step-by-step explanation:

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Use the Binomial expansion theorem to find and simplify each term.

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Man that was a lot to type out, hopefully i helped, and Gosh I hope I get a brainly for all that freaking typing. hehe~ ^.^

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